# Try to complete Factorization Factor the polynomial completely , and find all its zeros. State the multiplicity of each zero. P(x)=x^{4}-625

Try to complete Factorization Factor the polynomial completely , and find all its zeros. State the multiplicity of each zero.
$P\left(x\right)={x}^{4}-625$
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Concept used:
The multiplicity of zero of the polynomial having factor $\left(x—c\right)$ that appears k times in the factorization of the polynomial is k.
Calculation:
The given polynomial is $P\left(x\right)={x}^{4}—625$.
Factor the above polynomial to obtain the zeros
$P\left(x\right)={x}^{4}-625$
$=\left(\left({x}^{2}{\right)}^{2}-\left(25{\right)}^{2}\right)$
$=\left({x}^{2}-25\right)\left({x}^{2}+25\right)$
$=\left({x}^{2}-{5}^{2}\right)\left({x}^{2}-\left(5i{\right)}^{2}\right)$
Further factorize the above expression
$P\left(x\right)=\left({x}^{2}-{5}^{2}\right)\left({x}^{2}-\left(5i{\right)}^{2}\right)$
$=\left(x+5\right)\left(x-5\right)\left(x+5i\right)\left(x-5i\right)$
Substitute 0 for P(x) in the polynomial $P\left(x\right)={x}^{4}—625$ to obtain the zeros of the polynomial.
$\left(x+5\right)\left(x-5\right)\left(x+5i\right)\left(x-5i\right)=0$
Further solve for the value of x as,
$\left(x+5\right)=0,\left(x-5\right)=0,\left(x+5i\right)=0$ and $\left(x-5i\right)=0$
$x=-5,x=5,x=-5i$ and $x=5i$
All the zeros in the polynomial $P\left(x\right)={x}^{4}—625$ appear one times in the polynomial.
Therefore, the multiplicity of zeros 5i, —5i, 5 and —5 is 1.
Conclusion:
Thus, the factorization of the polynomial $P\left(x\right)={x}^{4}—625$ is
$P\left(x\right)=\left(x+5\right)\left(x—5\right)\left(x+5i\right)\left(x—5i\right)$, zeros of the polynomial are and the multiplicity of all the zeros is 1.