Question

question: if $f(x)=x^{3}-5x^{2}-31x - 45$ and $x - 9$ is a factor of $f(x)$, then find all of the zeros of $f(x)$ algebraically.

Explanation:

Step1: Use polynomial long - division

Since \(x - 9\) is a factor of \(f(x)=x^{3}-5x^{2}-31x - 45\), we divide \(f(x)\) by \(x - 9\) using polynomial long - division or synthetic division. Using synthetic division:
Set up the synthetic - division with \(c = 9\) and the coefficients \(1,-5,-31,-45\).
Bring down the first coefficient \(1\):
Multiply \(9\times1 = 9\), add to the second coefficient: \(-5 + 9=4\).
Multiply \(9\times4 = 36\), add to the third coefficient: \(-31+36 = 5\).
Multiply \(9\times5 = 45\), add to the fourth coefficient: \(-45 + 45=0\).
The quotient is \(x^{2}+4x + 5\).
So \(f(x)=(x - 9)(x^{2}+4x + 5)\).

Step2: Find the zeros of the quadratic factor

Set \(x^{2}+4x + 5 = 0\). Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for the quadratic equation \(ax^{2}+bx + c = 0\). Here \(a = 1\), \(b = 4\), \(c = 5\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(4)^{2}-4\times1\times5=16 - 20=-4\).
Then \(x=\frac{-4\pm\sqrt{-4}}{2}=\frac{-4\pm2i}{2}=-2\pm i\).
Set \(x - 9=0\), we get \(x = 9\).

Answer:

\(x = 9,x=-2 + i,x=-2 - i\)