 # A polynomial P is given. P(x)=x^{4}+16x^{2} (a) Find all zeros of P, eliaskidszs 2021-12-17 Answered
A polynomial P is given.
$P\left(x\right)={x}^{4}+16{x}^{2}$
(a) Find all zeros of P, real and complex.
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Step 1
Given polynomial, $P\left(x\right)={x}^{4}+16{x}^{2}$
Step 2
Zeroes of P(x), P(x)=0
${x}^{4}+16{x}^{2}=0$
${x}^{2}\left({x}^{2}+16\right)=0$
${x}^{2}=0$ & ${x}^{2}+16=0$
x=0,0 & ${x}^{2}=-16$
x=0,0 & $x=±4i$
Hence zeroes are $x=0,±4i$

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Sol: Given $P\left(x\right)={x}^{4}+16{x}^{2}$
(a) Then $P\left(x\right)={x}^{4}+16{x}^{2}=0$
$⇒{x}^{2}\left({x}^{2}+16\right)=0$
$⇒{x}^{2}\left(x+4i\right)\left(x-4i\right)=0$
$⇒x=0,0,x+4i=0,x-4i=0$
$⇒x=0,0,x=-4i,x=4i$
$⇒x=0,0,-4i,4i$
Zeroes are 0,0, -4i, 4i

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