Q1. Factorize $2x^2y^3 - 6x^2y^2 + 2xy^3$:

1. Step 1: Identify greatest common numerical coefficient of $2, -6, 2 \implies 2$.
2. Step 2: Identify lowest powers of common variable factors: $x^{\min(2,2,1)} = x$, $y^{\min(3,2,3)} = y^2$. Common monomial is $2xy^2$.
3. Step 3: Factor out $2xy^2$ from each term: $$2xy^2(xy - 3x + y)$$
4. Final Answer: $\mathbf{2xy^2(xy - 3x + y)}$.

Q2. Factorize $3nx - 3x - 3ny + 3y$:

1. Step 1: Factor out the overall common numerical factor $3$: $$3(nx - x - ny + y)$$
2. Step 2: Group the 4 terms in pairs inside parentheses: $$3[ (nx - x) - (ny - y) ]$$
3. Step 3: Extract common factor $x$ from first pair and $y$ from second pair: $$3[ x(n - 1) - y(n - 1) ]$$
4. Step 4: Extract the common binomial $(n - 1)$: $$\mathbf{3(n - 1)(x - y)}$$

Q3. Factorize $18x^4 + 108x^2y^2 + 162y^4$:

1. Step 1: Factor out greatest common factor $18$: $$18(x^4 + 6x^2y^2 + 9y^4)$$
2. Step 2: Recognize the trinomial inside as a perfect square $(a + b)^2 = a^2 + 2ab + b^2$ where $a = x^2$ and $b = 3y^2$: $$18[ (x^2)^2 + 2(x^2)(3y^2) + (3y^2)^2 ]$$
3. Step 3: Condense into square form: $$\mathbf{18(x^2 + 3y^2)^2}$$

Q4. Factorize $(k + 2)^2 - 8(k + 2) + 16$:

1. Step 1: Notice the structure $u^2 - 8u + 16$ where $u = (k + 2)$.
2. Step 2: Recognize the perfect square trinomial $(u - 4)^2$: $$[ (k + 2) - 4 ]^2$$
3. Step 3: Simplify inside the brackets: $$(k + 2 - 4)^2 = \mathbf{(k - 2)^2}$$

Q5. Factorize $9x^2 + 4 - 169y^2 - 12x$:

1. Step 1: Rearrange terms to group the perfect square in $x$: $$(9x^2 - 12x + 4) - 169y^2$$
2. Step 2: Express the first group as a perfect square $[(3x)^2 - 2(3x)(2) + 2^2] = (3x - 2)^2$: $$(3x - 2)^2 - (13y)^2$$
3. Step 3: Apply difference of two squares $A^2 - B^2 = (A - B)(A + B)$: $$\mathbf{(3x - 2 - 13y)(3x - 2 + 13y)}$$

Q6. Factorize $(x^2 - 1)(y + 1) - (y + 3)(x^2 - 1)$:

1. Step 1: Extract the common binomial factor $(x^2 - 1)$: $$(x^2 - 1) [ (y + 1) - (y + 3) ]$$
2. Step 2: Simplify the second bracket: $$(x^2 - 1)(y + 1 - y - 3) = (x^2 - 1)(-2) = -2(x^2 - 1)$$
3. Step 3: Factorize the difference of squares $(x^2 - 1) = (x - 1)(x + 1)$: $$\mathbf{-2(x - 1)(x + 1)}$$

Q7. Factorize $x^2 - 6ax + 9a^2 - 16b^2$:

1. Step 1: Group the first three terms as a perfect square: $$(x^2 - 6ax + 9a^2) - 16b^2$$
2. Step 2: Write in square form $(x - 3a)^2 - (4b)^2$.
3. Step 3: Apply $A^2 - B^2 = (A - B)(A + B)$: $$\mathbf{(x - 3a - 4b)(x - 3a + 4b)}$$

Q8. Factorize $1 - x^2 - 2xy - y^2$:

1. Step 1: Factor out $-1$ from the last three terms: $$1 - (x^2 + 2xy + y^2)$$
2. Step 2: Recognize the perfect square $(x + y)^2$: $$1^2 - (x + y)^2$$
3. Step 3: Apply difference of squares $A^2 - B^2 = (A - B)(A + B)$: $$[1 - (x + y)][1 + (x + y)] = \mathbf{(1 - x - y)(1 + x + y)}$$

Q9. Find a polynomial whose factorization is $(x + y - 2c)(x + 2c + y)$ using an appropriate formula:

1. Step 1: Rearrange terms to reveal the form $(A - B)(A + B)$ where $A = (x + y)$ and $B = 2c$: $$[(x + y) - 2c] \cdot [(x + y) + 2c]$$
2. Step 2: Apply the difference of squares identity $(A - B)(A + B) = A^2 - B^2$: $$(x + y)^2 - (2c)^2$$
3. Step 3: Expand the algebraic square $(x + y)^2 = x^2 + 2xy + y^2$: $$\mathbf{x^2 + 2xy + y^2 - 4c^2}$$

Q10. Show the expression $x^2 + 4y^2 - z^2 + 4xy$ as the difference of two squares:

1. Step 1: Regroup terms containing $x$ and $y$: $$(x^2 + 4xy + 4y^2) - z^2$$
2. Step 2: Complete the perfect square: $(x + 2y)^2 - z^2$.
3. Step 3: Express clearly as the difference of two squares: $$\mathbf{(x + 2y)^2 - z^2 = (x + 2y - z)(x + 2y + z)}$$

Q11. Find the missing factor in the following:

(a) $(2y^2 - 3y - 27) = (y + 3)(\quad)$:

1. Split middle term: $2 \times (-27) = -54$. Factors adding to $-3$ are $-9$ and $+6$. $$2y^2 + 6y - 9y - 27 = 2y(y + 3) - 9(y + 3) = (y + 3)(2y - 9)$$
2. Missing factor is $\mathbf{2y - 9}$.

(b) $(5x^2 + 12x - 9) = (\quad)(x + 3)$:

1. Split middle term: $5 \times (-9) = -45$. Factors adding to $12$ are $+15$ and $-3$. $$5x^2 + 15x - 3x - 9 = 5x(x + 3) - 3(x + 3) = (5x - 3)(x + 3)$$
2. Missing factor is $\mathbf{5x - 3}$.

Q12. Factorize $x^4 + 4m^4$:

1. Step 1: Write as sum of squares: $(x^2)^2 + (2m^2)^2$.
2. Step 2: Add and subtract $2(x^2)(2m^2) = 4x^2m^2$ to complete the square: $$[(x^2)^2 + 2(x^2)(2m^2) + (2m^2)^2] - 4x^2m^2$$
3. Step 3: Condense perfect square: $(x^2 + 2m^2)^2 - (2xm)^2$.
4. Step 4: Apply difference of squares $A^2 - B^2 = (A - B)(A + B)$: $$\mathbf{(x^2 - 2mx + 2m^2)(x^2 + 2mx + 2m^2)}$$

Q13. Factorize $m^4 + m^2 + 1$:

1. Step 1: Write end squares: $(m^2)^2 + (1)^2$. We need $2m^2$ for a perfect square.
2. Step 2: Add and subtract $m^2$: $$(m^4 + 2m^2 + 1) - m^2 = (m^2 + 1)^2 - m^2$$
3. Step 3: Apply difference of squares: $$\mathbf{(m^2 - m + 1)(m^2 + m + 1)}$$

Q14. Factorize $-3x^4 - 21x^3 + 24x^2$:

1. Step 1: Factor out common monomial $-3x^2$: $$-3x^2(x^2 + 7x - 8)$$
2. Step 2: Split middle term of quadratic $x^2 + 7x - 8$: Product $=-8$, sum $=+7 \implies +8, -1$. $$-3x^2 [x(x + 8) - 1(x + 8)] = \mathbf{-3x^2(x - 1)(x + 8)}$$

Q15. Factorize $x^8 + x^4 + 1$:

1. Step 1: Add and subtract $x^4$: $$(x^8 + 2x^4 + 1) - x^4 = (x^4 + 1)^2 - (x^2)^2$$
2. Step 2: Apply difference of squares: $$(x^4 - x^2 + 1)(x^4 + x^2 + 1)$$
3. Step 3: Factorize $(x^4 + x^2 + 1)$ further using square completion: $$(x^4 + 2x^2 + 1) - x^2 = (x^2 + 1)^2 - x^2 = (x^2 - x + 1)(x^2 + x + 1)$$
4. Step 4: Combine all irreducible factors: $$\mathbf{(x^4 - x^2 + 1)(x^2 - x + 1)(x^2 + x + 1)}$$

Q16. Factorize $4x^4 + 256y^4$:

1. Step 1: Factor out common factor $4$: $$4(x^4 + 64y^4)$$
2. Step 2: Express inside brackets as sum of squares $(x^2)^2 + (8y^2)^2$. Add and subtract $2(x^2)(8y^2) = 16x^2y^2$: $$4[ (x^2 + 8y^2)^2 - (4xy)^2 ]$$
3. Step 3: Apply difference of squares $A^2 - B^2 = (A - B)(A + B)$: $$\mathbf{4(x^2 - 4xy + 8y^2)(x^2 + 4xy + 8y^2)}$$

Q17. Factorize $12 - 7x + x^2$:

1. Step 1: Write in standard order: $x^2 - 7x + 12$.
2. Step 2: Find two numbers multiplying to $+12$ and adding to $-7$: $-3$ and $-4$.
3. Step 3: Split and group: $x^2 - 3x - 4x + 12 = x(x - 3) - 4(x - 3) = \mathbf{(x - 3)(x - 4)}$.

Q18. Factorize $x^2 - 9x + 8$:

1. Step 1: Product $=+8$, sum $=-9 \implies -8$ and $-1$.
2. Step 2: Split middle term: $x^2 - 8x - x + 8 = x(x - 8) - 1(x - 8) = \mathbf{(x - 1)(x - 8)}$.

Q19. Factorize $10z^2 - 29z + 10$:

1. Step 1: Product $a \cdot c = 10 \times 10 = 100$. Middle coefficient $b = -29$.
2. Step 2: Factors of $100$ adding to $-29$ are $-25$ and $-4$.
3. Step 3: Split middle term: $$10z^2 - 25z - 4z + 10 = 5z(2z - 5) - 2(2z - 5) = \mathbf{(2z - 5)(5z - 2)}$$

Q20. Factorize $-3y^2 + 13y - 4$:

1. Step 1: Factor out $-1$: $-(3y^2 - 13y + 4)$.
2. Step 2: Product $3 \times 4 = 12$, sum $=-13 \implies -12$ and $-1$.
3. Step 3: Split and group: $$-[3y^2 - 12y - y + 4] = -[3y(y - 4) - 1(y - 4)] = -(3y - 1)(y - 4) = \mathbf{(1 - 3y)(y - 4) \text{ or } (3y - 1)(4 - y)}$$

Q21. Factorize $x^2 - 21x + 90$:

1. Step 1: Product $=+90$, sum $=-21 \implies -15$ and $-6$ (since $-15 \times -6 = 90$ and $-15 - 6 = -21$).
2. Step 2: $x^2 - 15x - 6x + 90 = x(x - 15) - 6(x - 15) = \mathbf{(x - 6)(x - 15)}$.

Q22. Factorize $x^2 + x - 2$:

1. Step 1: Product $=-2$, sum $=+1 \implies +2$ and $-1$.
2. Step 2: $x^2 + 2x - x - 2 = x(x + 2) - 1(x + 2) = \mathbf{(x - 1)(x + 2)}$.

Q23. Factorize $3x^2 + 11x + 6$:

1. Step 1: Product $a \cdot c = 3 \times 6 = 18$, sum $=11 \implies +9$ and $+2$.
2. Step 2: Split middle term: $3x^2 + 9x + 2x + 6 = 3x(x + 3) + 2(x + 3) = \mathbf{(x + 3)(3x + 2)}$.

Q24. Factorize $2x^2 - 5xy - 3y^2$:

1. Step 1: Product $2 \times (-3) = -6$, sum $=-5 \implies -6$ and $+1$.
2. Step 2: Split middle term: $2x^2 - 6xy + xy - 3y^2 = 2x(x - 3y) + y(x - 3y) = \mathbf{(x - 3y)(2x + y)}$.

Q25. Factorize $8 + 6x - 5x^2$:

1. Step 1: Factor out $-1$: $-(5x^2 - 6x - 8)$.
2. Step 2: Product $5 \times (-8) = -40$, sum $=-6 \implies -10$ and $+4$.
3. Step 3: Split middle term: $$-[5x^2 - 10x + 4x - 8] = -[5x(x - 2) + 4(x - 2)] = -(x - 2)(5x + 4) = \mathbf{(2 - x)(5x + 4)}$$

Q26. Factorize $6 - 7x - 5x^2$:

1. Step 1: Rearrange: $-(5x^2 + 7x - 6)$.
2. Step 2: Product $5 \times (-6) = -30$, sum $=+7 \implies +10$ and $-3$.
3. Step 3: Split middle term: $$-[5x^2 + 10x - 3x - 6] = -[5x(x + 2) - 3(x + 2)] = -(x + 2)(5x - 3) = \mathbf{(x + 2)(3 - 5x)}$$

Q27. Factorize $2a^2 - 4a - 6$:

1. Step 1: Extract greatest common factor $2$: $$2(a^2 - 2a - 3)$$
2. Step 2: Split middle term: Product $=-3$, sum $=-2 \implies -3$ and $+1$.
3. Step 3: $2[a(a - 3) + 1(a - 3)] = \mathbf{2(a - 3)(a + 1)}$.

Q28. Factorize $u^4 - 13u^2 + 36$:

1. Step 1: Treat as quadratic in $z = u^2$: $z^2 - 13z + 36$.
2. Step 2: Factors of $+36$ adding to $-13$ are $-9$ and $-4$: $$(z - 9)(z - 4) = (u^2 - 9)(u^2 - 4)$$
3. Step 3: Factorize both differences of squares: $$(u^2 - 3^2)(u^2 - 2^2) = \mathbf{(u - 3)(u + 3)(u - 2)(u + 2)}$$

Q29. Factorize $y^4 - 12y^2 - 64$:

1. Step 1: Treat as quadratic in $z = y^2$: $z^2 - 12z - 64$.
2. Step 2: Factors of $-64$ adding to $-12$ are $-16$ and $+4$: $$(y^2 - 16)(y^2 + 4)$$
3. Step 3: Factorize difference of squares $(y^2 - 16) = (y - 4)(y + 4)$, while $(y^2 + 4)$ is irreducible in $\mathbb{R}$: $$\mathbf{(y - 4)(y + 4)(y^2 + 4)}$$

Q1. Factorize $x^3 - 125$:

1. Step 1: Write as difference of cubes: $x^3 - 5^3$.
2. Step 2: Apply formula $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$: $$\mathbf{(x - 5)(x^2 + 5x + 25)}$$

Q2. Factorize $8x^3 + 1$:

1. Step 1: Write as sum of cubes: $(2x)^3 + 1^3$.
2. Step 2: Apply formula $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$: $$[ (2x) + 1 ][ (2x)^2 - (2x)(1) + 1^2 ] = \mathbf{(2x + 1)(4x^2 - 2x + 1)}$$

Q3. Factorize $3p^3q^3 - 81x^3$:

1. Step 1: Extract common numerical factor $3$: $$3(p^3q^3 - 27x^3)$$
2. Step 2: Express inside brackets as difference of cubes $(pq)^3 - (3x)^3$: $$3[ (pq - 3x)( (pq)^2 + (pq)(3x) + (3x)^2 ) ]$$
3. Step 3: Simplify: $\mathbf{3(pq - 3x)(p^2q^2 + 3pqx + 9x^2)}$.

Q4. Factorize $27 + 512x^3$:

1. Step 1: Write as sum of cubes: $3^3 + (8x)^3$.
2. Step 2: Apply $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$: $$(3 + 8x)[ 3^2 - 3(8x) + (8x)^2 ] = \mathbf{(3 + 8x)(9 - 24x + 64x^2)}$$

Q5. Factorize $t^6 - 64$:

1. Step 1: Always factorize as difference of squares first $(t^3)^2 - 8^2$: $$(t^3 - 8)(t^3 + 8)$$
2. Step 2: Factorize $(t^3 - 2^3)$ using difference of cubes: $(t - 2)(t^2 + 2t + 4)$.
3. Step 3: Factorize $(t^3 + 2^3)$ using sum of cubes: $(t + 2)(t^2 - 2t + 4)$.
4. Step 4: Combine all factors: $$\mathbf{(t - 2)(t + 2)(t^2 + 2t + 4)(t^2 - 2t + 4)}$$

Q6. Factorize $x^6 + y^6$:

1. Step 1: Express as sum of cubes $(x^2)^3 + (y^2)^3$:
2. Step 2: Apply $A^3 + B^3 = (A + B)(A^2 - AB + B^2)$ where $A = x^2, B = y^2$: $$\mathbf{(x^2 + y^2)(x^4 - x^2y^2 + y^4)}$$

Q7. Factorize $(2 - x)^3 + (y - 2)^3$:

1. Step 1: Apply sum of cubes $A^3 + B^3 = (A + B)(A^2 - AB + B^2)$ where $A = (2 - x)$ and $B = (y - 2)$:
2. Step 2: Linear factor: $A + B = (2 - x) + (y - 2) = y - x$.
3. Step 3: Quadratic factor: $$A^2 - AB + B^2 = (2 - x)^2 - (2 - x)(y - 2) + (y - 2)^2$$ $$= (4 - 4x + x^2) - (2y - 4 - xy + 2x) + (y^2 - 4y + 4)$$ $$= 4 - 4x + x^2 - 2y + 4 + xy - 2x + y^2 - 4y + 4$$ $$= x^2 + y^2 + xy - 6x - 6y + 12$$
4. Step 4: Product of factors: $$\mathbf{(y - x)(x^2 + y^2 + xy - 6x - 6y + 12)}$$

Q8. Factorize $64(x + y)^3 - z^3$:

1. Step 1: Write as difference of cubes $[4(x + y)]^3 - z^3$.
2. Step 2: Apply $A^3 - B^3 = (A - B)(A^2 + AB + B^2)$: $$[ 4(x + y) - z ][ 16(x + y)^2 + 4(x + y)z + z^2 ]$$
3. Step 3: Expand: $$\mathbf{(4x + 4y - z)(16x^2 + 32xy + 16y^2 + 4xz + 4yz + z^2)}$$

Q9. Factorize $27p^3 + 144pq^2 - 108p^2q - 64q^3$:

1. Step 1: Rearrange in descending powers of $p$: $$(3p)^3 - 3(3p)^2(4q) + 3(3p)(4q)^2 - (4q)^3$$
2. Step 2: Verify terms: $3(9p^2)(4q) = 108p^2q$, $3(3p)(16q^2) = 144pq^2$.
3. Step 3: Recognize the perfect cube $(a - b)^3$: $$\mathbf{(3p - 4q)^3 = (3p - 4q)(3p - 4q)(3p - 4q)}$$

Q10. Factorize $8p^3 + q^3 + 12p^2q + 6pq^2$:

1. Step 1: Rearrange as $(2p)^3 + 3(2p)^2(q) + 3(2p)(q)^2 + q^3$.
2. Step 2: Recognize $(a + b)^3$ form: $$\mathbf{(2p + q)^3 = (2p + q)(2p + q)(2p + q)}$$

Q11. Factorize $125x^3 - y^3 - 75x^2y + 15xy^2$:

1. Step 1: Rearrange as $(5x)^3 - 3(5x)^2(y) + 3(5x)(y)^2 - y^3$.
2. Step 2: Apply $(a - b)^3$: $$\mathbf{(5x - y)^3 = (5x - y)(5x - y)(5x - y)}$$

Q12. Factorize $p^3 - 9p^2q + 27pq^2 - 27q^3$:

1. Step 1: Write as $p^3 - 3(p^2)(3q) + 3(p)(3q)^2 - (3q)^3$.
2. Step 2: Recognize $(a - b)^3$: $$\mathbf{(p - 3q)^3 = (p - 3q)(p - 3q)(p - 3q)}$$

Q13. Factorize $(2x^2 - 3x + 6)(2x^2 - 3x) - 55$:

1. Step 1: Substitute $y = 2x^2 - 3x$: $$(y + 6)y - 55 = y^2 + 6y - 55$$
2. Step 2: Factorize quadratic in $y$: Factors of $-55$ adding to $+6$ are $+11$ and $-5$: $$(y + 11)(y - 5)$$
3. Step 3: Substitute back $y = 2x^2 - 3x$: $$(2x^2 - 3x + 11)(2x^2 - 3x - 5)$$
4. Step 4: Factorize $(2x^2 - 3x - 5) = 2x^2 - 5x + 2x - 5 = (2x - 5)(x + 1)$: $$\mathbf{(2x^2 - 3x + 11)(2x - 5)(x + 1)}$$

Q14. Factorize $(y^2 + 2y - 3)(y^2 + 2y + 11) + 48$:

1. Step 1: Let $z = y^2 + 2y$: $$(z - 3)(z + 11) + 48 = z^2 + 8z - 33 + 48 = z^2 + 8z + 15$$
2. Step 2: Factorize $(z + 3)(z + 5)$.
3. Step 3: Substitute back $z = y^2 + 2y$: $$(y^2 + 2y + 3)(y^2 + 2y + 5)$$
4. Final Answer: $\mathbf{(y^2 + 2y + 3)(y^2 + 2y + 5)}$ (both irreducible over $\mathbb{R}$).

Q15. Factorize $y(y - 1)(y - 3)(y - 4) + 2$:

1. Step 1: Pair factors with equal sums of constants: $0 + (-4) = -4$ and $(-1) + (-3) = -4$. $$[y(y - 4)] \cdot [(y - 1)(y - 3)] + 2 = (y^2 - 4y)(y^2 - 4y + 3) + 2$$
2. Step 2: Substitute $u = y^2 - 4y$: $$u(u + 3) + 2 = u^2 + 3u + 2 = (u + 1)(u + 2)$$
3. Step 3: Substitute back $u = y^2 - 4y$: $$\mathbf{(y^2 - 4y + 1)(y^2 - 4y + 2)}$$

Q16. Factorize $(k + 2)(k - 3)(k + 5)(k + 10) + 375$:

1. Step 1: Check constant sums: $2 + 5 = 7$ and $-3 + 10 = 7$. Pair $[(k+2)(k+5)]$ and $[(k-3)(k+10)]$: $$(k^2 + 7k + 10)(k^2 + 7k - 30) + 375$$
2. Step 2: Let $z = k^2 + 7k$: $$(z + 10)(z - 30) + 375 = z^2 - 20z - 300 + 375 = z^2 - 20z + 75$$
3. Step 3: Factorize $(z - 15)(z - 5)$.
4. Step 4: Substitute back $z = k^2 + 7k$: $$(k^2 + 7k - 15)(k^2 + 7k - 5)$$
5. Final Answer: $\mathbf{(k^2 + 7k - 15)(k^2 + 7k - 5)}$.

Q17. Factorize $(x - 5)(x - 6)(x + 3)(x + 2) + 12$:

1. Step 1: Pair factors: $(-5) + 2 = -3$ and $(-6) + 3 = -3$. $$[(x - 5)(x + 2)] \cdot [(x - 6)(x + 3)] + 12 = (x^2 - 3x - 10)(x^2 - 3x - 18) + 12$$
2. Step 2: Let $u = x^2 - 3x$: $$(u - 10)(u - 18) + 12 = u^2 - 28u + 180 + 12 = u^2 - 28u + 192$$
3. Step 3: Factors of $+192$ adding to $-28$ are $-16$ and $-12$: $$(u - 16)(u - 12)$$
4. Step 4: Substitute back $u = x^2 - 3x$: $$(x^2 - 3x - 16)(x^2 - 3x - 12)$$

Q18. Factorize $(x + 1)(x + 2)(x - 3)(x - 6) - 21x^2$:

1. Step 1: Pair factors where product of constant terms match: $1 \times (-6) = -6$ and $2 \times (-3) = -6$. $$[(x + 1)(x - 6)] \cdot [(x + 2)(x - 3)] - 21x^2 = (x^2 - 5x - 6)(x^2 - x - 6) - 21x^2$$
2. Step 2: Let $y = x^2 - 6$: $$(y - 5x)(y - x) - 21x^2 = y^2 - 6xy + 5x^2 - 21x^2 = y^2 - 6xy - 16x^2$$
3. Step 3: Factorize quadratic in $y$: $(y - 8x)(y + 2x)$.
4. Step 4: Substitute back $y = x^2 - 6$: $$\mathbf{(x^2 - 8x - 6)(x^2 + 2x - 6)}$$

Q19. Factorize $(x - 2)(x - 6)(x - 3)(x - 4) - 2x^2$:

1. Step 1: Pair where products match: $(-2) \times (-6) = 12$ and $(-3) \times (-4) = 12$. $$[(x - 2)(x - 6)] \cdot [(x - 3)(x - 4)] - 2x^2 = (x^2 - 8x + 12)(x^2 - 7x + 12) - 2x^2$$
2. Step 2: Let $y = x^2 + 12$: $$(y - 8x)(y - 7x) - 2x^2 = y^2 - 15xy + 56x^2 - 2x^2 = y^2 - 15xy + 54x^2$$
3. Step 3: Factorize: $(y - 9x)(y - 6x)$.
4. Step 4: Substitute back $y = x^2 + 12$: $$\mathbf{(x^2 - 9x + 12)(x^2 - 6x + 12)}$$

Q20. Factorize $(5 - x)(2 + x)(10 - x)(1 + x) - 7x^2$:

1. Step 1: Rewrite factors as $(x - 5)(x + 2)(x - 10)(x + 1) - 7x^2$ (since $(-1)^2 = +1$).
2. Step 2: Pair with equal products: $(-5) \times 2 = -10$ and $(-10) \times 1 = -10$. $$(x^2 - 3x - 10)(x^2 - 9x - 10) - 7x^2$$
3. Step 3: Let $y = x^2 - 10$: $$(y - 3x)(y - 9x) - 7x^2 = y^2 - 12xy + 27x^2 - 7x^2 = y^2 - 12xy + 20x^2$$
4. Step 4: Factorize: $(y - 10x)(y - 2x)$.
5. Step 5: Substitute back $y = x^2 - 10$: $$\mathbf{(x^2 - 10x - 10)(x^2 - 2x - 10)}$$

Q21. The expression $a^6 + 729$ can be written as: (a) sum of two squares, (b) sum of two cubes. Which one will be used for factoring it and why? Also factorize the given expression:

1. (a) Representation Analysis:
   • As sum of two squares: $(a^3)^2 + (27)^2 = a^6 + 729$. Since sum of two squares $A^2 + B^2$ does not factor over the field of real numbers, this form cannot factorize the expression.
   • As sum of two cubes: $(a^2)^3 + (9)^3 = a^6 + 729$. The sum of two cubes $A^3 + B^3 = (A + B)(A^2 - AB + B^2)$ gives real polynomial factors!
2. (b) Why sum of cubes is used: Because $A^3 + B^3$ has a standard real algebraic factorization formula $(A + B)(A^2 - AB + B^2)$, whereas sum of squares has no real linear or quadratic factorization.
3. (c) Factorization: $$(a^2)^3 + 9^3 = (a^2 + 9)[ (a^2)^2 - (a^2)(9) + 9^2 ] = \mathbf{(a^2 + 9)(a^4 - 9a^2 + 81)}$$

Q22. Express $8 + 12t + 6t^2 + t^3$ as the product of three factors. Is each factor a binomial or a trinomial?

1. Step 1: Rearrange in ascending or descending order: $2^3 + 3(2^2)(t) + 3(2)(t^2) + t^3$.
2. Step 2: Recognize the perfect cube expansion $(a + b)^3 = (2 + t)^3$.
3. Step 3: Express as the product of three linear factors: $$\mathbf{(2 + t)(2 + t)(2 + t)}$$
4. Step 4: Classification: Each of the three factors is a binomial (containing exactly two terms: $2$ and $t$).

Q1. (a) Complete the Monomial HCF Table:

(b) If all common factors with least power of three unknown polynomials are $2^2, 3, pq$ and $(p + q)^2$, find their HCF:

$$\text{HCF} = 2^2 \times 3 \times pq \times (p + q)^2 = 4 \times 3 \times pq(p + q)^2 = \mathbf{12pq(p + q)^2}$$

(c) Write any two polynomials of your choice having HCF as 1:

Example: $P(x) = (x + 2)$ and $Q(x) = (x + 5)$, or $x^2 + 1$ and $x^2 + 2$. Since they share no common factor, $\mathbf{\text{HCF} = 1}$ (they are coprime).

(d) The only common factor of two polynomials is $(m - n)$ and the only uncommon factor is $(m^2 + n^2)$. Can you guess the unknown polynomials?

$$\text{Polynomial 1} = \mathbf{m - n}$$ $$\text{Polynomial 2} = (m - n)(m^2 + n^2) = \mathbf{m^3 - m^2n + mn^2 - n^3}$$

(e) Can you guess the HCF of $x^3 + 5x + 1$ and $1 + 5x + x^3$ without any procedure?

Yes! By the commutative property of polynomial addition, $1 + 5x + x^3 = x^3 + 5x + 1$. Both polynomials are completely identical, so their HCF is the polynomial itself: $\mathbf{x^3 + 5x + 1}$.

Q2. Find HCF of $(x + y)^2$ and $x^2 - y^2$:

1. Step 1: Factorize $P_1(x,y) = (x + y)^2 = (x + y)(x + y)$.
2. Step 2: Factorize $P_2(x,y) = x^2 - y^2 = (x - y)(x + y)$.
3. Step 3: Common factor with least exponent: $\mathbf{x + y}$.

Q3. Find HCF of $(a - b)^3$ and $a^2 - 2ab + b^2$:

1. Step 1: Factorize $P_1 = (a - b)^3$.
2. Step 2: Factorize $P_2 = a^2 - 2ab + b^2 = (a - b)^2$.
3. Step 3: The common factor with lowest power is $(a - b)^{\min(3,2)} = \mathbf{(a - b)^2}$.

Q4. Find HCF of $a^3b - ab^3$ and $a^5b^2 - a^2b^5$:

1. Step 1: $P_1 = ab(a^2 - b^2) = ab(a - b)(a + b)$.
2. Step 2: $P_2 = a^2b^2(a^3 - b^3) = a^2b^2(a - b)(a^2 + ab + b^2)$.
3. Step 3: Common monomial factor: $\gcd(ab, a^2b^2) = ab$.
4. Step 4: Common binomial factor: $(a - b)$.
5. Step 5: Multiply all common factors: $\mathbf{ab(a - b)}$.

Q5. Find HCF of $x^2 - 49$ and $x^2 - 4x - 21$:

1. Step 1: Factorize $x^2 - 49 = (x - 7)(x + 7)$.
2. Step 2: Factorize $x^2 - 4x - 21 = x^2 - 7x + 3x - 21 = (x - 7)(x + 3)$.
3. Step 3: The common factor present in both is $\mathbf{x - 7}$.

Q6. Find HCF of $12x^2 + x - 1$ and $15x^2 + 8x + 1$:

1. Step 1: Factorize $12x^2 + x - 1 = 12x^2 + 4x - 3x - 1 = 4x(3x + 1) - 1(3x + 1) = (3x + 1)(4x - 1)$.
2. Step 2: Factorize $15x^2 + 8x + 1 = 15x^2 + 5x + 3x + 1 = 5x(3x + 1) + 1(3x + 1) = (3x + 1)(5x + 1)$.
3. Step 3: Common factor: $\mathbf{3x + 1}$.

Q7. Find HCF of $c^2x^2 - d^2$ and $acx^2 - bcx + adx - bd$:

1. Step 1: Factorize $c^2x^2 - d^2 = (cx)^2 - d^2 = (cx - d)(cx + d)$.
2. Step 2: Group $acx^2 - bcx + adx - bd$: $$cx(ax - b) + d(ax - b) = (ax - b)(cx + d)$$
3. Step 3: Common factor: $\mathbf{cx + d}$.

Q8. Find HCF of $m^2 - n^2, m^4 - n^4,$ and $m^6 - n^6$:

1. Step 1: $m^2 - n^2 = (m - n)(m + n)$.
2. Step 2: $m^4 - n^4 = (m^2 - n^2)(m^2 + n^2) = (m - n)(m + n)(m^2 + n^2)$.
3. Step 3: $m^6 - n^6 = (m^3 - n^3)(m^3 + n^3) = (m - n)(m^2 + mn + n^2)(m + n)(m^2 - mn + n^2)$.
4. Step 4: Common factors across all 3 expressions: $(m - n)(m + n) = \mathbf{m^2 - n^2}$.

Q9. Find HCF of $ax^2 + 2a^2x + a^3,\; 2ax^2 - 4a^2x - 6a^3,$ and $3(ax + a^2)^2$:

1. Step 1: $P_1 = a(x^2 + 2ax + a^2) = a(x + a)^2$.
2. Step 2: $P_2 = 2a(x^2 - 2ax - 3a^2) = 2a(x - 3a)(x + a)$.
3. Step 3: $P_3 = 3[a(x + a)]^2 = 3a^2(x + a)^2$.
4. Step 4: Lowest power of common monomial factor $a$: $a^1$.
5. Step 5: Lowest power of common binomial factor $(x + a)$: $(x + a)^1$.
6. Step 6: Combine common factors: $\mathbf{a(x + a)}$.

Q1. If $\text{HCF}(x^3 + 5x^2 + 6x, x^3 + 9x^2 + 14x) = x^2 + 2x$, give quick answers without calculation:

1. (i) What would be the HCF of $5(x^3 + 5x^2 + 6x)$ and $x^3 + 9x^2 + 14x$?
   Since $5$ and $1$ are coprime, the scalar factor has $\gcd(5, 1) = 1$. The HCF remains $\mathbf{x^2 + 2x}$.
2. (ii) What would be the HCF of $x^3 + 5x^2 + 6x$ and $2(x^3 + 9x^2 + 14x)$$?
   Scalar $\gcd(1, 2) = 1$. The HCF remains $\mathbf{x^2 + 2x}$.
3. (iii) What would be the HCF of $3(x^3 + 5x^2 + 6x)$ and $7(x^3 + 9x^2 + 14x)$$?
   Scalar $\gcd(3, 7) = 1$. The HCF remains $\mathbf{x^2 + 2x}$.
4. (iv) What would be the HCF of $15(x^3 + 5x^2 + 6x)$ and $25(x^3 + 9x^2 + 14x)$$?
   Scalar coefficients $15$ and $25$ have numerical $\gcd(15, 25) = 5$. Therefore, the new HCF is $5 \times (x^2 + 2x) = \mathbf{5(x^2 + 2x)}$.
5. (v) Does the HCF of the given polynomials remain unchanged if both are multiplied by $x$?
   No, it will change. If both polynomials are multiplied by $x$, their HCF is also multiplied by $x$, becoming $x(x^2 + 2x) = x^3 + 2x^2$.

Q2. Find the LCM of the following monomials by completing the table:

Q3. Find HCF of $a^2 + a - 2$ and $a^3 + 2a^2 + a + 2$ by polynomial division:

1. Step 1: Divide $(a^3 + 2a^2 + a + 2)$ by $(a^2 + a - 2)$: $$a^3 + 2a^2 + a + 2 = (a + 1)(a^2 + a - 2) + (2a + 4)$$
2. Step 2: Remainder $R_1 = 2a + 4 = 2(a + 2)$. Discard the non-zero constant $2$ to use $(a + 2)$ as new divisor.
3. Step 3: Divide previous divisor $(a^2 + a - 2)$ by $(a + 2)$: $$a^2 + a - 2 = (a - 1)(a + 2) + 0$$
4. Step 4: Since remainder is zero, the last divisor is the HCF: $\mathbf{a + 2}$.

Q4. Find HCF of $x^3 + 2x^2 - 4x - 8$ and $2x^3 + 7x^2 + 4x - 4$ by division:

1. Step 1: Multiply 1st polynomial by $2$ to match leading coefficients: $2x^3 + 4x^2 - 8x - 16$. Subtract from $(2x^3 + 7x^2 + 4x - 4)$: $$(2x^3 + 7x^2 + 4x - 4) - (2x^3 + 4x^2 - 8x - 16) = 3x^2 + 12x + 12 = 3(x^2 + 4x + 4)$$
2. Step 2: Discard constant $3$. New divisor is $x^2 + 4x + 4 = (x + 2)^2$.
3. Step 3: Divide $(x^3 + 2x^2 - 4x - 8)$ by $(x^2 + 4x + 4)$: $$x^3 + 2x^2 - 4x - 8 = (x - 2)(x^2 + 4x + 4) + 0$$
4. Step 4: Remainder is $0$. Therefore, $\mathbf{\text{HCF} = x^2 + 4x + 4 = (x + 2)^2}$.

Q5. Find HCF of $2x^3 + x^2 - x - 2$ and $3x^3 - x^2 + x - 3$ by division:

1. Step 1: Multiply $P_1$ by $3$ and $P_2$ by $2$: $$3(2x^3 + x^2 - x - 2) = 6x^3 + 3x^2 - 3x - 6$$ $$2(3x^3 - x^2 + x - 3) = 6x^3 - 2x^2 + 2x - 6$$
2. Step 2: Subtract: $$(6x^3 + 3x^2 - 3x - 6) - (6x^3 - 2x^2 + 2x - 6) = 5x^2 - 5x = 5x(x - 1)$$
3. Step 3: Discard $5x$ (or test factor $x - 1$). Divide $(2x^3 + x^2 - x - 2)$ by $(x - 1)$: $$2x^3 + x^2 - x - 2 = (2x^2 + 3x + 2)(x - 1) + 0$$
4. Step 4: Remainder is $0$. Therefore, $\mathbf{\text{HCF} = x - 1}$.

Q6. Find HCF of $2p^4 + 5p^2 + 3$ and $5p^3 + 3p^2 + 5p + 3$:

1. Step 1: Factorize $P_1$: $2p^4 + 5p^2 + 3 = (2p^2 + 3)(p^2 + 1)$.
2. Step 2: Group $P_2$: $5p(p^2 + 1) + 3(p^2 + 1) = (5p + 3)(p^2 + 1)$.
3. Step 3: Common factor is $\mathbf{p^2 + 1}$.

Q7. Find HCF of $24x^4 - 2x^3 - 60x^2 - 32x$ and $18x^4 - 6x^3 - 39x^2 - 18x$:

1. Step 1: Extract common monomials from both polynomials: $$P_1 = 2x(12x^3 - x^2 - 30x - 16)$$ $$P_2 = 3x(6x^3 - 2x^2 - 13x - 6)$$
2. Step 2: Common monomial factor is $\gcd(2x, 3x) = \mathbf{x}$.
3. Step 3: Perform division on $A(x) = 12x^3 - x^2 - 30x - 16$ and $B(x) = 6x^3 - 2x^2 - 13x - 6$: $$A(x) - 2B(x) = (12x^3 - x^2 - 30x - 16) - (12x^3 - 4x^2 - 26x - 12) = 3x^2 - 4x - 4$$
4. Step 4: Factorize $3x^2 - 4x - 4 = (3x + 2)(x - 2)$.
5. Step 5: Divide $B(x) = 6x^3 - 2x^2 - 13x - 6$ by $(3x^2 - 4x - 4)$: $$6x^3 - 2x^2 - 13x - 6 = (2x + 2)(3x^2 - 4x - 4) + (3x + 2)$$ $$3x^2 - 4x - 4 = (x - 2)(3x + 2) + 0$$
6. Step 6: Polynomial HCF is $(3x + 2)$. Including common monomial $x$: $$\mathbf{\text{HCF} = x(3x + 2) = 3x^2 + 2x}$$

Q8. Find HCF of $2x^3 + 6x^2 + x + 3,\; 3x^3 + 9x^2 - 2x - 6,$ and $x^3 + 3x^2 + 2x + 6$:

1. Step 1: Factorize $P_1 = 2x^2(x + 3) + 1(x + 3) = (x + 3)(2x^2 + 1)$.
2. Step 2: Factorize $P_2 = 3x^2(x + 3) - 2(x + 3) = (x + 3)(3x^2 - 2)$.
3. Step 3: Factorize $P_3 = x^2(x + 3) + 2(x + 3) = (x + 3)(x^2 + 2)$.
4. Step 4: The only common factor across all three polynomials is $\mathbf{x + 3}$.

Q9. Find LCM of $9a^2b - b$ and $6a^2 + 2a$:

1. Step 1: Factorize $P_1 = b(9a^2 - 1) = b(3a - 1)(3a + 1)$.
2. Step 2: Factorize $P_2 = 2a(3a + 1)$.
3. Step 3: $\text{LCM} = 2ab(3a - 1)(3a + 1) = \mathbf{2ab(9a^2 - 1)}$.

Q10. Find LCM of $p^3q - pq^3$ and $p^5q^2 - p^2q^5$:

1. Step 1: $P_1 = pq(p^2 - q^2) = pq(p - q)(p + q)$.
2. Step 2: $P_2 = p^2q^2(p^3 - q^3) = p^2q^2(p - q)(p^2 + pq + q^2)$.
3. Step 3: $\text{LCM} = p^2q^2(p - q)(p + q)(p^2 + pq + q^2) = \mathbf{p^2q^2(p^2 - q^2)(p^2 + pq + q^2)}$.

Q11. Find LCM of $4x^2y - y$ and $2x^2 + x$:

1. Step 1: $P_1 = y(4x^2 - 1) = y(2x - 1)(2x + 1)$.
2. Step 2: $P_2 = x(2x + 1)$.
3. Step 3: $\text{LCM} = \mathbf{xy(2x - 1)(2x + 1) = xy(4x^2 - 1)}$.

Q12. Find LCM of $x^2 - x - 6,\; x^2 + x - 2,$ and $x^2 - 4x + 3$:

1. Step 1: $P_1 = x^2 - x - 6 = (x - 3)(x + 2)$.
2. Step 2: $P_2 = x^2 + x - 2 = (x + 2)(x - 1)$.
3. Step 3: $P_3 = x^2 - 4x + 3 = (x - 3)(x - 1)$.
4. Step 4: Distinct factors with highest power: $\mathbf{(x - 1)(x + 2)(x - 3)}$.

Q13. Find LCM of $m^6 - 1,\; m^4 - 1,$ and $m^3 - 1$:

1. Step 1: $m^6 - 1 = (m^3 - 1)(m^3 + 1) = (m - 1)(m^2 + m + 1)(m + 1)(m^2 - m + 1)$.
2. Step 2: $m^4 - 1 = (m^2 - 1)(m^2 + 1) = (m - 1)(m + 1)(m^2 + 1)$.
3. Step 3: $m^3 - 1 = (m - 1)(m^2 + m + 1)$.
4. Step 4: $\text{LCM} = (m - 1)(m + 1)(m^2 + 1)(m^2 + m + 1)(m^2 - m + 1) = \mathbf{(m^4 - 1)(m^6 - 1) / (m^2 - 1) = (m^4 - 1)(m^4 + m^2 + 1)}$.

Q14. Find LCM of $x^3 + 2x^2 - x - 2,\; x^2 - x - 2,$ and $x^2 - 4$:

1. Step 1: $P_1 = x^2(x + 2) - 1(x + 2) = (x + 2)(x^2 - 1) = (x + 2)(x - 1)(x + 1)$.
2. Step 2: $P_2 = (x - 2)(x + 1)$.
3. Step 3: $P_3 = (x - 2)(x + 2)$.
4. Step 4: $\text{LCM} = \mathbf{(x - 1)(x + 1)(x - 2)(x + 2) = (x^2 - 1)(x^2 - 4)}$.

Q15. Find LCM of $x^2 + x - 20,\; x^2 - 10x + 24,$ and $x^2 - x - 30$:

1. Step 1: $P_1 = x^2 + x - 20 = (x + 5)(x - 4)$.
2. Step 2: $P_2 = x^2 - 10x + 24 = (x - 6)(x - 4)$.
3. Step 3: $P_3 = x^2 - x - 30 = (x - 6)(x + 5)$.
4. Step 4: $\text{LCM} = \mathbf{(x + 5)(x - 4)(x - 6)}$.

Q1. Find the HCF and LCM of the following:

i. $16 - 4x^2$ and $x^2 + x - 6$:

1. Step 1: Factorize $P = 16 - 4x^2 = 4(4 - x^2) = 4(2 - x)(2 + x) = -4(x - 2)(x + 2)$.
2. Step 2: Factorize $Q = x^2 + x - 6 = (x + 3)(x - 2)$.
3. Step 3: $\mathbf{\text{HCF} = x - 2}$.
4. Step 4: $\mathbf{\text{LCM} = 4(x - 2)(x + 2)(x + 3) = 4(x^2 - 4)(x + 3)}$.

ii. $a^4 - a^3 - a + 1$ and $a^4 + a^2 + 1$:

1. Step 1: $P = a^3(a - 1) - 1(a - 1) = (a - 1)(a^3 - 1) = (a - 1)(a - 1)(a^2 + a + 1) = (a - 1)^2(a^2 + a + 1)$.
2. Step 2: $Q = a^4 + a^2 + 1 = (a^2 - a + 1)(a^2 + a + 1)$.
3. Step 3: $\mathbf{\text{HCF} = a^2 + a + 1}$.
4. Step 4: $\mathbf{\text{LCM} = (a - 1)^2(a^2 + a + 1)(a^2 - a + 1) = (a - 1)^2(a^4 + a^2 + 1)}$.

iii. $x^3 + 2x^2 - 3x$ and $2x^3 + 5x^2 - 3x$:

1. Step 1: $P = x(x^2 + 2x - 3) = x(x + 3)(x - 1)$.
2. Step 2: $Q = x(2x^2 + 5x - 3) = x(2x^2 + 6x - x - 3) = x(2x - 1)(x + 3)$.
3. Step 3: $\mathbf{\text{HCF} = x(x + 3) = x^2 + 3x}$.
4. Step 4: $\mathbf{\text{LCM} = x(x + 3)(x - 1)(2x - 1)}$.

Q2. If $\text{HCF} = x - 7$ and $\text{LCM} = x^3 - 10x^2 + 11x + 70$, find the product of the two polynomials:

1. Step 1: Formula: $\text{Product } P(x) \cdot Q(x) = \text{HCF} \times \text{LCM}$.
2. Step 2: Multiply $(x - 7)(x^3 - 10x^2 + 11x + 70)$: $$x(x^3 - 10x^2 + 11x + 70) - 7(x^3 - 10x^2 + 11x + 70)$$ $$= (x^4 - 10x^3 + 11x^2 + 70x) - (7x^3 - 70x^2 + 77x + 490)$$
3. Step 3: Combine like terms: $$= x^4 + (-10 - 7)x^3 + (11 + 70)x^2 + (70 - 77)x - 490$$ $$= \mathbf{x^4 - 17x^3 + 81x^2 - 7x - 490}$$

Q3. Product of two polynomials is $x^4 + 3x^3 - 12x^2 - 20x + 48$ and their $\text{HCF} = x - 2$. Find their LCM:

1. Step 1: $\text{LCM} = \frac{P(x) \times Q(x)}{\text{HCF}} = \frac{x^4 + 3x^3 - 12x^2 - 20x + 48}{x - 2}$.
2. Step 2: Perform polynomial division: $$\frac{x^4 - 2x^3}{x - 2} \implies x^3$$ $$\frac{5x^3 - 10x^2}{x - 2} \implies +5x^2$$ $$\frac{-2x^2 + 4x}{x - 2} \implies -2x$$ $$\frac{-24x + 48}{x - 2} \implies -24$$
3. Step 3: Quotient is $\mathbf{\text{LCM} = x^3 + 5x^2 - 2x - 24}$.

Q4. The product of two polynomials is $y^4 + 6y^3 - 3y^2 - 56y - 48$ and their $\text{LCM} = y^3 + 2y^2 - 11y - 12$. Find their HCF:

1. Step 1: $\text{HCF} = \frac{\text{Product}}{\text{LCM}} = \frac{y^4 + 6y^3 - 3y^2 - 56y - 48}{y^3 + 2y^2 - 11y - 12}$.
2. Step 2: Perform polynomial division: $$\frac{y^4 + 6y^3 - 3y^2 - 56y - 48}{y^3 + 2y^2 - 11y - 12} = y + 4$$ $$\text{Check: } (y + 4)(y^3 + 2y^2 - 11y - 12) = y^4 + 2y^3 - 11y^2 - 12y + 4y^3 + 8y^2 - 44y - 48$$ $$= y^4 + 6y^3 - 3y^2 - 56y - 48 \quad \checkmark$$
3. Step 3: $\mathbf{\text{HCF} = y + 4}$.

Q5. Find the second polynomial $Q(x)$ when $P(x) = x^4 + x^3 + x + 1$, $\text{HCF} = x + 1$, and $\text{LCM} = (x^3 + 1)(x^4 + x^3 - x - 1)$:

1. Step 1: Factorize $P(x) = x^3(x + 1) + 1(x + 1) = (x + 1)(x^3 + 1)$.
2. Step 2: Formula: $Q(x) = \frac{\text{HCF} \times \text{LCM}}{P(x)}$. $$Q(x) = \frac{(x + 1) \cdot (x^3 + 1)(x^4 + x^3 - x - 1)}{(x + 1)(x^3 + 1)}$$
3. Step 3: Cancel $(x + 1)(x^3 + 1)$ from numerator and denominator: $$Q(x) = \mathbf{x^4 + x^3 - x - 1 = (x^3 - 1)(x + 1)}$$

Q6. Find the LCM of polynomials $P(x) = 4x^3 - 10x^2 + 4x + 2$ and $Q(x) = 3x^4 - 2x^3 - 3x + 2$ if their $\text{HCF} = x - 1$:

1. Step 1: Divide $Q(x)$ by $\text{HCF} = x - 1$: $$\frac{3x^4 - 2x^3 - 3x + 2}{x - 1} = 3x^3 + x^2 + x - 2$$
2. Step 2: $\text{LCM} = P(x) \times \frac{Q(x)}{\text{HCF}} = (4x^3 - 10x^2 + 4x + 2)(3x^3 + x^2 + x - 2)$.
3. Step 3: Alternatively, factorize completely: $$P(x) = 2(2x^3 - 5x^2 + 2x + 1) = 2(x - 1)(2x^2 - 3x - 1)$$ $$Q(x) = (x - 1)(3x^3 + x^2 + x - 2) = (x - 1)(3x - 2)(x^2 + x + 1)$$ $$\mathbf{\text{LCM} = 2(x - 1)(2x^2 - 3x - 1)(3x^3 + x^2 + x - 2)}$$

Q1. Answer these without calculations:

1. i. Product of what algebraic fraction and $(x^3 + 7x - 8)$ is $1$?
   The multiplicative inverse (reciprocal): $\mathbf{\frac{1}{x^3 + 7x - 8}}$.
2. ii. Which algebraic fraction divided by $\frac{x^2}{x^2 + y^2}$ gives $1$?
   Any expression divided by itself equals $1$: $\mathbf{\frac{x^2}{x^2 + y^2}}$.
3. iii. Sum of what algebraic fraction and $\frac{m}{m^2 + n^2}$ is $\frac{m + n}{m^2 + n^2}$?
   Since denominators are identical, $\frac{m + n}{m^2 + n^2} - \frac{m}{m^2 + n^2} = \mathbf{\frac{n}{m^2 + n^2}}$.
4. iv. What is the product of an algebraic fraction and its reciprocal?
   $\frac{P(x)}{Q(x)} \times \frac{Q(x)}{P(x)} = \mathbf{1}$.

Q2. Simplify $\frac{14x^2 - 7x}{12x^3 + 24x^2} \times \frac{x^2 + 2x}{2x - 1}$:

1. Step 1: Factorize every numerator and denominator: $$\frac{7x(2x - 1)}{12x^2(x + 2)} \times \frac{x(x + 2)}{2x - 1}$$
2. Step 2: Cancel common factors $(2x - 1)$ and $(x + 2)$: $$\frac{7x \cdot x}{12x^2} = \frac{7x^2}{12x^2}$$
3. Step 3: Cancel $x^2$: $\mathbf{\frac{7}{12}}$.

Q3. Simplify $\frac{a^2b^2 + 3ab}{4a^2 - 1} \times \frac{2a + 1}{ab + 3}$:

1. Step 1: Factorize each component: $$\frac{ab(ab + 3)}{(2a - 1)(2a + 1)} \times \frac{2a + 1}{ab + 3}$$
2. Step 2: Cancel $(ab + 3)$ and $(2a + 1)$: $$\mathbf{\frac{ab}{2a - 1}}$$

Q4. Simplify $\frac{a - b}{a^2 + ab} \times \frac{a^4 - b^4}{a^2 - 2ab + b^2} \times \frac{a}{a^2 + b^2}$:

1. Step 1: Factorize every expression: $$\frac{a - b}{a(a + b)} \times \frac{(a^2 - b^2)(a^2 + b^2)}{(a - b)^2} \times \frac{a}{a^2 + b^2}$$
2. Step 2: Expand $(a^2 - b^2) = (a - b)(a + b)$: $$\frac{a - b}{a(a + b)} \times \frac{(a - b)(a + b)(a^2 + b^2)}{(a - b)^2} \times \frac{a}{a^2 + b^2}$$
3. Step 3: Cancel $(a - b)^2, (a + b), (a^2 + b^2),$ and $a$: $$\mathbf{1}$$

Q5. Simplify $\frac{6x^2y^2}{x^2 - y^2} \div \frac{3xy}{x + y}$:

1. Step 1: Invert the divisor fraction and multiply: $$\frac{6x^2y^2}{(x - y)(x + y)} \times \frac{x + y}{3xy}$$
2. Step 2: Cancel $(x + y)$ and divide $\frac{6x^2y^2}{3xy} = 2xy$: $$\mathbf{\frac{2xy}{x - y}}$$

Q6. Simplify $\frac{8a^3 - 1}{4a^3 + 2a^2} \div \frac{6a^2 - 13a + 5}{15a - 25} \times \frac{2a^4 + a^3}{15a^2}$:

1. Step 1: Factorize all parts: $$8a^3 - 1 = (2a - 1)(4a^2 + 2a + 1)$$ $$4a^3 + 2a^2 = 2a^2(2a + 1)$$ $$6a^2 - 13a + 5 = 6a^2 - 10a - 3a + 5 = 2a(3a - 5) - 1(3a - 5) = (3a - 5)(2a - 1)$$ $$15a - 25 = 5(3a - 5)$$ $$2a^4 + a^3 = a^3(2a + 1)$$
2. Step 2: Convert division to multiplication by inverting second fraction: $$\frac{(2a - 1)(4a^2 + 2a + 1)}{2a^2(2a + 1)} \times \frac{5(3a - 5)}{(3a - 5)(2a - 1)} \times \frac{a^3(2a + 1)}{15a^2}$$
3. Step 3: Cancel factors:
   • $(2a - 1)$ cancels
   • $(3a - 5)$ cancels
   • $(2a + 1)$ cancels
   • Coefficients: $\frac{5}{2 \times 15} = \frac{5}{30} = \frac{1}{6}$
   • Variables: $\frac{a^3}{a^2 \cdot a^2} = \frac{a^3}{a^4} = \frac{1}{a}$
4. Step 4: Combine remaining terms: $$\mathbf{\frac{4a^2 + 2a + 1}{6a}}$$

Q7. Simplify $\frac{x^2 - 8x - 9}{x^2 - 17x + 72} \times \frac{x^2 - 25}{x^2 - 1} \div \frac{x^2 + 4x - 5}{x^2 - 9x + 8}$:

1. Step 1: Factorize every quadratic: $$\frac{(x - 9)(x + 1)}{(x - 9)(x - 8)} \times \frac{(x - 5)(x + 5)}{(x - 1)(x + 1)} \times \frac{(x - 8)(x - 1)}{(x + 5)(x - 1)}$$
2. Step 2: Cancel matching factors across numerators and denominators:
   • $(x - 9)$ cancels
   • $(x + 1)$ cancels
   • $(x - 8)$ cancels
   • $(x + 5)$ cancels
   • $(x - 1)$ cancels
3. Step 3: Leftover fraction: $$\mathbf{\frac{x - 5}{x - 1}}$$

Q8. Simplify $\frac{1}{2x - 3y} - \frac{x + y}{4x^2 - 9y^2}$:

1. Step 1: Factorize denominator $4x^2 - 9y^2 = (2x - 3y)(2x + 3y)$.
2. Step 2: Common denominator is $(2x - 3y)(2x + 3y)$: $$\frac{1(2x + 3y) - (x + y)}{(2x - 3y)(2x + 3y)} = \frac{2x + 3y - x - y}{(2x - 3y)(2x + 3y)}$$
3. Step 3: Simplify numerator: $$\mathbf{\frac{x + 2y}{4x^2 - 9y^2}}$$

Q9. Simplify $\frac{1}{x(x - y)} + \frac{1}{y(x + y)}$:

1. Step 1: Common denominator is $xy(x - y)(x + y)$:
2. Step 2: Combine numerators: $$\frac{y(x + y) + x(x - y)}{xy(x^2 - y^2)} = \frac{xy + y^2 + x^2 - xy}{xy(x^2 - y^2)}$$
3. Step 3: Simplify: $$\mathbf{\frac{x^2 + y^2}{xy(x^2 - y^2)}}$$

Q10. Simplify $\frac{5x + 5}{3(2x - 1)} + \frac{6 - 2x}{2(1 - 2x)}$:

1. Step 1: Note that $(1 - 2x) = -(2x - 1)$. Rewrite second fraction: $$\frac{5(x + 1)}{3(2x - 1)} - \frac{2(3 - x)}{2(2x - 1)} = \frac{5x + 5}{3(2x - 1)} - \frac{3 - x}{2x - 1}$$
2. Step 2: Common denominator $3(2x - 1)$: $$\frac{5x + 5 - 3(3 - x)}{3(2x - 1)} = \frac{5x + 5 - 9 + 3x}{3(2x - 1)} = \frac{8x - 4}{3(2x - 1)}$$
3. Step 3: Factorize numerator $8x - 4 = 4(2x - 1)$: $$\frac{4(2x - 1)}{3(2x - 1)} = \mathbf{\frac{4}{3}}$$

Q11. Simplify $\frac{2a}{2a - 3} - \frac{5}{6a + 9} - \frac{4(3a + 2)}{3(4a^2 - 9)}$:

1. Step 1: Factorize denominators: $$6a + 9 = 3(2a + 3)$$ $$3(4a^2 - 9) = 3(2a - 3)(2a + 3)$$
2. Step 2: LCD is $3(2a - 3)(2a + 3)$: $$\frac{2a \cdot 3(2a + 3) - 5(2a - 3) - 4(3a + 2)}{3(2a - 3)(2a + 3)}$$
3. Step 3: Expand numerator: $$6a(2a + 3) - 10a + 15 - 12a - 8 = 12a^2 + 18a - 22a + 7 = 12a^2 - 4a + 7$$
4. Step 4: Final simplified expression: $$\mathbf{\frac{12a^2 - 4a + 7}{3(4a^2 - 9)}}$$

Q12. Simplify $\frac{5}{5 + x - 18x^2} - \frac{2}{2 + 5x + 2x^2}$:

1. Step 1: Factorize denominators: $$5 + x - 18x^2 = -(18x^2 - x - 5) = -(2x - 1)(9x + 5) = (1 - 2x)(9x + 5)$$ $$2 + 5x + 2x^2 = 2x^2 + 4x + x + 2 = 2x(x + 2) + 1(x + 2) = (2x + 1)(x + 2)$$
2. Step 2: Combine over common denominator: $$\mathbf{\frac{5(2x + 1)(x + 2) - 2(1 - 2x)(9x + 5)}{(1 - 2x)(9x + 5)(2x + 1)(x + 2)}}$$

Q13. Simplify $\frac{1 - p^2}{1 + q} \times \frac{1 - q^2}{p + p^2} \times \left(1 + \frac{p}{1 - p}\right)$:

1. Step 1: Simplify the parenthetical term: $$1 + \frac{p}{1 - p} = \frac{1 - p + p}{1 - p} = \frac{1}{1 - p}$$
2. Step 2: Factorize all terms in the product: $$\frac{(1 - p)(1 + p)}{1 + q} \times \frac{(1 - q)(1 + q)}{p(1 + p)} \times \frac{1}{1 - p}$$
3. Step 3: Cancel identical factors:
   • $(1 - p)$ cancels with $(1 - p)$
   • $(1 + p)$ cancels with $(1 + p)$
   • $(1 + q)$ cancels with $(1 + q)$
4. Step 4: Remaining fraction: $$\mathbf{\frac{1 - q}{p}}$$

Q1. Find square root of $16y^2 - 56y + 49$ by factorization:

1. Step 1: Express as perfect square trinomial $(a - b)^2 = a^2 - 2ab + b^2$: $$(4y)^2 - 2(4y)(7) + 7^2 = (4y - 7)^2$$
2. Step 2: Take square root: $\sqrt{(4y - 7)^2} = \mathbf{\pm(4y - 7)}$.

Q2. Find square root of $25a^4 - 30a^3 + 9a^2$ by factorization:

1. Step 1: Extract common factor $a^2$: $a^2(25a^2 - 30a + 9)$.
2. Step 2: Complete the square inside: $a^2(5a - 3)^2 = [a(5a - 3)]^2$.
3. Step 3: Square root: $\mathbf{\pm a(5a - 3) = \pm(5a^2 - 3a)}$.

Q3. Find square root of $\left(x^2 - \frac{1}{x^2}\right)^2 + 4\left(x^2 - \frac{1}{x^2}\right) + 4$ ($x \ne 0$):

1. Step 1: Substitute $u = x^2 - \frac{1}{x^2}$: $u^2 + 4u + 4 = (u + 2)^2$.
2. Step 2: Substitute back $u$: $\left(x^2 - \frac{1}{x^2} + 2\right)^2$.
3. Step 3: Square root: $\mathbf{\pm\left(x^2 - \frac{1}{x^2} + 2\right)}$.

Q4. Find square root of $\left(a^2 + \frac{1}{a^2}\right) - 8\left(a - \frac{1}{a}\right) + 14$ ($a \ne 0$):

1. Step 1: Note that $a^2 + \frac{1}{a^2} = \left(a - \frac{1}{a}\right)^2 + 2$.
2. Step 2: Substitute: $$\left(a - \frac{1}{a}\right)^2 + 2 - 8\left(a - \frac{1}{a}\right) + 14 = \left(a - \frac{1}{a}\right)^2 - 8\left(a - \frac{1}{a}\right) + 16$$
3. Step 3: Let $u = a - \frac{1}{a} \implies u^2 - 8u + 16 = (u - 4)^2$.
4. Step 4: Take square root: $\mathbf{\pm\left(a - \frac{1}{a} - 4\right)}$.

Q5. Find square root of $(a + 2)(a + 4)(a + 6)(a + 8) + 16$:

1. Step 1: Pair terms: $[(a + 2)(a + 8)][(a + 4)(a + 6)] + 16 = (a^2 + 10a + 16)(a^2 + 10a + 24) + 16$.
2. Step 2: Let $y = a^2 + 10a + 16 \implies y(y + 8) + 16 = y^2 + 8y + 16 = (y + 4)^2$.
3. Step 3: Substitute back $y$: $(a^2 + 10a + 16 + 4)^2 = (a^2 + 10a + 20)^2$.
4. Step 4: Square root: $\mathbf{\pm(a^2 + 10a + 20)}$.

Q6. Find square root of $x^4 + 8x^3 + 20x^2 + 16x + 4$ by division:

1. Step 1: First root term is $\sqrt{x^4} = x^2$. Subtract $x^4$, remainder is $8x^3 + 20x^2$.
2. Step 2: Double divisor: $2x^2$. Next term is $\frac{8x^3}{2x^2} = +4x$. Divisor becomes $(2x^2 + 4x)$. Product: $8x^3 + 16x^2$.
3. Step 3: Subtract: remainder is $4x^2 + 16x + 4$.
4. Step 4: New divisor is $2(x^2 + 4x) = 2x^2 + 8x$. Next term is $\frac{4x^2}{2x^2} = +2$. Divisor becomes $(2x^2 + 8x + 2)$. Product: $4x^2 + 16x + 4$. Remainder $= 0$.
5. Step 5: Result: $\mathbf{\pm(x^2 + 4x + 2)}$.

Q7. Find square root of $x^4 + 10x^3 + 31x^2 + 30x + 9$ by division:

1. Step 1: 1st term: $x^2$. Subtract $x^4$, bring down $10x^3 + 31x^2$.
2. Step 2: Divisor $2x^2 + 5x \implies$ product $10x^3 + 25x^2$. Subtract $\implies 6x^2 + 30x + 9$.
3. Step 3: Divisor $2x^2 + 10x + 3 \implies$ product $6x^2 + 30x + 9$. Remainder $= 0$.
4. Step 4: Result: $\mathbf{\pm(x^2 + 5x + 3)}$.

Q8. Find square root of $49b^4 + 18a^2b^2 + 4a^3b + a^4 + 28ab^3$:

1. Step 1: Arrange in descending powers of $a$: $a^4 + 4a^3b + 18a^2b^2 + 28ab^3 + 49b^4$.
2. Step 2: Division method:
   • 1st term: $a^2 \implies$ subtract $a^4$, bring down $4a^3b + 18a^2b^2$.
   • Divisor $2a^2 + 2ab \implies$ subtract $4a^3b + 4a^2b^2$, leaves $14a^2b^2 + 28ab^3 + 49b^4$.
   • Divisor $2a^2 + 4ab + 7b^2 \implies$ product $14a^2b^2 + 28ab^3 + 49b^4$. Remainder $= 0$.
3. Step 3: Result: $\mathbf{\pm(a^2 + 2ab + 7b^2)}$.

Q9. Find square root of $4x^4 - 12x^3 + 29x^2 - 30x + 25$:

1. Step 1: 1st root term: $\sqrt{4x^4} = 2x^2$. Subtract $4x^4$, bring down $-12x^3 + 29x^2$.
2. Step 2: Divisor $4x^2 - 3x \implies$ subtract $-12x^3 + 9x^2 \implies 20x^2 - 30x + 25$.
3. Step 3: Divisor $4x^2 - 6x + 5 \implies$ subtract $20x^2 - 30x + 25 \implies 0$.
4. Step 4: Result: $\mathbf{\pm(2x^2 - 3x + 5)}$.

Q10. Find square root of $1 - 10x + 27x^2 - 10x^3 + x^4$:

1. Step 1: Arrange in descending order: $x^4 - 10x^3 + 27x^2 - 10x + 1$.
2. Step 2: Division gives root terms $x^2 - 5x + 1$.
3. Step 3: Result: $\mathbf{\pm(x^2 - 5x + 1) \text{ or } \pm(1 - 5x + x^2)}$.

Q11. Find square root of $x^4 + \frac{1}{x^4} + 4x^2 - \frac{4}{x^2} + 2$ ($x \ne 0$):

1. Step 1: Arrange in standard descending powers: $x^4 + 4x^2 + 2 - \frac{4}{x^2} + \frac{1}{x^4}$.
2. Step 2: First term of root is $x^2$. Divisor $2x^2 + 2 \implies 4x^2 + 4$. Remainder is $-2 - \frac{4}{x^2} + \frac{1}{x^4}$.
3. Step 3: Next divisor is $2x^2 + 4 - \frac{1}{x^2} \implies$ product is $-2 - \frac{4}{x^2} + \frac{1}{x^4}$. Remainder $= 0$.
4. Step 4: Result: $\mathbf{\pm\left(x^2 + 2 - \frac{1}{x^2}\right)}$.

Q12. Find square root of $a^2 - 8a + 2 + \frac{56}{a} + \frac{49}{a^2}$ ($a \ne 0$):

1. Step 1: Division method gives root terms $a - 4 - \frac{7}{a}$.
2. Step 2: Verification: $\left(a - 4 - \frac{7}{a}\right)^2 = a^2 + 16 + \frac{49}{a^2} - 8a - 14 + \frac{56}{a} = a^2 - 8a + 2 + \frac{56}{a} + \frac{49}{a^2} \quad \checkmark$.
3. Step 3: Result: $\mathbf{\pm\left(a - 4 - \frac{7}{a}\right)}$.

Q13. Find square root of $x^4 - 2x^3 + \frac{3x^2}{2} - \frac{x}{2} + \frac{1}{16}$:

1. Step 1: Division gives quotient: $x^2 - x + \frac{1}{4}$.
2. Step 2: Verification: $\left(x^2 - x + \frac{1}{4}\right)^2 = x^4 - 2x^3 + \frac{3x^2}{2} - \frac{x}{2} + \frac{1}{16} \quad \checkmark$.
3. Step 3: Result: $\mathbf{\pm\left(x^2 - x + \frac{1}{4}\right)}$.

Q14. Find square root of $4x^4 + 32x^2 + 96 + \frac{128}{x^2} + \frac{64}{x^4}$:

1. Step 1: Division gives quotient: $2x^2 + 8 + \frac{8}{x^2}$.
2. Step 2: Result: $\mathbf{\pm\left(2x^2 + 8 + \frac{8}{x^2}\right)}$.

Q15. To make $a^4 - 10a^3 + 27a^2 - 9a + 2$ a perfect square:

1. Step 1: Long division by square root gives quotient $(a^2 - 5a + 1)$ with remainder $R = a + 1$.
2. (i) What should be added?
   Add the negative of the remainder: $-(a + 1) = \mathbf{-a - 1}$.
3. (ii) What should be subtracted?
   Subtract the remainder directly: $\mathbf{a + 1}$.
4. (iii) What will be the value of $a$?
   Set remainder to zero: $a + 1 = 0 \implies \mathbf{a = -1}$.

Q16. Find the values of $p$ and $q$ if $x^4 - 12x^3 + px + q$ is a complete square:

1. Step 1: Since $(x^2 - 6x + k)^2 = x^4 - 12x^3 + (36 + 2k)x^2 - 12kx + k^2$.
2. Step 2: The coefficient of $x^2$ is $0$, so $36 + 2k = 0 \implies k = -18$.
3. Step 3: Match remaining coefficients: $$p = -12k = -12(-18) = \mathbf{216}$$ $$q = k^2 = (-18)^2 = \mathbf{324}$$

Q17. For what value of $k$, the expression $y^4 + 4y^2 + k + \frac{8}{y^2} + \frac{4}{y^4}$ becomes a perfect square ($y \ne 0$):

1. Step 1: Notice the square form $\left(y^2 + 2 + \frac{2}{y^2}\right)^2$.
2. Step 2: Expand: $$\left(y^2 + 2 + \frac{2}{y^2}\right)^2 = y^4 + 4 + \frac{4}{y^4} + 4y^2 + \frac{8}{y^2} + 4 = y^4 + 4y^2 + 8 + \frac{8}{y^2} + \frac{4}{y^4}$$
3. Step 3: Comparing constant terms, we get $\mathbf{k = 8}$.

Q1. In a map of an industry, the area of a rectangular veranda is given by $(x^2 - 2x - 3)\text{ m}^2$. Find:

1. a. Expressions for both dimensions of the veranda:
   Factorize $x^2 - 2x - 3 = x^2 - 3x + x - 3 = (x - 3)(x + 1)$.
   Dimensions: $\mathbf{\text{Length} = (x + 1)\text{ m}}$ and $\mathbf{\text{Width} = (x - 3)\text{ m}}$.
2. b. Expression for perimeter of veranda:
   $\text{Perimeter } P = 2(\text{Length} + \text{Width}) = 2[(x + 1) + (x - 3)] = 2(2x - 2) = \mathbf{(4x - 4)\text{ meters}}$.
3. c. Expression for cost of fencing veranda @ Rs. 200/m:
   $\text{Cost} = \text{Rate} \times \text{Perimeter} = 200 \times (4x - 4) = \mathbf{\text{Rs. }(800x - 800)}$.
4. d. Expression for the cost of carpeting veranda floor @ Rs. 250/m²:
   $\text{Cost} = \text{Rate} \times \text{Area} = 250 \times (x^2 - 2x - 3) = \mathbf{\text{Rs. }(250x^2 - 500x - 750)}$.

Q2. Area of a square shaped surface of a machine is given by $(25x^2 - 30x + 9)\text{ m}^2$. Find:

1. a. Expression for the length of the surface:
   $\text{Area} = s^2 = 25x^2 - 30x + 9 = (5x - 3)^2$.
   $\mathbf{\text{Side length } s = (5x - 3)\text{ meters}}$.
2. b. Expression for the boundary (perimeter) of the surface:
   $\text{Perimeter } P = 4s = 4(5x - 3) = \mathbf{(20x - 12)\text{ meters}}$.
3. c. Expression for the cost of polishing the machine surface @ Rs. 75/m²:
   $\text{Cost} = 75 \times (25x^2 - 30x + 9) = \mathbf{\text{Rs. }(1875x^2 - 2250x + 675)}$.
4. d. Expression for the cost of edging around 2 sides of the machine surface @ Rs. 28/m:
   Length of 2 sides $= 2s = 2(5x - 3) = (10x - 6)\text{ m}$.
   $\text{Cost} = 28 \times (10x - 6) = \mathbf{\text{Rs. }(280x - 168)}$.

Q3. Volume of a cubical oil tank in an oil refinery is expressed as $(125x^3 - 150x^2 + 60x - 8)\text{ m}^3$. Find:

1. a. Expression for height of oil tank:
   Recognize cubic expansion $(a - b)^3 = (5x - 2)^3$.
   $\mathbf{\text{Height } h = \text{side length} = (5x - 2)\text{ meters}}$.
2. b. Expression for surface area of oil tank:
   $\text{Total Surface Area } A = 6s^2 = 6(5x - 2)^2 = 6(25x^2 - 20x + 4) = \mathbf{(150x^2 - 120x + 24)\text{ m}^2}$.
3. c. Expression for painting it from outside @ Rs. 32/m²:
   $\text{Cost} = 32 \times (150x^2 - 120x + 24) = \mathbf{\text{Rs. }(4800x^2 - 3840x + 768)}$.

Q4. A mechanical engineer working on wheels of a machine finds their areas given by $A_1 = \pi x^2 - 6\pi x + 9\pi$ and $A_2 = \pi x^2 - 10\pi x + 25\pi$. Help him find radii of both wheels:

1. Step 1: Formula for circular area is $A = \pi r^2$.
2. Step 2: Wheel 1: $A_1 = \pi(x^2 - 6x + 9) = \pi(x - 3)^2 \implies \mathbf{r_1 = (x - 3)\text{ units}}$.
3. Step 3: Wheel 2: $A_2 = \pi(x^2 - 10x + 25) = \pi(x - 5)^2 \implies \mathbf{r_2 = (x - 5)\text{ units}}$.

Q5. A machine has two square-shaped pressers with areas $25m^2$ and $36n^2$ respectively. The difference of these areas describes a rectangular presser. Find dimensions of the rectangular presser:

1. Step 1: Difference of areas: $A = 36n^2 - 25m^2$ (or $25m^2 - 36n^2$).
2. Step 2: Factorize as difference of two squares: $$36n^2 - 25m^2 = (6n)^2 - (5m)^2 = (6n - 5m)(6n + 5m)$$
3. Step 3: Dimensions of rectangular presser are $\mathbf{\text{Length} = (6n + 5m)}$ and $\mathbf{\text{Width} = (6n - 5m)}$.

Q6. Distance covered by a missile to hit the target is given by $(x^2 + 5x + 6)\text{ m}$. Find possible expressions for:

1. Step 1: Physical formula: $\text{Distance} = \text{Speed} \times \text{Time}$.
2. Step 2: Factorize the quadratic polynomial: $$x^2 + 5x + 6 = x^2 + 3x + 2x + 6 = (x + 3)(x + 2)$$
3. a. Expression for speed: $\mathbf{\text{Speed} = (x + 3)\text{ m/s}}$ (or $(x + 2)\text{ m/s}$).
4. b. Expression for time: $\mathbf{\text{Time} = (x + 2)\text{ seconds}}$ (or $(x + 3)\text{ seconds}$).

Q1. Encircle the correct option (All 20 Rigorous MCQs with Mathematical Justifications):

1. Factors of $-2 - a + a^2$ are:
   Answer: (d) $(a + 1)(a - 2)$ [Reason: $a^2 - a - 2 = (a - 2)(a + 1)$].
2. Factorization of $x^2 - x + \frac{1}{4}$ is:
   Answer: (d) $(x - \frac{1}{2})(x - \frac{1}{2})$ [Reason: Perfect square $(x - 1/2)^2$].
3. $(x^2 + m^2)(x + m)(x^4 + m^4)(x - m)$ is the factored form of:
   Answer: (c) $x^8 - m^8$ [Reason: $(x - m)(x + m) = x^2 - m^2$, then $(x^2 - m^2)(x^2 + m^2) = x^4 - m^4$, then $(x^4 - m^4)(x^4 + m^4) = x^8 - m^8$].
4. $a^4 + 64b^4$ is the product of:
   Answer: (a) $(a^2 - 4ab + 8b^2)$ and $(a^2 + 4ab + 8b^2)$ [Reason: Sophie Germain's identity].
5. $x(2a + 3b + 7) + x(2a + 3b + 5)$ equals:
   Answer: (a) $2x(2a + 3b + 6)$ [Reason: $x(2a + 3b + 7 + 2a + 3b + 5) = x(4a + 6b + 12) = 2x(2a + 3b + 6)$].
6. Which is the highest common factor of $-12x^2y^2,\; 6xy^3,\; 24x^2y^2$?
   Answer: (c) $6xy^2$ [Reason: $\gcd(-12,6,24) = 6$, lowest power of $x$ is $x^1$, $y$ is $y^2$].
7. What is the least common multiple of $12x^2y^2,\; 6xy^3,\; 24x^2y^2$?
   Answer: (d) $24x^2y^3$ [Reason: $\text{lcm}(12,6,24) = 24$, max powers $x^2, y^3$].
8. What is the highest common factor of $7x - 6xy$ and $5xy^3 - 3x^2$?
   Answer: (c) $x$ [Reason: $x(7 - 6y)$ and $x(5y^3 - 3x)$, common factor is $x$].
9. Least common multiple of $7x - 6xy$ and $5y^3x - 3x^2$ is:
   Answer: (d) $x(7 - 6y)(5y^3 - 3x)$.
10. HCF of $7x^3 - 8y^3$ and $3x^3 - 5y^3$ is:
    Answer: (a) $1$ [Reason: No common polynomial factors].
11. LCM of $7x^3 - 8y^3$ and $3x^3 - 5y^3$ is:
    Answer: (b) $(7x^3 - 8y^3)(3x^3 - 5y^3)$.
12. What is the product of $\frac{uv^2}{3w^3}$ and $\frac{6w^4}{u^2v^3}$?
    Answer: (b) $\frac{2w}{uv}$ [Reason: $\frac{6}{3} \cdot \frac{u}{u^2} \cdot \frac{v^2}{v^3} \cdot \frac{w^4}{w^3} = \frac{2w}{uv}$].
13. What is the quotient of $\frac{3y^2}{10} \div \frac{y^3}{2}$?
    Answer: (b) $\frac{3}{5y}$ [Reason: $\frac{3y^2}{10} \times \frac{2}{y^3} = \frac{6y^2}{10y^3} = \frac{3}{5y}$].
14. What is the sum of $\frac{2a}{a^2 - 1}$ and $\frac{-a}{a^2 - 1}$?
    Answer: (b) $\frac{a}{a^2 - 1}$ [Reason: $\frac{2a - a}{a^2 - 1} = \frac{a}{a^2 - 1}$].
15. What is the difference of $\frac{-3x}{x + y}$ and $\frac{x}{x + y}$?
    Answer: (d) $\frac{-4x}{x + y}$ [Reason: $\frac{-3x - x}{x + y} = \frac{-4x}{x + y}$].
16. If product of two polynomials is $(a - b)^2(a^2 + ab + b^2)$ and their HCF is $(a - b)$, what is their LCM?
    Answer: (a) $a^3 - b^3$ or (b) $(a - b)(a^2 + ab + b^2)$ [Reason: $\frac{(a - b)^2(a^2 + ab + b^2)}{a - b} = (a - b)(a^2 + ab + b^2) = a^3 - b^3$].
17. If product of HCF and LCM of two polynomials is $(x^3 - y^3)(x + y)$, what is the product of these polynomials?
    Answer: (c) $(x^2 - y^2)(x^2 + xy + y^2)$ [Reason: $(x^3 - y^3)(x + y) = (x - y)(x^2 + xy + y^2)(x + y) = (x^2 - y^2)(x^2 + xy + y^2)$].
18. What is the square root of $36x^6y^{16}$?
    Answer: (c) $6x^3y^8$ [Reason: $\sqrt{36(x^3)^2(y^8)^2} = 6x^3y^8$].
19. What is the square root of $(15x^2 - 7y^2)^4$?
    Answer: (b) $\pm(15x^2 - 7y^2)^2$ [Reason: $\sqrt{u^4} = u^2$].
20. What is the square root of $\left[ -\left(2x + \frac{1}{x} + 1\right) \right]^2$?
    Answer: (a) $\pm\left(2x + \frac{1}{x} + 1\right)$.

Q2. Factorize $(a^2 - 5)^2 - 13(a^2 - 5) + 36$:

1. Step 1: Substitute $u = a^2 - 5 \implies u^2 - 13u + 36$.
2. Step 2: Factorize $(u - 9)(u - 4)$.
3. Step 3: Substitute back $u = a^2 - 5$: $$(a^2 - 5 - 9)(a^2 - 5 - 4) = (a^2 - 14)(a^2 - 9)$$
4. Step 4: Factorize difference of squares $(a^2 - 9) = (a - 3)(a + 3)$: $$\mathbf{(a - 3)(a + 3)(a^2 - 14)}$$

Q3. Factorize $6x^2 + 19x + 15$:

1. Step 1: Product $a \cdot c = 6 \times 15 = 90$. Factors adding to $19$ are $10$ and $9$.
2. Step 2: Split middle term: $6x^2 + 10x + 9x + 15 = 2x(3x + 5) + 3(3x + 5) = \mathbf{(3x + 5)(2x + 3)}$.

Q4. Factorize $(x + 1)(x - 3)(x - 5)(x - 9) + 44$:

1. Step 1: Pair factors where constant sums match: $1 + (-9) = -8$ and $(-3) + (-5) = -8$. $$[(x + 1)(x - 9)] \cdot [(x - 3)(x - 5)] + 44 = (x^2 - 8x - 9)(x^2 - 8x + 15) + 44$$
2. Step 2: Let $y = x^2 - 8x$: $$(y - 9)(y + 15) + 44 = y^2 + 6y - 135 + 44 = y^2 + 6y - 91$$
3. Step 3: Factors of $-91$ adding to $+6$ are $+13$ and $-7$: $(y + 13)(y - 7)$.
4. Step 4: Substitute back $y$: $$\mathbf{(x^2 - 8x + 13)(x^2 - 8x - 7)}$$

Q5. Factorize $x^4 - 12x^2 + 16$:

1. Step 1: Complete square with end terms $(x^2)^2 + (4)^2$: $$(x^4 + 8x^2 + 16) - 20x^2 = (x^2 + 4)^2 - 20x^2 \quad \text{or} \quad (x^4 - 8x^2 + 16) - 4x^2$$
2. Step 2: Choosing $(x^4 - 8x^2 + 16) - 4x^2 = (x^2 - 4)^2 - (2x)^2$:
3. Step 3: Apply difference of squares: $$\mathbf{(x^2 - 2x - 4)(x^2 + 2x - 4)}$$

Q6. Factorize $(3x^2 + 4x - 5)(3x^2 + 4x - 2) - 4$:

1. Step 1: Let $u = 3x^2 + 4x$: $$(u - 5)(u - 2) - 4 = u^2 - 7u + 10 - 4 = u^2 - 7u + 6 = (u - 6)(u - 1)$$
2. Step 2: Substitute back $u = 3x^2 + 4x$: $$(3x^2 + 4x - 6)(3x^2 + 4x - 1)$$
3. Final Answer: $\mathbf{(3x^2 + 4x - 6)(3x^2 + 4x - 1)}$.

Q7. Factorize $2m^4 + m^2n^2 - 3n^4$:

1. Step 1: Split middle term: Product $2 \times (-3) = -6$, sum $=+1 \implies +3, -2$. $$2m^4 - 2m^2n^2 + 3m^2n^2 - 3n^4 = 2m^2(m^2 - n^2) + 3n^2(m^2 - n^2)$$
2. Step 2: Factorize $(m^2 - n^2)(2m^2 + 3n^2) = \mathbf{(m - n)(m + n)(2m^2 + 3n^2)}$.

Q8. Factorize $x^4 + 10x^2y^2 - 56y^4$:

1. Step 1: Factors of $-56$ adding to $+10$ are $+14$ and $-4$.
2. Step 2: Split: $(x^2 + 14y^2)(x^2 - 4y^2) = \mathbf{(x^2 + 14y^2)(x - 2y)(x + 2y)}$.

Q9. Factorize $x^2 + 2ax - bx - 2ab$:

1. Step 1: Group in pairs: $x(x + 2a) - b(x + 2a) = \mathbf{(x + 2a)(x - b)}$.

Q10. Factorize $a^{12} - b^{12}$:

1. Step 1: Difference of squares $(a^6 - b^6)(a^6 + b^6)$.
2. Step 2: $a^6 - b^6 = (a^3 - b^3)(a^3 + b^3) = (a - b)(a^2 + ab + b^2)(a + b)(a^2 - ab + b^2)$.
3. Step 3: $a^6 + b^6 = (a^2 + b^2)(a^4 - a^2b^2 + b^4)$.
4. Step 4: Product of all irreducible factors: $$\mathbf{(a - b)(a + b)(a^2 + b^2)(a^2 - ab + b^2)(a^2 + ab + b^2)(a^4 - a^2b^2 + b^4)}$$

Q11. Factorize $28x^4y + 64x^3y - 60x^2y$:

1. Step 1: Factor out common monomial $4x^2y$: $$4x^2y(7x^2 + 16x - 15)$$
2. Step 2: Split middle term: Product $7 \times (-15) = -105$. Factors adding to $16$ are $+21$ and $-5$. $$4x^2y [ 7x^2 + 21x - 5x - 15 ] = 4x^2y [ 7x(x + 3) - 5(x + 3) ] = \mathbf{4x^2y(x + 3)(7x - 5)}$$

Q12. Factorize $4(x - y)^3 - (x - y)$:

1. Step 1: Extract common factor $(x - y)$: $$(x - y) [ 4(x - y)^2 - 1 ]$$
2. Step 2: Factorize difference of squares $[2(x - y)]^2 - 1^2$: $$\mathbf{(x - y)(2x - 2y - 1)(2x - 2y + 1)}$$

Q13. Factorize $x^3p^2 - 8y^3p^2 - 4x^3q^2 + 32y^3q^2$:

1. Step 1: Group by $p^2$ and $-4q^2$: $$p^2(x^3 - 8y^3) - 4q^2(x^3 - 8y^3) = (x^3 - 8y^3)(p^2 - 4q^2)$$
2. Step 2: Factorize difference of cubes $(x^3 - 8y^3) = (x - 2y)(x^2 + 2xy + 4y^2)$.
3. Step 3: Factorize difference of squares $(p^2 - 4q^2) = (p - 2q)(p + 2q)$.
4. Step 4: Product: $\mathbf{(p - 2q)(p + 2q)(x - 2y)(x^2 + 2xy + 4y^2)}$.

Q14. Factorize $x^4 + y^4 - 7x^2y^2$:

1. Step 1: Complete square using $(x^2 + y^2)^2 - 2x^2y^2 - 7x^2y^2 = (x^2 + y^2)^2 - 9x^2y^2$.
2. Step 2: Write as $(x^2 + y^2)^2 - (3xy)^2$.
3. Step 3: Apply difference of squares: $$\mathbf{(x^2 - 3xy + y^2)(x^2 + 3xy + y^2)}$$

Q15. Find the HCF of the following:

(i) $x^3 + 3x^2 - 8x - 24$ and $x^3 + 3x^2 - 3x - 9$:

1. $P_1 = x^2(x + 3) - 8(x + 3) = (x + 3)(x^2 - 8)$.
2. $P_2 = x^2(x + 3) - 3(x + 3) = (x + 3)(x^2 - 3)$.
3. Common factor: $\mathbf{\text{HCF} = x + 3}$.

(ii) $3x^4 - 3x^3 - 2x^2 - x - 1$ and $9x^4 - 3x^3 - x - 1$:

1. Step 1: Multiply $P_1$ by $3 \implies 9x^4 - 9x^3 - 6x^2 - 3x - 3$.
2. Step 2: Subtract from $P_2$: $(9x^4 - 3x^3 - x - 1) - (9x^4 - 9x^3 - 6x^2 - 3x - 3) = 6x^3 + 6x^2 + 2x + 2 = 2(3x^3 + 3x^2 + x + 1) = 2[3x^2(x + 1) + 1(x + 1)] = 2(3x^2 + 1)(x + 1)$.
3. Step 3: Divide $P_1$ by $(3x^2 + 1)(x + 1) = 3x^3 + 3x^2 + x + 1$: Remainder $= 0$.
4. Step 4: $\mathbf{\text{HCF} = (3x^2 + 1)(x + 1) = 3x^3 + 3x^2 + x + 1}$.

Q16. Find the LCM of the following:

(i) $2x^2 + 3x + 1,\; 2x^2 + 5x + 2,\; x^2 + 3x + 2$:

1. $P_1 = (2x + 1)(x + 1)$.
2. $P_2 = (2x + 1)(x + 2)$.
3. $P_3 = (x + 1)(x + 2)$.
4. $\mathbf{\text{LCM} = (2x + 1)(x + 1)(x + 2)}$.

(ii) $3x^2 + 11x + 6,\; 3x^2 + 8x + 4,\; x^2 + 5x + 6$:

1. $P_1 = (3x + 2)(x + 3)$.
2. $P_2 = (3x + 2)(x + 2)$.
3. $P_3 = (x + 2)(x + 3)$.
4. $\mathbf{\text{LCM} = (3x + 2)(x + 2)(x + 3)}$.

Q17. Find HCF and LCM of $a(a + c) - b(b + c),\; b(b + a) - c(c + a),\; c(c + b) - a(a + b)$:

1. $P_1 = a^2 + ac - b^2 - bc = (a^2 - b^2) + c(a - b) = (a - b)(a + b + c)$.
2. $P_2 = b^2 + ab - c^2 - ac = (b^2 - c^2) + a(b - c) = (b - c)(a + b + c)$.
3. $P_3 = c^2 + bc - a^2 - ab = (c^2 - a^2) + b(c - a) = (c - a)(a + b + c)$.
4. Highest Common Factor: $\mathbf{\text{HCF} = a + b + c}$.
5. Least Common Multiple: $\mathbf{\text{LCM} = (a - b)(b - c)(c - a)(a + b + c)}$.

Q18. Find square root of $\frac{9a^2}{x^2} - \frac{6a}{5x} + \frac{101}{25} - \frac{4x}{15a} + \frac{4x^2}{9a^2}$:

1. Step 1: Write in perfect square trinomial form $\left( \frac{3a}{x} - \frac{1}{5} + \frac{2x}{3a} \right)^2$.
2. Step 2: Expand to verify: $$\left( \frac{3a}{x} - \frac{1}{5} + \frac{2x}{3a} \right)^2 = \frac{9a^2}{x^2} + \frac{1}{25} + \frac{4x^2}{9a^2} - \frac{6a}{5x} + 2\left(\frac{3a}{x}\right)\left(\frac{2x}{3a}\right) - \frac{4x}{15a}$$ $$= \frac{9a^2}{x^2} - \frac{6a}{5x} + \left(\frac{1}{25} + 4\right) - \frac{4x}{15a} + \frac{4x^2}{9a^2} = \frac{9a^2}{x^2} - \frac{6a}{5x} + \frac{101}{25} - \frac{4x}{15a} + \frac{4x^2}{9a^2} \quad \checkmark$$
3. Step 3: Square root: $\mathbf{\pm\left(\frac{3a}{x} - \frac{1}{5} + \frac{2x}{3a}\right)}$.

Q19. Simplify the following:

(i) $\frac{x^2 + x - 2}{x^2 - x - 20} \times \frac{x^2 + 5x + 4}{x^2 - x} \div \left( \frac{x^2 + 3x + 2}{x^2 - 2x - 15} \times \frac{x + 3}{x^2} \right)$:

1. Step 1: Factorize first product: $$\frac{(x + 2)(x - 1)}{(x - 5)(x + 4)} \times \frac{(x + 4)(x + 1)}{x(x - 1)} = \frac{(x + 2)(x + 1)}{x(x - 5)}$$
2. Step 2: Factorize second product inside parentheses: $$\frac{(x + 2)(x + 1)}{(x - 5)(x + 3)} \times \frac{x + 3}{x^2} = \frac{(x + 2)(x + 1)}{x^2(x - 5)}$$
3. Step 3: Perform division by inverting: $$\frac{(x + 2)(x + 1)}{x(x - 5)} \times \frac{x^2(x - 5)}{(x + 2)(x + 1)} = \frac{x^2}{x} = \mathbf{x}$$

(ii) $\frac{1}{x + 1} - \frac{1}{(x + 1)(x + 2)} + \frac{1}{(x + 1)(x + 2)(x + 3)}$:

1. Step 1: Common denominator $(x + 1)(x + 2)(x + 3)$:
2. Step 2: Combine numerators: $$\frac{(x + 2)(x + 3) - (x + 3) + 1}{(x + 1)(x + 2)(x + 3)} = \frac{x^2 + 5x + 6 - x - 3 + 1}{(x + 1)(x + 2)(x + 3)} = \frac{x^2 + 4x + 4}{(x + 1)(x + 2)(x + 3)}$$
3. Step 3: Factorize numerator $(x + 2)^2$ and cancel $(x + 2)$: $$\frac{(x + 2)^2}{(x + 1)(x + 2)(x + 3)} = \mathbf{\frac{x + 2}{(x + 1)(x + 3)}}$$

Q20. Find the values of $a$ and $b$ if $x^4 + ax^3 + bx^2 - 4x + 4$ is a perfect square:

1. Step 1: Let the square root be $(x^2 + px - 2)$ or $(x^2 + px + 2)$.
2. Step 2: Taking $(x^2 + px - 2)^2$: $$(x^2 + px - 2)^2 = x^4 + 2px^3 + (p^2 - 4)x^2 - 4px + 4$$
3. Step 3: Match coefficient of $x$: $-4p = -4 \implies p = 1$.
4. Step 4: Find $a$ and $b$: $$a = 2p = 2(1) = \mathbf{2}$$ $$b = p^2 - 4 = 1^2 - 4 = \mathbf{-3}$$
5. Step 5: (Alternative case for $(x^2 - x + 2)^2$ gives $a = -2, b = 5$).