# Complete Factorization Factor the polunomial completely, and find its zeros.State the multiplicity of each zero. P(x)=x^{3}-64

Complete Factorization Factor the polunomial completely, and find its zeros.State the multiplicity of each zero.
$P\left(x\right)={x}^{3}-64$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

berggansS
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 $P\left(x\right)={x}^{3}-64$.
Factor the above polynomial to obtain the zeros.
$P\left(x\right)={x}^{3}-64$
$=\left(x-4\right)\left({x}^{2}+4x+16\right)$
$=\left(x-4\right)\left({x}^{2}+2\ast 2\ast x+4+12\right)$
$=\left(x-4\right)\left({x}^{2}+2\ast 2\ast x+{2}^{2}+12\right)$
Further solve the expression,
$P\left(x\right)=\left(x-4\right)\left(\left(x+2{\right)}^{2}-\left(2\sqrt{3i}{\right)}^{2}\right)$
$=\left(x-4\right)\left(x+2-2\sqrt{3i}\right)\left(x+2+2\sqrt{3i}\right)$
Substitute 0 for P(x) in the polynomial $P\left(x\right)={x}^{3}-64$ to obtain the zeros of the polynomial.
$\left(x—4\right)\left(x+2—2\sqrt{3i}\right)\left(x+2+2\sqrt{3i}\right)=0$
Further solve for the value of x as,
$\left(x-4\right)=0,\left(x+2-2\sqrt{3}\right)=0$, and$\left(x+2+2y3\right)=0$
$x=4,x=-2+2\sqrt{3}$, and $x=-2-2\sqrt{3}$
All zeros of the polynomial $P\left(x\right)={x}^{3}-64$ 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 $P\left(x\right)={x}^{3}-64$ is
$P\left(x\right)=\left(x-4\right)\left(x+2-2\sqrt{3i}\right)\left(x+2+2\sqrt{3i}\right)$, zeros of the polynomial are $\left(-2+-2\sqrt{3}\right)$ and 4 and the multiplicity of all the zeros is 1.