How do you factor #x^3+9x^2+15x-25#?

1 Answer
Nov 11, 2016

#x^3+9x^2+15x-25 = (x-1)(x+5)^2#

Explanation:

Given:

#x^3+9x^2+15x-25#

First notice that the sum of the coefficients is #0#. That is:

#1+9+15-25 = 0#

Hence #x=1# is a zero and #(x-1)# a factor:

#x^3+9x^2+15x-25 = (x-1)(x^2+10x+25)#

To factor the remaining quadratic note that both #x^2# and #25 = 5^2# are perfect squares, with #10x = 2(5)x#. So this quadratic is a perfect square trinomial:

#x^2+10x+25 = x^2+2(5)x+5^2 = (x+5)^2#

Putting it all together:

#x^3+9x^2+15x-25 = (x-1)(x+5)^2#