Find x intercepts of a higher degree polynomial $2x^4+6x^2-8$ I am to factor and then find the x intercepts (roots?) of $2x^4+6x^2-8$
The solutions are provided as 1 and -1 and I am struggling to get to this.
My working:
$2x^4+6x^2-8$ =
$2(x^4+3x^2-4)$
Focus on just the right term $(x^4+3x^2-4)$:
Let $u$ = $x^2$, then:
$u^2+3u-4$ =
master term is 1 * -4 = -4. Factors that give minus 4 and sum to 3 are 4 and -1...
$(u^2-u)+(4u-4)$ =
$u(u-1)+4(u-1)$ =
$(u+4)(u-1)$
I don't know where to go from here. If I write $u$ back into it's original $x^2$ I get:
$(x^2+4)(x^2-1)$
Where do I go from here to arrive at x intercepts of 1 and -1?

$$2x^4+6x^2-8=2(x^2+4)(x^2-1)=2(x^2+4)(x-1)(x+1)=0$$
is true when either $$x^2+4=0$$ or $$x+1=0$$ or $$x-1=0.$$
The first condition is not possible in the reals as $x^2+4\ge4$.
