Factoring algebraic expressions

Question

I would like to know how to factor the following algebraic expressions:

1. $x^6-7x^3-8$
2. $4x^2+4-8x-12y-12xy$

Answer 1

For $x^6-7x^3-8$, set $t=x^3$. Then

$x^6-7x^3-8 \\=t^2-7t-8 \\=(t+1)(t-8) \\=(x^3+1)(x^3-8)\\=(x + 1) (x^2 - x + 1)(x - 2) (x^2 + 2 x + 4)$

Answer 2

for 1) we get $$x^6-7x^3-8=(x-2) (x+1) \left(x^2-x+1\right) \left(x^2+2 x+4\right)$$ since $$2^6-7\cdot 2^3-8=0$$ we can do polynomial Long Division and since $$(-1)^6-7(-1)^3-8=0$$ we have also the solution $$x=-1$$ for 2) we have $$4(x^2+1-2x-3y-3xy)=4((x-1)^2-3y(1+x))$$

Answer 3

Hint:

For 1), solve as a quadratic equation

$$(x^3)^2-7(x^3)-8=0=(x^3+1)(x^3-8).$$

Then you factor using the cubic roots of $-1$ and $2$, giving the linear factors $(x+1)(x-2)$ and two quadratic ones related to the complex conjugate roots.

Answer 4

What if I did it like this $$x^6-7x^3-8$$ The relationship between $x^6$ and $7x^3$ is $x^3$. So let $P=x^3$,thus $$P^2-7P-8=(P-8)(P+1).$$ But since $P=x^3$, the answer is $(x^3-8)(x^3+1)$