How do you factor completely #x^3 - 8x^2 + 5x + 50 = 0#?

1 Answer
KillerBunny
Nov 13, 2015

#(x+2)(x-5)^2#.

Explanation:

We know that if a number solves such an equation, it must divide its last coefficient, i.e. #50#. So, we can try with some of its divisors: if #f(x)=x^3-8x^2+5x+50#, then:

  • #f(1)=48 ne 0#;
  • #f(-1)= 36 ne 0#
  • #f(2) =36 ne 0#
  • #f(-2) = 0#.

So, #-2# is a solution, which means that #f(x)# can be divided by #(x+2)#. Do the long division, and you have that

#x^3-8x^2+5x+50 = (x+2)(x^2-10x+25)#

Now we must factor the second parenthesis, but you can see that it is the square of #(x-5)#, and so #f(x)# is completely factored.

Related questions
See all questions in Factoring Completely
Impact of this question
5161 views around the world
You can reuse this answer
Creative Commons License