Class 9, Exercise 6, Factorization (Advanced)

Class 9: Factorization – Exercise 6.1

Date: June 15, 2018Author: ICSE CBSE ISC Board Mathematics Portal for Students 1 Comment

Question 1: Factorize:

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

Question 2: Factorize

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

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

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

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

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

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

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

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

$\displaystyle = x(x-y) -(x-y)$

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

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

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

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

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

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

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

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

$\displaystyle \\$

Question 3: Factorize

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

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

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

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

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

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

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

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

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

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

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

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

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

(v) $\displaystyle 4(x-y)^2-12(x-y)(x+y)+9(x+y)^2$

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

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

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

$\displaystyle \\$

Question 4: Factorize:

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

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

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

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

(ii) $\displaystyle 3x^4 - 243$

$\displaystyle = 3(x^4-81)$

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

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

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

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

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

(iv) $\displaystyle x^4 - 625$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

Question 5: Factorize:

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

(i) $\displaystyle 27a^3+8$

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

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

(ii) $\displaystyle 1-27a^3$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

Question 6: Factorize:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

Question 7: Factorize:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Let $\displaystyle a+b = x$

$\displaystyle = x^2-5x+4$

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

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

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

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

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

Let $\displaystyle x+y = a$

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

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

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

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

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

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

Let $\displaystyle p^2+4p = x$

$\displaystyle = x^2 + 21x + 98$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle = a^2-19a+84$

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

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

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

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

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

$\displaystyle \\$

Question 8: Factorize

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Let $\displaystyle a-2 = x$

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

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

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

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

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

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

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

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

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

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

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

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

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

Let $\displaystyle x^3 = a$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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