Question

find the zeros and their multiplicities. consider using descartes rule of signs and the upper and lower bound theorem to limit your search for rational zeros. write numbers as integers or simplified fractions. (h(x)=3x^{4}-16x^{3}+38x^{2}-56x + 15). part 1 of 2: if there is more than one answer, separate them with a comma. select
one\ if applicable. the zeros of (h(x)): (3,\frac{1}{3},1pm2i). part 2 of 2: the multiplicity of 3 is (square). the multiplicity of (\frac{1}{3}) is (square). the multiplicity of (1 - 2i) is (square). the multiplicity of (1 + 2i) is (square).

Explanation:

Step1: Recall polynomial zero - multiplicity concept

If \(x = a\) is a zero of a polynomial \(h(x)\), and \((x - a)^k\) is a factor of \(h(x)\) but \((x - a)^{k + 1}\) is not a factor of \(h(x)\), then the multiplicity of the zero \(x=a\) is \(k\). Complex zeros of polynomials with real - coefficients come in conjugate pairs. Given \(h(x)=3x^{4}-16x^{3}+38x^{2}-56x + 15\), if \(1 + 2i\) is a zero, then its conjugate \(1-2i\) is also a zero.
We can use synthetic division or factor the polynomial. Let's assume \(h(x)=(x-(1 + 2i))(x-(1 - 2i))q(x)\). First, \((x-(1 + 2i))(x-(1 - 2i))=x^{2}-2x + 5\).
We perform polynomial long - division: \(\frac{3x^{4}-16x^{3}+38x^{2}-56x + 15}{x^{2}-2x + 5}=3x^{2}-10x+3\).

Step2: Factor the quotient polynomial

Factor \(3x^{2}-10x + 3=(3x - 1)(x - 3)\). So \(h(x)=(x^{2}-2x + 5)(3x - 1)(x - 3)\).
The zeros of \(h(x)\) are \(x = 3,\frac{1}{3},1 + 2i,1 - 2i\).
The multiplicity of a zero is determined by the exponent of the corresponding factor.
For \(x = 3\), since \((x - 3)\) is a factor and \((x - 3)^2\) is not a factor of \(h(x)\), the multiplicity of \(3\) is \(1\).
For \(x=\frac{1}{3}\), since \((3x - 1)\) is a factor and \((3x - 1)^2\) is not a factor of \(h(x)\), the multiplicity of \(\frac{1}{3}\) is \(1\).
For \(x=1 + 2i\), since \((x-(1 + 2i))\) is a factor and \((x-(1 + 2i))^2\) is not a factor of \(h(x)\), the multiplicity of \(1 + 2i\) is \(1\).
For \(x=1 - 2i\), since \((x-(1 - 2i))\) is a factor and \((x-(1 - 2i))^2\) is not a factor of \(h(x)\), the multiplicity of \(1 - 2i\) is \(1\).

Answer:

The multiplicity of \(3\) is \(1\).
The multiplicity of \(\frac{1}{3}\) is \(1\).
The multiplicity of \(1 + 2i\) is \(1\).
The multiplicity of \(1 - 2i\) is \(1\).