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

1 Answer
Oct 12, 2015

Answer:

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

Explanation:

Your starting expression looks like this

#3x^3 + 2x^2 - 2x - 3#

Notice that you can group the term according to their coefficients to get

#3x^3 - 3 + 2x^2 - 2x#

Use #3# and #2x# as common factors for their respective terms to get

#3(x^3 - 1) + 2x * (x - 1)#

Notice that you're dealing with the difference of two cubes, for which you know that

#color(blue)(a^3 - b^3 = (a-b) * (a^2 + ab + b^2))#

In your case, you have

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

The expression can thus be written as

#3 * (x-1) * (x^2 + x + 1) + 2x * (x-1)#

Use #(x-1)# as a common factor

#(x-1) * [3 * (x^2 + x + 1) + 2x]#

Finally, expand the paranthesis and group like terms to get

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