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. (f(x)=x^{6}-4x^{5}+13x^{4}) part: 0 / 2 part 1 of 2 if there is more than one answer, separate them with commas. select
one\ if applicable. the zeros of (f(x)):

Explanation:

Step1: Factor out the common factor

First, factor out $x^{4}$ from the polynomial $f(x)=x^{6}-4x^{5}+13x^{4}$. We get $f(x)=x^{4}(x^{2}-4x + 13)$.

Step2: Find zeros of $x^{4}$

Set $x^{4}=0$. Solving for $x$, we have $x = 0$ with multiplicity 4 since the exponent of $x$ is 4.

Step3: Find zeros of $x^{2}-4x + 13$

Use the quadratic - formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 1$, $b=-4$, and $c = 13$. Then $\Delta=b^{2}-4ac=(-4)^{2}-4\times1\times13=16 - 52=-36$. So $x=\frac{4\pm\sqrt{-36}}{2}=\frac{4\pm6i}{2}=2\pm3i$. These are non - rational zeros.

Answer:

0 (multiplicity 4)