floymdiT

2021-03-05

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

$P(x)={x}^{3}-64$

berggansS

Skilled2021-03-06Added 91 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$P(x)={x}^{3}-64$ .

Factor the above polynomial to obtain the zeros.

$P(x)={x}^{3}-64$

$=(x-4)({x}^{2}+4x+16)$

$=(x-4)({x}^{2}+2\ast 2\ast x+4+12)$

$=(x-4)({x}^{2}+2\ast 2\ast x+{2}^{2}+12)$

Further solve the expression,

$P(x)=(x-4)((x+2{)}^{2}-(2\sqrt{3i}{)}^{2})$

$=(x-4)(x+2-2\sqrt{3i})(x+2+2\sqrt{3i})$

Substitute 0 for P(x) in the polynomial$P(x)={x}^{3}-64$ to obtain the zeros of the polynomial.

$(x\u20144)(x+2\u20142\sqrt{3i})(x+2+2\sqrt{3i})=0$

Further solve for the value of x as,

$(x-4)=0,(x+2-2\sqrt{3})=0$ , and$(x+2+2y3)=0$

$x=4,x=-2+2\sqrt{3}$ , and $x=-2-2\sqrt{3}$

All zeros of the polynomial$P(x)={x}^{3}-64$ appears one times in the polynomial therefore, the multiplicity of zeros 4, -2 + 2 $\sqrt{3i}$ , and
-2-2 $\sqrt{3i}$ is 1.

Conclusion:

Thus, the factorization of the polynomial$P(x)={x}^{3}-64$ is

$P(x)=(x-4)(x+2-2\sqrt{3i})(x+2+2\sqrt{3i})$ , zeros of the polynomial are $(-2+-2\sqrt{3})$ and 4 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 solve the expression,

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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