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 1 of 2 if there is more than one answer, separate them with commas. select
one\ if applicable. the zeros of f(x): 0,0,0,0. correct answer 0, 2 + 3i, 2 - 3i. part: 1 / 2 part 2 of 2 the multiplicity of 0 is the multiplicity of 2 + 3i is the multiplicity of 2 - 3i is

Explanation:

Step1: Recall polynomial - zero relationship

For a polynomial \(f(x)=a_nx^n + a_{n - 1}x^{n-1}+\cdots+a_1x + a_0\), if \(r\) is a zero of \(f(x)\), then \((x - r)\) is a factor of \(f(x)\). Complex - zeros of polynomials with real coefficients occur in conjugate pairs. Given \(f(x)=x^6-4x^5 + 13x^4\), we can factor out \(x^4\) first: \(f(x)=x^4(x^2-4x + 13)\).

Step2: Find zeros of \(x^4\)

Set \(x^4 = 0\). Solving for \(x\), we get \(x = 0\). Since the factor is \(x^4\), the multiplicity of \(x = 0\) 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 \(x=\frac{4\pm\sqrt{(-4)^2-4\times1\times13}}{2\times1}=\frac{4\pm\sqrt{16 - 52}}{2}=\frac{4\pm\sqrt{- 36}}{2}=\frac{4\pm6i}{2}=2\pm3i\). Since the factor is \(x^2-4x + 13=(x-(2 + 3i))(x-(2 - 3i))\), the multiplicity of \(2 + 3i\) and \(2 - 3i\) is \(1\) each.

Answer:

The zeros of \(f(x)\) are \(0\) with multiplicity \(4\), \(2 + 3i\) with multiplicity \(1\), and \(2 - 3i\) with multiplicity \(1\).