How do you factor #2x^3 - 3x^2 - 5x#?

1 Answer
sente
Nov 14, 2015

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

Explanation:

First, note that each term has a factor of #x#, and so we have
#2x^3 - 3x^2 - 5x = x(2x^2 - 3x - 5)#

Now, we can use the quadratic formula to find the remaining factors, but first let's see if there are easy integer solutions by looking for #a,b,c,d# where #2x^2 - 3x - 5 = (ax+b)(cx+d)#

We know that #ac = 2# and so we can look at #(2x+b)(x+d)#

We also know #bd = -5# and so our possible choices are #(2x+1)(x-5)# and #(2x-5)(x+1)#

Multiplying these out shows that #2x^2 - 3x - 5 = (2x-5)(x+1)#

So our final result is
#2x^3 - 3x^2 - 5x = x(2x-5)(x+1)#

Related questions
See all questions in Monomial Factors of Polynomials
Impact of this question
2981 views around the world
You can reuse this answer
Creative Commons License