Try to complete Factorization Factor the polynomial completely , and find all its zeros. State the multiplicity of each zero.

$P(x)={x}^{4}-625$

Brittney Lord
2020-10-20
Answered

Try to complete Factorization Factor the polynomial completely , and find all its zeros. State the multiplicity of each zero.

$P(x)={x}^{4}-625$

You can still ask an expert for help

SkladanH

Answered 2020-10-21
Author has **80** answers

Concept used:

The multiplicity of zero of the polynomial having factor$(x\u2014c)$ that appears k times in the factorization of the polynomial is k.

Calculation:

The given polynomial is$P(x)={x}^{4}\u2014625$ .

Factor the above polynomial to obtain the zeros

$P(x)={x}^{4}-625$

$=(({x}^{2}{)}^{2}-(25{)}^{2})$

$=({x}^{2}-25)({x}^{2}+25)$

$=({x}^{2}-{5}^{2})({x}^{2}-(5i{)}^{2})$

Further factorize the above expression

$P(x)=({x}^{2}-{5}^{2})({x}^{2}-(5i{)}^{2})$

$=(x+5)(x-5)(x+5i)(x-5i)$

Substitute 0 for P(x) in the polynomial$P(x)={x}^{4}\u2014625$ to obtain the zeros of the polynomial.

$(x+5)(x-5)(x+5i)(x-5i)=0$

Further solve for the value of x as,

$(x+5)=0,(x-5)=0,(x+5i)=0$ and $(x-5i)=0$

$x=-5,x=5,x=-5i$ and $x=5i$

All the zeros in the polynomial$P(x)={x}^{4}\u2014625$ 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$P(x)={x}^{4}\u2014625$ is

$P(x)=(x+5)(x\u20145)(x+5i)(x\u20145i)$ , zeros of the polynomial are $\pm 5\text{}and\text{}\pm 5i$ and the multiplicity of all the zeros is 1.

The multiplicity of zero of the polynomial having factor

Calculation:

The given polynomial is

Factor the above polynomial to obtain the zeros

Further factorize the above expression

Substitute 0 for P(x) in the polynomial

Further solve for the value of x as,

All the zeros in the polynomial

Therefore, the multiplicity of zeros 5i, —5i, 5 and —5 is 1.

Conclusion:

Thus, the factorization of the polynomial

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