Rui Baldwin

2020-11-09

Complete Factorization Factor the polynomial completely, and find all its zeros.State the multiplicity of each zero.
$P\left(x\right)={x}^{4}+2{x}^{2}+1$

2abehn

Concept used:
The multiplicity of zero of the polynomial having factor (x — c) that appears k times in the factorization of the polynomial is k.
Calculation:
The given polynomial is $P\left(x\right)={x}^{4}+2{x}^{2}?+1$.
Factor the above polynomial to obtain the zeros.
$P\left(x\right)={x}^{4}+2{x}^{2}+1=\left(\left({x}^{2}{\right)}^{2}+2\ast 1\ast x+1\right)$
$=\left({x}^{2}+1{\right)}^{2}=\left({x}^{2}+1\right)\left({x}^{2}+1\right)$
$P\left(x\right)=\left({x}^{2}+1\right)\left({x}^{2}+1\right)$
$=\left({x}^{2}-{i}^{2}\right)\left({x}^{2}-{i}^{2}\right)$
$=\left(x+i\right)\left(x-i\right)\left(x+i\right)\left(x-i\right)$
Substitute 0 for P (x) in the polynomial $P\left(x\right)={x}^{4}+2{x}^{2}+1$ to obtain the zeros of the polynomial.
$\left(x+i\right)\left(x-i\right)\left(x+i\right)\left(x-i\right)=0$
Further solve for the value of x as,
$\left(x-i\right)=0$ and $\left(x+i\right)=0$
$x=i$ and $x=-i$
All zeros of the polynomial $P\left(x\right)={x}^{4}+2{x}^{2}+1$ appears two times in the polynomial therefore, the multiplicity of zeros -i, and i is 2.
Conclusion:
Thus, the factorization of the polynomial $P\left(x\right)={x}^{4}+{x}^{2}+1$ is $P\left(x\right)=\left(x-i{\right)}^{2}\left(x+i{\right)}^{2}$,zeros of the polynomial are +i and the multiplicity of the zeros is 2.

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