Question

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

Answer 1

`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)`