Is it possible to factor #y= 3x^3+8x^2+3x-2 #? If so, what are the factors?

1 Answer
Feb 12, 2016

Answer:

#3x^3+8x^2+3x-2 = (x+1)(3x-1)(x+2)#

Explanation:

Let #f(x) = 3x^3+8x^2+3x-2#

First notice that #f(-1) = -3+8-3-2 = 0#, so #x=-1# is a zero and #(x+1)# is a factor:

#y = 3x^3+8x^2+3x-2 = (x+1)(3x^2+5x-2)#

We can factor the remaining quadratic by grouping, with the help of an AC Method.

Look for a pair of factors of #AC = 3*2 = 6# which differ by #B=5#. The pair #6, 1# works, so use that to split the middle term before factoring by grouping:

#3x^2+5x-2#

#=3x^2+6x-x-2#

#=(3x^2+6x)-(x+2)#

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

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

Putting it all together:

#3x^3+8x^2+3x-2 = (x+1)(3x-1)(x+2)#