How do you factor completely 3x^3+9x^2+x+33x3+9x2+x+3?

1 Answer
Steve M
Dec 16, 2016

3x^3+9x^2+x+3 = (3x^2+1)(x + 3) 3x3+9x2+x+3=(3x2+1)(x+3)

Explanation:

This factors by grouping, as:
3x^3+9x^2+x+3 = 3x^2*x + 3x^2*3+x+3 3x3+9x2+x+3=3x2⋅x+3x2⋅3+x+3
" " = 3x^2(x + 3)+(x+3) =3x2(x+3)+(x+3)
" " = (3x^2+1)(x + 3) =(3x2+1)(x+3)

The quadratic term does not factorise further, so this is the fully factorised form.

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