Rui Baldwin

2020-11-09

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

$P(x)={x}^{4}+2{x}^{2}+1$

2abehn

Skilled2020-11-10Added 88 answers

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$P(x)={x}^{4}+2{x}^{2}?+1$ .

Factor the above polynomial to obtain the zeros.

$P(x)={x}^{4}+2{x}^{2}+1=(({x}^{2}{)}^{2}+2\ast 1\ast x+1)$

$=({x}^{2}+1{)}^{2}=({x}^{2}+1)({x}^{2}+1)$

$P(x)=({x}^{2}+1)({x}^{2}+1)$

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

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

Substitute 0 for P (x) in the polynomial$P(x)={x}^{4}+2{x}^{2}+1$ to obtain the zeros of the polynomial.

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

Further solve for the value of x as,

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

$x=i$ and $x=-i$

All zeros of the polynomial$P(x)={x}^{4}+2{x}^{2}+1$ appears two times in the polynomial therefore, the multiplicity of zeros -i, and i is 2.

Conclusion:

Thus, the factorization of the polynomial$P(x)={x}^{4}+{x}^{2}+1$ is $P(x)=(x-i{)}^{2}(x+i{)}^{2}$ ,zeros of the polynomial are +i and the multiplicity of the zeros is 2.

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

Factor the above polynomial to obtain the zeros.

Substitute 0 for P (x) in the polynomial

Further solve for the value of x as,

All zeros of the polynomial

Conclusion:

Thus, the factorization of the polynomial

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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