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)]

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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)

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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

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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)

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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)

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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

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Exercise 6(a) continued …

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