[4 marks sum]

Question 14 Marks

Factorise $a^2b - b^3.$ Using this result, find the value of $101^2 \times 100 - 100^3.$

Answer

$a^2b - b^3b(a^2 - b^2)$
$b(a + b)(a - b)$
Now,
$101^2 \times 100 - 100^3$
$= 100 (1012 - 100^2)$
$= 100(101 + 100)(101 - 100)$
$= 100(201)(1)$
$= 20100$

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Question 24 Marks

factorise :$16(5x + 4)^2 -9(3x - 2)^2$

Answer

$16(5x + 4)^2 -9(3x - 2)^2= (4(5x + y))^2 - (3(3x - 2))^2$
$= [4 (5x + 4) - 3 (3x - 2)] [4(5x + 4) +3(3x - 2)]$
$.....[a^2 - b^2 = (a - b)(a + b)]$
$= (20x + 16 - 9x + 6)(20x + 16 + 9x - 6)$
$= (11x + 22)(29x + 10)$
$= 11 (x + 2)(29x + 10)$

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Question 34 Marks

factorise : $(ax + by)^2 + (bx - ay)^2$

Answer

$ (a x+b y)^2+(b x-a y)^2$
$ =(a x)^2+2 \times a x \times b y+(b y)^2+(b x)^2-2 \times b x \times a y+(a y)^2$
$ =a^2 x^2+2 a b x y+b^2 y^2+b^2 x^2-2 a b x y+a^2 y^2$
$ =a^2 x^2+2 a b x y+b^2 y^2+b^2 x^2-2 a b x y+a^2 y^2$
$ =a^2 x^2+b^2 y^2+b^2 x^2+a^2 y^2$
$ =a^2 x^2+a^2 y^2+b^2 x^2+b^2 y^2$
$ =a^2\left(x^2+y^2\right)+b^2\left(x^2+y^2\right)$
$ =\left(x^2+y^2\right)\left(a^2+b^2\right)$

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Question 44 Marks

Factorise : $2(x + 2y)^2 - 5(x + 2y) + 2$

Answer

$2(x + 2y)^2 - 5(x + 2y) + 2$ Let $x + 2y = a,$ then
$2a^2 - 5a + 2$
$\Rightarrow 2a^2 - a - 4a + 2$
$= a(2a - 1) - 2(2a - 1)$
$= (2a - 1)(a - 2)$
$= \{2(x + 2y - 1)\} \{(x + 2y) - 2\} ...(a = x + 2y)$
$= (2x + 4y - 2)(x + 2y - 2)$

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Question 54 Marks

factorise: $8 + 6(a + b) - 5(a + b)^2$

Answer

$8 + 6(a + b) - 5(a + b)^2$ Let $a + b = x$
$(a + b)^2 = x^2$
$8 + 6 (a+b) - 5(a + b)^2$
$= 8 + 6x - 5x^2$
$= 8 + 10x - 4x - 5x^2$
$= 2(4 + 5x) - x (4 + 5x)$
$= (4 + 5x)(2 - x)$
$= [4 + 5(a + b)] [2 - (a + b)]$
$($Substituting the value of $x)$
$= [4 + 5a + 5b][2 - a - b]$

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Question 64 Marks

factorise : $(x - 2y)^2 - 12(x - 2y) + 32$

Answer

$(x - 2y)^2 - 12(x - 2y) + 32$ Let $(x - 2y) = a$
$\therefore (x - 2y)^2 = a^2$
$\therefore (x - 2y)^2 - 12(x - 2y) + 32$
$= a^2 - 12a + 32$
$= a^2 - 8a - 4a + 32$
$= a(a - 8) - 4 (a - 8)$
$= (a - 8)(a - 4)$
$= (x - 2y - 8)(x - 2y - 4)$
$($Substituting the value of $a)$

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Question 74 Marks

factorise : $1 - (2x + 3y) - 6(2x + 3y)^2$

Answer

$1 - (2x + 3y) - 6(2x + 3y)^2$ Let $(2x + 3y) = a$
$\therefore (2x + 3y)^2 = a^2$
$\therefore 1 - (2x + 3y) - 6(2x + 3y)^2$
$= 1 - a - 6a^2$
$= 1 - 3a + 2a - 6a^2$
$= 1(1 - 3a) + 2a (1 - 3a)$
$= (1 - 3a) (1 + 2a)$
$= [1 - 3 (2x + 3y)][1 + 2(2x + 3y)]$
$($Substituting the value of $a)$
$= (1 - 6x - 9y) (1 + 4x + 6y)$

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Question 84 Marks

factorise: $(2a + b)^2 + 5(2a + b) + 6$

Answer

$(2a + b)^2 + 5(2a + b) + 6$ Let $(2a + b) = x$
$(2a + b)^2 = x^2$
$(2a + b)^2 + 5 (2a + b) + 6$
$= x^2 + 5x + 6$
$= x^2 + 3x + 2x + 6$
$= x(x + 3)+2(x + 3)$
$= (x + 3)(x + 2)$
$= (2a + b + 3)(2a + b + 2)$
$($Substituting the value of $x)$

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Question 94 Marks

Factorise: $75(x + y)^2 - 48(x - y)^2$

Answer

$75(x + y)^2 - 48(x - y)^2= 3[25 (x + y)^2 - 16 (x - y)^2]$
$= 3[\{5 (x + y)^2\} - \{4 (x - y)\}^2]$
Using $a^2– b^2 = (a + b) (a – b)$
$= 3 [5(x + y) + 4(x - y)] [5(x + y) - 4(x - y)]$
$= 3 [5x + 5y + 4x - 4y] [5x + 5y - 4x + 4y]$
$= 3 (9x + y) (x + 9y)$

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Question 104 Marks

Factorise: $49(x - y)^2 -9(2x + y)^2$

Answer

$49(x - y)^2 -9(2x + y)^2$
$= [7(x-y)]^2 - [3(2x + y)]^2$
$= (7x - 7y)^2 - (6x + 3y)^2$
$= (7x - 7y + 6x + 3y) (7x - 7y - 6x - 3y)$
$= (13x - 4y) (x - 10y)$

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Question 114 Marks

Factorise : $4(2a + b)^2 - (a - b)^2$

Answer

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

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Question 124 Marks

Factorise : $15(2x - y)^2 - 16(2x - y) - 15 $

Answer

$15(2x - y)^2 - 16(2x - y) - 15 =$ Let $2x - y = a,$
then $15a^2 - 16a - 15$
$= 15a^2 - 25a + 9a - 15$
$...{15 \times (-15) = -225 , -225 = -25 \times 9 , -16 = -25 + 9}$
$= 5a (3a - 5) + 3(3a - 5)$
$= (3a - 5)(5a + 3)$
$= [3 (2x - y) - 5][5 (2x - y) + 3]$
$= (6x - 3y - 5)(10x - 5y + 3)$

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Question 134 Marks

Factorise : $x^4 - 5x^2 - 36$

Answer

$x^4 - 5x^2 - 36$
$= (x^2)^2 - 5x^2 - 36$
$= (x^2)^2 - 9x^2 + 4x^2 - 36 ...$
${\because -36 = -9 \times 4 }$
${-5 = -9 + 4}$
$= x^2 (x^2 - 9) + (x^2 - 9)$
$= (x^2 - 9)(x^2 + 4)$
$= [x^2 - (3)^2] [x^2 + 4]$
$= (x + 3)(x - 3)(x^2 + 4)$
$= (x^2 +4) (x + 3)(x - 3)$

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Question 144 Marks

Factorise: $ 25(2x + y)^2 - 16(x - y)^2 $

Answer

$25(2x + y)^2 - 16(x - y)^2$
$ = [5(2x + y)]^2 - [4(x-y)]^2$
$= (10x + 5y)^2 - (4x - 4y)^2$
$= (10x + 5y + 4x - 4y) (10x + 5y - 4x + 4y)$
$= (14x + y)(6x + 9y)$
$= (14x + y) 3(2x + 3y)$
$= 3 (14x + y)(2x + 3y)$

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