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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?

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2 Answers 2

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$$2x^4+6x^2-8=2(x^2+4)(x^2-1)=2(x^2+4)(x-1)(x+1)=0$$

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 $x^2+4\ge4$.

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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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