Factorize the following:

1. $x^2+18x+81=(x+9)^2$
2. $a^2-14a+49=(a-7)^2$
3. $4x^2+12x+9=(2x+3)^2$
4. $9x^2-24x+16=(3x-4)^2$
5. $49x^2+56x+16=(7x+4)^2$
6. $25z^2-30z+9=(5z-3)^2$
7. $9q^4 r^4-6p^4 q^2 r^2+p^8=(3q^2 r^2-p^4 )^2$
8. $\frac{9p^2}{q^2} +\frac{16r^2}{m^2} +\frac{24pr}{qm}=(\frac{3p}{q}+\frac{4r}{m})^2$
9. $\frac{1}{4} z^6+9a^2-3az^3=( \frac{1}{2} z^3-3a)^2$
10. $\frac{9}{4} a^2+ \frac{49}{9} p^2-7ap=( \frac{3}{2} a- \frac{7}{3} p)^2$

11.  $t^2+22t+85 = 0$

Find two numbers with sum = 22 and product=85

$a+b=22$

$ab=85$

Two numbers are 17, 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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1. $x^2-10x+24$

$a+b=-10$

$ab=24$

$a=-6, b=-4$

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

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1. $m^2-3m-40$

$a+b=-3$

$ab=-40$

$a=-8, b=+5$

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

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1. $x^2+x-72$

$a+b=1$

$ab=-72$

$a=9, b=-8$

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

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1. $p^2-7p-120$

$a+b=-7$

$ab=-120$

$a=8, b=-15$

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

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1. $16-17z+z^2$

$a+b=-17$

$ab=16$

$a=-1, b=-16$

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

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1. $a^2+5a-104$

$x+y=5$

$xy=-104$

$x=13, y=-8$

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

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1. $3x^2+11x+10$

$a+b=11$

$ab=30$

$a=5, b=6$

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

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

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

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1. $6x^2+7x-3$

$a+b=7$

$ab=-18$

$a=9, b=-2$

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

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

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

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1. $3z^2-4z-4$

$a+b=-4$

$ab=-12$

$a=-6, b=2$

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

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

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

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1. $72-x-x^2$

$a+b=-1$

$ab=-72$

$a=-8, b=-9$

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

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

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

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

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1. $3x^2 y+11xy+6y=y(3x^2+11x+6)$

$a+b=11$

$ab=18$

Therefore $a=9, b=2$

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

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

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

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1. $(a-b)^2-5(a-b)+6$

$Let a-b=x$

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

Substituting Back

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

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

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

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

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

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1. $3(y-2)^2-(y-2)-44$

$Let (y-2)=x$

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

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

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

Substituting Back

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

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

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

$Let (x+y)=a$

Therefore we have

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

$m+n=10$

$mn=-56$

$m=14 \ \ n=-4$

Hence

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

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

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

Substituting Back

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

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