How do you factor #x^2 + 5x - 36 #?

1 Answer
George C.
May 23, 2016

#x^2+5x-36=(x+9)(x-4)#

Explanation:

Find a pair of factors of #36# which differ by #5#.

The pair #9, 4# works in that #9xx4=36# and #9-4=5#

Hence we find:

#x^2+5x-36=(x+9)(x-4)#

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Alternative method

Alternatively, we can complete the square and use the difference of squares identity:

#a^2-b^2=(a-b)(a+b)#

with #a=(2x+5)# and #b=13# as follows.

First multiply by #2^2 = 4# to cut down on arithmetic involving fractions. Remember to divide by it at the end...

#4(x^2+5x-36)#

#=4x^2+20x-144#

#=(2x)^2+2(2x)(5)-144#

#=(2x+5)^2-25-144#

#=(2x+5)^2-169#

#=(2x+5)^2-13^2#

#=((2x+5)-13)((2x+5)+13)#

#=(2x-8)(2x+18)#

#=(2(x-4))(2(x+9))#

#=4(x-4)(x+9)#

Dividing both ends by #4# we find:

#x^2+5x-36 = (x-4)(x+9)#

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