Question

complete the following problems. please submit them at the end of this module along with any additional practice problems contained in this module. problem 7. find all the zeros given function and one zero of the function. (f(x)=x^{4}+3x^{3}-x^{2}+27x - 90) with knowledge that (x = 2) and (x=-5) are zeros of the polynomial function.

Explanation:

Step1: Use factor - theorem

Since \(x = 2\) and \(x=-5\) are zeros, \((x - 2)\) and \((x + 5)\) are factors of \(f(x)=x^{4}+3x^{3}-x^{2}+27x - 90\). Multiply \((x - 2)(x + 5)=x^{2}+3x - 10\).

Step2: Perform polynomial long - division

Divide \(x^{4}+3x^{3}-x^{2}+27x - 90\) by \(x^{2}+3x - 10\).
\[

$$\begin{align*} x^{4}+3x^{3}-x^{2}+27x - 90&=(x^{2}+3x - 10)(x^{2}+9) \end{align*}$$

\]

Step3: Find zeros of the quadratic factors

For \(x^{2}+3x - 10 = 0\), we already know the roots \(x = 2\) and \(x=-5\) from the given information.
For \(x^{2}+9=0\), we solve \(x^{2}=-9\), so \(x = 3i\) and \(x=-3i\).

Answer:

The zeros of the function \(f(x)\) are \(x = 2\), \(x=-5\), \(x = 3i\), \(x=-3i\)