# Find the roots of the polynomial: f(x)=x^{4}+4x^{2}-32

Find the roots of the polynomial: $f\left(x\right)={x}^{4}+4{x}^{2}-32$
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Step 1
We have to find the roots of polynomial:
$f\left(x\right)={x}^{4}+4{x}^{2}-32$
To find the roots of given polynomial we need to put that polynomial equals to zero.
Therefore,
f(x)=0
${x}^{4}+4{x}^{2}-32=0$
Solving by factorization method,
${x}^{4}+8{x}^{2}-4{x}^{2}-32=0$
${x}^{2}\left({x}^{2}+8\right)-4\left({x}^{2}+8\right)=0$
$\left({x}^{2}+8\right)\left({x}^{2}-4\right)=0$
Step 2
Either,
${x}^{2}+8=0$
${x}^{2}=-8$
$x=±\sqrt{-8}$
$=±\sqrt{8}i$ (since $i=\sqrt{-1}$, imaginary unit)
$=±2\sqrt{2}i$
$=2\sqrt{2}i,-2\sqrt{2}i$
or,
${x}^{2}-4=0$
${x}^{2}=4$
$x=±\sqrt{4}$
$=±2$
=2, -2
Hence, roots of the polynomial are $x=-2\sqrt{2}i,-2,2,2\sqrt{2}i$.

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Dawn Neal
Step 1
Explanation:
Given that,
$f\left(x\right)={x}^{4}+4{x}^{2}-32$
Let us suppose
${x}^{2}=u$
Function becomes
${u}^{2}+4u-32$
Step 2
Now factorize the quadratic
${u}^{2}+4u-32$
(u+8)(u-4)
Recall the value of u
$\left({x}^{2}+8\right)\left({x}^{2}-4\right)$
Further, factorize the $\left({x}^{2}-4\right)=\left(x-2\right)\left(x+2\right)$
We get,
$f\left(x\right)=\left({x}^{2}+8\right)\left(x-2\right)\left(x+2\right)$

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