Factoring the polynomial $3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4x^2 - 4$

Question

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.

Answer 1

at "Cross Multiplication" it should be $$ p(3p+4)(p+1) $$

Answer 2

Hint: $y = x^2-1$:

$3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4(x^2 - 1) = 3y^3+7y^2+4y = y(3y^2+7y+4) = y(y+1)(3y+4)$

Answer 3

It's $$(x^2-1)(3(x^2-1)^2+7(x^2-1)+4)=(x^2-1)(3(x^2-1)+4)(x^2-1+1)=$$ $$=x^2(x^2-1)(3x^2+1)=x^2(x-1)(x+1)(3x^2+1).$$

By your way.

It should be $$3p^2+7p+4=(p+1)(3p+4).$$

Answer 4

it is $$(x^2-1)(3(x^2-1)^2+7(x^2-1)+4)=(x^2-1)x^2(3x^2+1)$$

Answer 5

Taking $ (x^2 - 1) $ as common, you get:

$$ (x^2 - 1) (3(x^2 - 1)^2 + 7(x^2 - 1) + 4) $$ $$ = (x+1)(x-1)( 3 (x^2-1)^2 + 3(x^2 - 1) + 4(x^2 - 1) + 4 )$$ $$ = (x+1)(x-1)( 3(x^2 - 1) + 4 ) ( x^2 - 1 + 1 ) $$ $$ = (x+1)(x-1)( 3x^2 - 3 + 4)(x^2) $$ $$ = x^2(x+1)(x-1)(3x^2 + 1) $$