How do you factor #6x^3+21x^2+15x #?

1 Answer
Jun 26, 2015

Answer:

#6x^3+21x^2+15x = 3x(2x^2+7x+5)#

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

Explanation:

Given #6x^3+21x^2+15x#

First notice that all of the terms are divisible by #3x#, so separate out that as a factor first:

#6x^3+21x^2+15x = 3x(2x^2+7x+5)#

Let #f(x) = 2x^2+7x+5#

First notice that #2-7+5 = 0#, so #f(-1) = 0#, so #(x+1)# is a factor of #f(x)#

The remaining factor must be #(2x+5)# in order that the coefficient #2# of #x^2# and the constant term #5# result:

#2x^2+7x+5 = (2x+5)(x+1)#

Putting it all together:

#6x^3+21x^2+15x = 3x(2x+5)(x+1)#