Factoring the polynomial $3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4x^2 - 4$ I'm trying to factor the following polynomial:
$$3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4x^2 - 4$$
What I've done:
$$3(x^2 -1)^3 + 7(x^2-1)^2 + 4(x^2 -1)$$
Then I set $p=x^2 -1$ so the polynomial is: 
$$3p^3 + 7p^2 + 4p$$
Therefore: $$p(3p^2 + 7p + 4)$$
I apply Cross Multiplication Method: $$p(p+3)(p+4)$$
I substitute $p$ with $x^2-1$:
$$(x^2-1)(x^2-1+3)(x^2-1+4)$$
$$(x-1)(x+1)(x^2-2)(x^2-3)$$
I don't know if I've done something wrong or if I have to proceed further and how. The result has to be: $x^2(3x^2+1)(x+1)(x-1)$. Can you help me? Thanks.

Taking $ (x^2 - 1) $ as common, you get: 
$$ (x^2 - 1) (3(x^2 - 1)^2 + 7(x^2 - 1) + 4) $$
$$ = (x+1)(x-1)( 3 (x^2-1)^2 + 3(x^2 - 1) + 4(x^2 - 1) + 4 )$$
$$ = (x+1)(x-1)( 3(x^2 - 1) + 4 ) ( x^2 - 1 + 1 ) $$
$$ = (x+1)(x-1)( 3x^2 - 3 + 4)(x^2) $$
$$ = x^2(x+1)(x-1)(3x^2 + 1) $$
