Question 1: Factorize:

(i) $xyz-2xy = xy(z-2)$                 $[HCF = xy]$

(ii) $35x^3y-49xy^2 = 7xy(5x^2-7y)$                 $[HCF = 7xy]$

(iii) $p(2a-1)+q(1-2a)$

$= p(2a-1)-q(2a-1)$

$= (2a-1)(p-q) c$                  $[HCF = (2a-1)]$

(iv) $a(a+b)^3-3a^2b(a+b)$

$= a(a+b) \Big[ (a+b)^2 - 3ab \Big]$

$= a(a+b)(a^2+2ab+b^2-3ab)$

$= a(a+b)(a^2+b^2-ab )$                 $[HCF = a(a+b)]$

(v) $10x(2a+b)^3-15y(2a+b)^2 + 35(2a+b)$

$5(2a+b) \Big[ 2x(2a+b)^2 - 3y(2a+b) + 7 \Big]$                   $[HCF = 5(2a+b)]$

$\\$

Question 2: Factorize

(i) $x^3 -2x^2y+3xy^2-6y^3$

$= x^2(x-2y)+3y^2(x-2y)$

$= (x-2y)(x^2+3y^2)$

(ii) $a^2x^2+(ax^2+1)x+a$

$= a^2x^2+ax^3+x+a$

$= ax^2(a+x)+(a+x)$

$= (a+x)(ax^2+1)$

(iii) $x^2+y-xy-x$

$= x(x-y) -(x-y)$

$= (x-y)(x-1)$

(iv) $ab(x^2+y^2)+xy(a^2+b^2)$

$= abx^2+aby^2+xya^2+xyb^2$

$= ax(bx+ay)+by(ay+bx)$

$= (bx+ay)(ax+by)$

(v) $a^3+ab(1-2a)-2b^2$

$= a^2(a-2b)+b(a-2b)$

$= (a-2b)(a^2+b)$

$\\$

Question 3: Factorize

(i)   $\frac{a^2}{4b^2}$ $-$ $\frac{1}{3}$ $+$ $\frac{b^2}{9a^2}$

$= ($ $\frac{a}{2b})^2$ $- 2 \times$ $\frac{a}{2b}$ $\times$ $\frac{b}{3a}$ $+ ($ $\frac{b}{3a})^2$

$= \Big($ $\frac{a}{2b}$ $- \frac{b}{3a}$ $\Big)^2$

(ii) $\Big( x-$ $\frac{1}{x}$ $\Big)^2 + 6 \Big( x-$ $\frac{1}{x}$ $\Big) + 9$

$=$ $\Big( x-$ $\frac{1}{x}$ $\Big)^2 + 2 \times 3 \times \Big( x-$ $\frac{1}{x}$ $\Big) + (3)^3$

$=$ $\Big( x-$ $\frac{1}{x}$ $+ 3 \Big)^2$

(iii)  $\Big( x^2 +$ $\frac{1}{x^2}$ $\Big)^2 -4 \Big( x+$ $\frac{1}{x}$ $\Big) + 6$

$=$ $\Big( x+$ $\frac{1}{x}$ $\Big)^2 -2 - 4 \Big( x+$ $\frac{1}{x}$ $\Big) + 6$

$=$ $\Big( x+$ $\frac{1}{x}$ $\Big)^2 - 4 \Big( x+$ $\frac{1}{x}$ $\Big) + 4$

$=$ $\Big( x +$ $\frac{1}{x}$ $-2 \Big)^2$

(iv) $2a^2 + 2 \sqrt{6}ab + 3b^2$

$= (\sqrt{2}a)^2 + 2 \sqrt{2} \sqrt{3} ab + (\sqrt{3}b)^2$

$= (\sqrt{2}a + \sqrt{3}b)^2$

(v)  $4(x-y)^2-12(x-y)(x+y)+9(x+y)^2$

$= [2(x-y)]^2 - 2 . 2(x-y) . [3(x+y)]^2 + [3(x+y)]^2$

$= \Big( 2(x-y) - 3(x+y) \Big)$

$= (-x-5y)^2 = (x+5y)^2$

$\\$

Question 4: Factorize:

Note: We are going to use the following formulas in this set of exercise: ${(a+b)}^2=\ a^2+2ab+b^2$ and ${(a-b)}^2=\ a^2-2ab+b^2$

(i) $p^2q^2-p^4q^4$

$= p^2q^2(1-p^2q^2)$

$= p^2q^2(1-pq)(1+pq)$

(ii) $3x^4 - 243$

$= 3(x^4-81)$

$= 3(x^2-9)(x^2+9)$

$= 3(x-3)(x+3)(x^2+9)$

(iii) $25x^2-10x+1-36y^2$

$= (5x-1)^2 - (6y)^2$

$= (5x-1-6y)(5x-1+6y)$

(iv) $x^4 - 625$

$= (x^2-25)(x^2+25)$

$= (x-5)(x+5)(x^2+25)$

(v) $16(2x-1)^2 - 25y^2$

$= \Big[ 4(2x-1) \Big]^2 - (5y)^2$

$= \Big[ 4(2x-1) - 5y \Big] \Big[ 4(2x-1) + 5y \Big]$

$= (8x-4-5y)(8x-4+5y)$

(vi) $x^4-(2y-3z)^2$

$= (x^2)^2 - (2y-3z)^2$

$= (x^2-2y+3z)(x^2+2y-3z)$

(vii) $x^4-14x^2+1$

$= (x^2+1)^2 - 2x^2 - 14x^2$

$= (x^2+1)^2 - 16x^2$

$= (x^2+1-4x)(x^2+1+4x)$

(viii) $x^2-y^2-4xz +4z^2$

$= x^2 - 4zx+4z^2 - y^2$

$= (x-2z)^2 - y^2$

$= (x-2z-y)(x-2z+y)$

(ix) $a^2(b+c)-(b+c)^3$

$= (b+c) \Big[ a^2 - (b+c)^2 \Big]$

$= (b+c) (a-b-c)(a+b+c)$

(x) $x^4+y^4-7x^2y^2$

$= x^4 +y^4 - 2x^2y^2 - 9x^2y^2$

$= (x^2-y^2)^2 - (3xy)^2$

$= (x^2-y^2 - 3xy)(x^2-y^2+3xy)$

$\\$

Question 5: Factorize:

Note: We are going to use the following formulas in this set of exercise: $a^3+b^3 = (a+b)(a^2 -ab + b^2)$ and $a^3-b^3 = (a-b)(a^2 +ab + b^2)$

(i) $27a^3+8$

$= (3a)^3+(2)^3$

$= (3a+2)(9a^2-6a+4)$

(ii) $1-27a^3$

$= (1)^3-(3a)^3$

$= (1-3a)(1+3a+9a^2)$

(iii) $a-b -a^3+b^3$

$= (a-b) - (a^3-b^3)$

$= (a-b) -(a-b)(a^2+ab+b^2)$

$= (a-b)(1-a^2-ab-b^2)$

(iv) $(2x+3y)^3 - (2x-3y)^3$

$= (2x+3y-2x+3y) \Big[ (2x+3y)^2 + (2x+3y)(2x-3y) + (2x-3y)^2 \Big]$

$= 6y(4x^2+12xy+9y^2 + 4x^2 - 9y^2 + 4x^2-12xy+9y^2 )$

$= 6y(12x^2+9y^2) = 18y(4x^2+3y^2)$

(v) $(a+b)^3 - 8(a-b)^3$

$= (a+b)^3 - 2^3(a-b)^3$

$=(a+b-2a+2b) \big[ (a+b)^2 + 2(a+b)(a-b) + 4(a-b)^2 \Big]$

$= (3b-a)(a^2 + 2ab + b^2 + 2a^2 - 2b^2 + 4a^2 -8ab + 4b^2 )$

$= (3b-a)(7a^2-6ab+3b^2)$

(vi) $a^{12} + b^{12}$

$= (a^4)^3+(b^4)^3$

$= (a^4+b^4)(a^8-a^4b^4+b^8)$

(vii) $x^3-12x(x-4)-64$

$= x^3 - 12x(x-4)-4^3$

$= (x-4)(x^2+4x+16)-12x(x-4)$

$= (x-4)(x^2-8x+16)$

$=(x-4)(x-4)^2 = (x-4)^3$

(viii) $x^3+x^2-$ $\frac{1}{x^2}$ $+$ $\frac{1}{x^3}$

$= x^3 +$ $\frac{1}{x^3}$ $+ x^2 -$ $\frac{1}{x^2}$

$= (x+$ $\frac{1}{x}$ $)(x^2 +$ $\frac{1}{x^2}$ $-1) + (x-$ $\frac{1}{x}$ $)(x+$ $\frac{1}{x}$ $)$

$= (x +$ $\frac{1}{x}$ $)(x^2 +$ $\frac{1}{x^2}$ $-1+x -$ $\frac{1}{x}$ $)$

$= (x +$ $\frac{1}{x}$ $) \Big[ (x -$ $\frac{1}{x}$ $) ((x -$ $\frac{1}{x}$ $+1) + 1 \Big]$

(ix) $8a^3-b^3-4ax+2bx$

$= (2a)^2 - b^3 -4ax+2bx$

$= (2a-b)(4a^2+2ab+b^2) - 2x(2a-b)$

$= (2a-b)(4a^2+2ab+b^2-2x)$

(x) $a^3-$ $\frac{1}{a^3}$ $-2a+2$ $\frac{1}{a}$

$= (a -$ $\frac{1}{a}$ $)(a^2+1+$ $\frac{1}{a^2}$ $)-2(a-$ $\frac{1}{a}$ $)$

$= (a -$ $\frac{1}{a}$ $)(a^2+1+$ $\frac{1}{a^2}$ $-2)$

$= (a -$ $\frac{1}{a}$ $)(a^2+$ $\frac{1}{a^2}$ $-1)$

(xi) $a^3+b^3+a+b$

$= (a+b)(a^2-ab+b^2)+(a+b)$

$= (a+b)(a^2-ab+b^2+1)$

$\\$

Question 6: Factorize:

(i) $64a^3+125b^3+240a^2b+300ab^2$

$= (4a)^3+(5b)^3+60ab(4a+5b)$

$= (4a+5b)(16a^2-20ab+25b^2)+60ab(4a+5b)$

$= (4a+5b)(16a^2-20ab+25b^2 + 60ab)$

$= (4a+5b)(16a^2+40ab+25b^2)$

$= (4a+5b)(4a+5b)^2 = (4a+5b)^3$

(ii) $8x^3+27y^3+36x^2y+54xy^2$

$=(2x)^3+(3y)^3+18xy(2x+3y)$

$= (2x+3y)(4x^2+9y^2-6xy)+18xy(2x+3y)$

$= (2x+3y)(4x^2+9y^2-6xy+18xy)$

$= (2x+3y)(4x^2+9y^2+12xy)$

$= (2x+3y)(2x+3y)^2 = (2x+3y)^3$

(iii) $a^3 - 3a^2b+3ab^2 - b^3 + 8$

$= (a-b)^3 +2^3$

$= (a-b+2)\Big[ (a-b)^2 -2(a-b)+4 \Big]$

$= (a-b+2)(a^2+b^2 - 2ab -2a+2b+4)$

(iv) $8x^3+y^3+12x^2y+6xy^2$

$= (2x)^3+y^3 + 6xy(2x+y)$

$= (2x+y)(4x^2-2xy+y^2) + 6xy(2x+y)$

$= (2x+y)(4x^2-2xy+y^2 +6xy)$

$= (2x+y)(4x^2+4xy + y^2)$

$= (2x+y)(2x+y)^2 = (2x+y)^3$

(v) $a^3x^3 - 3a^2bx^2+3ab^2x-b^3$

$= (ax)^3 - 3abx(ax-b) - b^3$

$= (ax-b)(a^2x^2+abx+b^2) - 3abx(ax-b)$

$= (ax-b)(a^2x^2+abx+b^2-3abx)$

$= (ax-b)(a^2x^2+b^2-2abx) = (ax-b)(ax-b)^2 = (ax-b)^3$

$\\$

Question 7: Factorize:

(i) $x^2+12x-45$

$= x^2+15x-3x-45$

$= x(x+15)-3(x+15)$

$= (x+15)(x-3)$

(ii) $x^2-22x+120$

$= x^2 - 12x - 10x + 120$

$= x(x-12)-10(x-12)$

$= (x-10)(x-12)$

(iii) $x^2-11x-42$

$= x^2-14x+3x-42$

$= x(x-14)+3(x-14)$

$= (x+3)(x-14)$

(iv) $y^2+5y-36$

$= y^2 +9y-4y-36$

$= y(y+9)-4(y+9)$

$=(y-4)(y+9)$

(v) $(a+b)^2-5(a+b)+4$

Let $a+b = x$

$= x^2-5x+4$

$= x^2-4x-x+ 4$

$= x(x-4)-1(x-4)$

$= (x-4)(x-1)$

$= (a+b-4)(a+b-4)$

(vi) $3(x+y)^2-5(x+y)+2$

Let $x+y = a$

$= 3a^2-5a+2$

$= 3a^2 - 3a - 2a + 2$

$= 3a(a-1)-2(a-1)$

$=(a-1)(3a-2)$

$= (x+y-1)(3x+3y-2)$

(vii) $(p^2+4p)^2+21(p^2+4p)+98$

Let $p^2+4p = x$

$= x^2 + 21x + 98$

$= x^2 + 14x + 7x + 98$

$= x(x+14) + 7( x + 14)$

$= (x+14)(x+7)$

$= (p^2+4p+14)(p^2+4p+7)$

(viii) $x^2-\sqrt{3}x-6$

$= x^2 -2\sqrt{3}x+\sqrt{3}x-6$

$= x(x+\sqrt{3})-2\sqrt{3}(x+\sqrt{3})$

$= (x+\sqrt{3})(x-2\sqrt{3})$

(ix) $x^2+5\sqrt{5}x+30$

$= x^2 + 3\sqrt{5} x + 2\sqrt{5} x + 3\sqrt{5} \times 2\sqrt{5}$

$= x(x+3\sqrt{5}) + 2\sqrt{5}(x+3\sqrt{5})$

$= (x+3\sqrt{5})(x+2\sqrt{3})$

(x) $(2x^2+5x)(2x^2+5x-19)+84$

Let $2x^2+5x=a$

$= (a)(a-19)+84$

$= a^2-19a+84$

$= a^2-12a-7a+84$

$= a(a-12)-7(a-12)$

$= (a-12)(a-7)$

$= (2x^2+5x-12)(2x^2+5x-7)$

$= (2x-3)(x+4)(2x+7)(x-1)$

$\\$

Question 8: Factorize

(i) $5x^2-32x+12$

$= 5x^2-30x-2x+12$

$= 5x(x-6)-2(x-6)$

$= (x-6)(5x-2)$

(ii) $30x^2+7x-15$

$= 30x^2+25x-18x-15$

$= 5x(6x+5)-3(6x+5)$

$=(6x+5)(5x-3)$

(iii) $6x^2-\sqrt{5}x-5$

$= 6x^2-3\sqrt{5}x+2\sqrt{5}x-5$

$= 3x(2x-\sqrt{5})+\sqrt{5}(2x-\sqrt{5})$

$= (2x-\sqrt{5})(3x+\sqrt{5})$

(iv) $\frac{1}{2}$ $x^2-3x+4$

$=$ $\frac{1}{2}$ $x^2-2x-x+4$

$=$ $\frac{1}{2}$ $x(x-4)-1(x-4)$

$= (x-4)($ $\frac{1}{2}$ $x -1)$

(v) $4\sqrt{3}x^2+5x-2\sqrt{3}$

$= 4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3}$

$=4x(\sqrt{3}x+2) - \sqrt{3}(\sqrt{3}x+2)$

$= (\sqrt{3}x+2)(4x-\sqrt{3})$

(vi) $\frac{a}{b}$ $x^2+($ $\frac{a}{b}$ $+$ $\frac{c}{d})$ $x +$ $\frac{c}{d}$

$=$ $\frac{a}{b}$ $x (x+1) +$ $\frac{c}{d}$ $(x+1)$

$= (x+1) ($ $\frac{a}{b}$ $x+$ $\frac{c}{d}$ $)$

(vii) $px^2+(4p^2-3q)x-12pq$

$= px^2 + 4p^2x-3qx-12pq$

$= px(x+4p)-3q(x+4p)$

$= (x+4p)(px-3q)$

(viii) $3(a-2)^2-2(a-2)-8$

Let $a-2 = x$

$= 3x^2-2x-8$

$= 3x^2 - 6x+4x-8$

$= 3x(x-2)+4(x-2)$

$= (x-2)(3x+4)$

(ix) $12(a+1)^2-25(a+1)(b+2)+12(b+2)^2$

Let $a+1 = x$ and $b+2 = y$

$= 12x^2 - 25xy + 12y^2$

$= 12x^2 - 9xy - 16xy + 12y^2$

$= 3x(4x-3y) + 4y(4x-3y)$

$= (4x-3y)(3x+4y)$

$= \Big( 4(a+1)-3(b+2) \Big) \Big( 3(a+1)+4(b+2) \Big)$

$= (4a-3b-2)(3a-4b-5)$

(x) $5x^6-7x^3-6$

Let $x^3 = a$

$= 5a^2-7a-6$

$= 5a^2-10a+3a-6$

$= 5a^2(a-2)+3(a-2)$

$= (a-2)(5a+3)$

$=(x^3-2)(5x^3+3)$

(xi) $x^2+$ $\frac{12}{35}$ $x+$ $\frac{1}{35}$

$= x^2 +$ $\frac{1}{7}$ $x +$ $\frac{1}{5}$ $x +$ $\frac{1}{35}$

$= x(x+$ $\frac{1}{7}$ $)+$ $\frac{1}{5}$ $(x+$ $\frac{1}{7}$ $)$

$= (x+$ $\frac{1}{7}$ $)(x+$ $\frac{1}{5}$ $)$

(xii) $2x^2+3\sqrt{3}x+3$

$= 2x^2 + 2\sqrt{3} x + \sqrt{3} x + 3$

$= 2x(x+\sqrt{3}) + \sqrt{3}(x+\sqrt{3})$

$= (x+\sqrt{3})(2x+\sqrt{3})$

(xiii) $5\sqrt{5}x^2+20x+3\sqrt{5}$

$= 5\sqrt{5} x^2 + 15x + 5 x + 3\sqrt{5}$

$= \sqrt{5}x(5x+\sqrt{5}) + 3 (5x+\sqrt{5})$

$= (5x+\sqrt{5})(\sqrt{5}x+3)$

(xiv) $2x^2+3\sqrt{5}x+5$

$= 2x^2+2\sqrt{5}x+\sqrt{5}x+5$

$= 2x(x+\sqrt{5}) + \sqrt{5}(x+\sqrt{5})$

$= (x+\sqrt{5})(2x+\sqrt{5})$

(xv) $7x^2+2\sqrt{14}x+2$

$= 7x^2+\sqrt{14} x + \sqrt{14} x + 2$

$= 7x^2+\sqrt{2 \times 7} x + \sqrt{2 \times 7} x + 2$

$= \sqrt{7}x(\sqrt{7}x+\sqrt{2})+\sqrt{2}(\sqrt{7}x+\sqrt{2})$

$= (\sqrt{7}x + \sqrt{2})(\sqrt{7}x+\sqrt{2})$

$= (\sqrt{7}x + \sqrt{2})^2$