# 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 $$x^2+4\ge4$$.
you are almost done $$(x^2+4)(x^2-1)=0$$ $$x^2=-4, \ x^2=1$$ $$x=\pm2 i, \ x=\pm 1$$ considering the real values, the x-intercepts are $$x=-1, y=0$$ and $$x=1, y=0$$