How do you factor completely #9x^2+9x-10#?

1 Answer
Apr 17, 2016

Answer:

#9x^2+9x-10 = (3x-2)(3x+5)#

Explanation:

One method involves completing the square, then using the difference of squares identity:

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

First multiply through by #4# to make the arithmetic simpler, not forgetting to divide through by #4# at the end.

#4(9x^2+9x-10)#

#=36x^2+36x-40#

#=(6x+3)^2-9-40#

#=(6x+3)^2-7^2#

#=((6x+3)-7)((6x+3)+7)#

#=(6x-4)(6x+10)#

#=(2(3x-2))(2(3x+5))#

#=4(3x-2)(3x+5)#

Dividing both ends by #4# we get:

#9x^2+9x-10 = (3x-2)(3x+5)#

#color(white)()#
Why did I multiply by #4# to start?

The middle term of the original quadratic expression is odd, so if you do not multiply through by some multiple of #4# first then you end up doing arithmetic with #1/2#'s and #1/4#'s. It was not necessary to multiply through by anything else to avoid fractions since the leading term #9x^2 = (3x)^2# was already a perfect square.

Without premultiplying, it looks like this:

#9x^2+9x-10#

#=(3x+3/2)^2-9/4-10#

#=(3x+3/2)^2-49/4#

#=(3x+3/2)^2-(7/2)^2#

#=((3x+3/2)-7/2)((3x+3/2)+7/2)#

#=(3x-2)(3x+5)#