<s>[INST] 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?
 [/INST] 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 $x=1, y=0$
x-intercepts in point form: (-1,0) and (1,0)
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