Factorize the following:

Question 1:  $x^2+18x+81$

$x^2+18x+81 = (x+9)^2$

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Question 2:  $a^2-14a+49$

$a^2-14a+49 = (a-7)^2$

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Question 3:  $4x^2+12x+9$

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

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Question 4:  $9x^2-24x+16$

$9x^2-24x+16 = (3x-4)^2$

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Question 5:  $49x^2+56x+16$

$49x^2+56x+16 = (7x+4)^2$

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Question 6:  $25z^2-30z+9$

$25z^2-30z+9 = (5z-3)^2$

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Question 7:  $9q^4 r^4-6p^4 q^2 r^2+p^8$

$9q^4 r^4-6p^4 q^2 r^2+p^8 = (3q^2 r^2-p^4 )^2$

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Question 8:  $\displaystyle \frac{9p^2}{q^2} +\frac{16r^2}{m^2} +\frac{24pr}{qm}$

$\displaystyle \frac{9p^2}{q^2} +\frac{16r^2}{m^2} +\frac{24pr}{qm} = (\frac{3p}{q}+\frac{4r}{m})^2$

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Question 9:  $\displaystyle \frac{1}{4} z^6+9a^2-3az^3$

$\displaystyle \frac{1}{4} z^6+9a^2-3az^3 = ( \frac{1}{2} z^3-3a)^2$

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Question 10:  $\displaystyle \frac{9}{4} a^2+ \frac{49}{9} p^2-7ap$

$\displaystyle \frac{9}{4} a^2+ \frac{49}{9} p^2-7ap = ( \frac{3}{2} a- \frac{7}{3} p)^2$

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Question 11:  $t^2+22t+85 = 0$

Find two numbers with sum = 22 and product = 85

$a+b = 22$   &  $ab = 85$  $\Rightarrow a =17, b = 5$

Hence the factors are

$t^2+22t+85 = (t+17)(t+5)$

Note: This technique would be used all across this exercise.

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Question 12:  $x^2-10x+24$

$a+b = -10$  & $ab = 24$   $\Rightarrow a = -6, b = -4$

Therefore $\ \ x^2-10x+24 = (x-6)(x-4)$

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Question 13:  $m^2-3m-40$

$a+b = -3$$ab = -40$  $\Rightarrow a = -8, b = +5$

Therefore $\ \ m^2-3m-40 = (m-8)(m+5)$

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Question 14:  $x^2+x-72$

$a+b = 1$ & $ab = -72$  $\Rightarrow a = 9, b = -8$

Therefore $\ \ x^2+x-72 = (x+9)(x-8)$

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Question 15:  $p^2-7p-120$

$a+b = -7$ & $ab = -120$  $\Rightarrow a = 8, b = -15$

Therefore $\ \ p^2-7p-120 = (p+8)(p-15)$

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Question 16:  $16-17z+z^2$

$a+b = -17$ & $ab = 16$  $\Rightarrow a = -1, b = -16$

Therefore $16-17z+z^2 = (z-1)(z-16)$

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Question 17:  $a^2+5a-104$

$x+y = 5$ & $xy = -104$   $\Rightarrow x = 13, y = -8$

Therefore $\ \ q^2+5a-104 = (a+13)(a-8)$

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Question 18:  $3x^2+11x+10$

$a+b = 11$ & $ab = 30$ $\Rightarrow a = 5, b = 6$

Therefore $3x^2+11x+10 = 3x^2+6x+5x+10$

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

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

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Question 19:  $6x^2+7x-3$

$a+b = 7$ & $ab = -18$  $\Rightarrow a = 9, b = -2$

Therefore $6x^2+7x-3 = 6x^2+9x-2x-3$

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

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

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Question 20:  $3z^2-4z-4$

$a+b = -4$ & $ab = -12$   $\Rightarrow a = -6, b = 2$

Therefore $3z^2-4z-4 = 3z^2-6z+2z-4$

$= 3z(z-2)+2(z-2)$

$= (z-2)(3z+2)$

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Question 21:  $72-x-x^2$

$a+b = -1$ & $ab = -72$  $\Rightarrow a = -8, b = -9$

Therefore $72-x-x^2 = -x^2-9x+8x+72$

$= -x(x-8)-9(x-8)$

$= (-x-9)(x-8)$

$= (x+9)(8-x)$

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Question 22:  $x^2-3xy-40y^2$

$x^2-3xy-40y^2$

$= x^2-8xy+5xy-40y^2$

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

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

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Question 23:  $3x^2 y+11xy+6y$

$a+b = 11$ & $ab = 18$  $\Rightarrow a = 9, b = 2$

$3x^2 y+11xy+6y$

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

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

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

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Question 24:  $(a-b)^2-5(a-b)+6$

$Let a-b = x$

$Hence (a-b)^2-5(a-b)+6 = x^2-5x+6 = (x-2)(x-3)$

Substituting Back

$= (a-b)^2-5(a-b)+6 = (a-b-2)(a-b-3)$

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Question 25:  $(a-3b)^2-4(a-3b)-21$

$(a-3b)^2-4(a-3b)-21$

$= (a-3b)^2-7(a-3b)+3(a-3b)-21$

$= (a-3b)[(a-3b)-7]+3[(a-3b)-7]$

$= (a-3b-7)(a-3b)$

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Question 26:  $3(y-2)^2-(y-2)-44$

Let $(y-2) = x$

$3(y-2)^2-(y-2)-44$

$= 3x^2-x-44 = 3x^2-12x+11x-44$

$= 3x(x-4)+11(x-4)$

$= (3x+11)(x-4)$

Substituting Back

$= (3(y-2)+11)(y-2-4)$

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

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Question 27:  $7+10(x+y)-8(x+y)^2$

$Let (x+y) = a$

Therefore $7+10(x+y)-8(x+y)^2$

$= 7+10a-8a^2$

$m+n = 10$  &  $mn = -56$  $\Rightarrow m = 14 \ \ n = -4$

Therefore $= 7+10a-8a^2+14a-4a+7$

$= -4a(2a+1)+7(2a+1)$

$= (7-4a)(2a+1)$

Substituting Back

$= (7-4x+4y)(8x+8y+1)$

$= 7+10(x+y)-8(x+y)^2 = (7-4x-4y)(8x+8y+1)$