Factorize

Question 1:  $x^2-81$

$x^2-81 = (x-9)(x+9)$

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Question 2:  $9a^2-25$

$9a^2-25 = (3a-5)(3a+5)$

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Question 3:  $36y^2-121$

$36y^2-121 = (6y-11)(6y+11)$

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Question 4:  $49a^2-100b^2$

$49a^2-100b^2 = (7a-10b)(7a+10b)$

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Question 5:  $(a+b^2 )-36$

$(a+b^2 )-36 = (a+b-6)(a+b+6)$

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Question 6:  $16c^2-1$

$16c^2-1 = (4c-1)(4c+1)$

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Question 7:  $1-64b^2$

$1-64b^2 = (1-8b)(1+8b)$

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Question 8:  $\displaystyle \frac{9}{16} -25x^2$

$\displaystyle \frac{9}{16} -25x^2 = \Big( \frac{3}{4} -5x \Big) \Big( \frac{3}{4} +5x \Big)$

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Question 9:  $\displaystyle z^2- \frac{1}{144}$

$\displaystyle z^2- \frac{1}{144} = \Big(z- \frac{1}{12} \Big) \Big(z+ \frac{1}{12} \Big)$

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Question 10:  $1-(a-b)^2$

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

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Question 11:  $(3m-n)^2-(m-2n)^2$

$(3m-n)^2-(m-2n)^2$

$= [(3m-n)-(m-2n)][(3m-n)+(m-2n)]$

$= (3m-n-m+2n)(3m-n+m-2n)$

$= (2m+n)(4m-3n)$

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Question 12:  $(3x+2y)^2-(2x-3y)^2$

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

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

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

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Question 13:  $16(a+b)^2-9(a-b)^2$

$16(a+b)^2-9(a-b)^2$

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

$= (a+7b)(7a+b)$

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Question 14:  $9(x+y)^2-16(x-2y)^2$

$9(x+y)^2-16(x-2y)^2$

$= [3(x+y)-4(x-2y)][3(x+y)+4(x-2y)]$

$= (11y-x)(7x-5y)$

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Question 15: $36(a-b)^2-25(a+b)^2$

$36(a-b)^2-25(a+b)^2$

$= [6(a-b)-5(a+b)][6(a-b)+5(a+b)]$

$= (6a-6b-5a-5b)(6a-6b+5a+5b)$

$= (a-11b)(11a-b)$

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Question 16:  $9(3x+1)^2-4(x-1)^2$

$9(3x+1)^2-4(x-1)^2$

$= [3(3x+1)-2(x-1)][3(3x+1)+(x-1)]$

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

$= (7x+5)(11x+1)$

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Question 17:  $a^2-2ab+b^2-c^2$

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

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Question 18:  $x^2-a^2-2a-1$

$x^2-a^2-2a-1$

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

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

$= (x-a-1)(x+a+1)$

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Question 19:  $x^2-m^2+6mn-9n^2$

$x^2-m^2+6mn-9n^2$

$= x^2-(m^2-6mn+9n^2)$

$= x^2-(m-3n)^2$

$= (x-m+3n)(x+m-3n)$

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Question 20:  $a^4-b^4$

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

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Question 21: $16a^4-81b^4$

$16a^4-81b^4$  $= (4a^2-9b^2)(4a^2+9b^2)$  $= (2a-3b)(2a+3b)(4a^2+9b^2 )$

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Question 22: $3-75z^2$

$3-75z^2$  $= 3(1-25z^2 )$  $= 3(1-5z)(1+5z)$

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Question 23: $48a^2 b^2-3$

$48a^2 b^2-3$  $= 3(16a^2 b^2-1)$  $= 3(4ab-1)(4ab+1)$

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Question 24: $4x^3-81x$

$4x^3-81x$  $= x(4x^2-81)$  $= x(2x-9)(2x+9)$

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Question 25: $9b^3-144b$

$9b^3-144b$  $= b(9b^2-144)$  $= b(3b-12)(3b+12)$

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Question 26: $32x^2-72y^2$

$32x^2-72y^2$  $= 2(16x^2-36y^2 )$  $= 2(4x-6y)(4x-6y)$

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Question 27: $50x^2-32y^2$

$50x^2-32y^2$  $= 2y(25x^2-16y^2 )$  $= 2y(5x-4y)(5x+4y)$

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Question 28: $a^3-4ab^2$

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

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Question 29: $ab^3 c-abc^3$

$ab^3 c-abc^3$  $= abc(b^2-c^2 )$  $= abc(b-c)(b+c)$

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Question 30: $9(x+y)^3-16(x+y)$

$9(x+y)^3-16(x+y)$

$= (x+y)[9(x+y)^2-16]$

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

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Question 31: $1-0.49c^6$

$1-0.49c^6$  $= (1-0.7c^3 )(1+0.7c^3 )$

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Question 32: $x^2-y^2-8yz-16z^2$

$x^2-y^2-8yz-16z^2$

$= x^2-(y^2+8yz+16z^2 )$

$= x^2-(y+4z)^2$

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

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Question 33: $\displaystyle x^3 y^3- \frac{25xy}{z^2}$

$\displaystyle x^3 y^3- \frac{25xy}{z^2} = xy \Big(x^2 y^2- \frac{25}{z^2} \Big) = xy \Big( xy+ \frac{5}{2} \Big) \Big( xy- \frac{5}{2} \Big)$

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Question 34: $324x^4-0.0064b^4$

$324x^4-0.0064b^4$

$= 1/1000 (324x^4-64b^4 )$

$= 4/100 (81x^4-16b^4 )$

$= 4/100 (9x^2-4b^2 )(9x^2+4b^2 )$

$= 0.0004(3x-2b)(3x+2b)(9x^2+4b^2 )$

$= (0.09x^2+0.04b^2 )(0.3x+0.2b)(0.3x-0.2b)$

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Question 35:  Using the identity $(a^2 - b^2 ) = (a+b)(a-b)$, evaluate each of the following:

i)  $82^2-18^2$     ii)  $8^2-9.2^2$     iii)  $8^2-0.2^2$

$\displaystyle \text{iv) } (7 \frac{3}{4})^2-(2 \frac{1}{4})^2$     $\displaystyle \text{v) } (6 \frac{4}{11})^2-(4 \frac{7}{11})^2$     $\displaystyle \text{vi) } \frac{(7.3 \times 7.3-2.7 \times 2.7)}{(7.3-2.7)}$

i)  $82^2-18^2 = (82+18)(82-18) = 100 \times 64 = 6400$
ii)  $8^2-9.2^2 = (15.8+9.2)(15.8-9.2) = 25 \times 6.6 = 165$
iii)  $8^2-0.2^2 = (0.8+0.2)(0.8-0.2) = 1 \times (.06) = 0.6$
$\displaystyle \text{iv) } \Big(7 \frac{3}{4} \Big)^2-(2 \frac{1}{4} \Big)^2 = \Big(7 \frac{3}{4} +2 \frac{1}{4} \Big) \Big(7 \frac{3}{4} -2 \frac{1}{4} \Big) = 10 \times 5 \frac{1}{2} = 55$
$\displaystyle \text{v) }\Big(6 \frac{4}{11} \Big)^2-\Big(4 \frac{7}{11} \Big)^2 = \Big(6 \frac{4}{11} +4 \frac{7}{11} \Big) \Big(6 \frac{4}{11} -4 \frac{7}{11} \Big) = 11 \times 1 \frac{4}{11} = 19$
$\displaystyle \text{vi) }\frac{(7.3 \times 7.3-2.7 \times 2.7)}{(7.3-2.7)} = \frac{(7.3+2.7)(7.3-2.7)}{(7.3-2.7)} = 10$