Question

question if ( f(x)=2x^{3}+22x^{2}+40x + 36) and ( f(-9)=0), then find all of the zeros of ( f(x)) algebraically.

Explanation:

Step1: Use factor - theorem

Since \(f(-9)=0\), \((x + 9)\) is a factor of \(f(x)=2x^{3}+22x^{2}+40x + 36\).

Step2: Perform polynomial long - division

Divide \(2x^{3}+22x^{2}+40x + 36\) by \(x + 9\).
\[

$$\begin{align*} \frac{2x^{3}+22x^{2}+40x + 36}{x + 9}&=2x^{2}+4x + 4 \end{align*}$$

\]

Step3: Factor the quotient

Factor \(2x^{2}+4x + 4\). First, factor out the greatest common factor 2: \(2(x^{2}+2x + 2)\).
The quadratic formula for \(ax^{2}+bx + c\) is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). For \(x^{2}+2x + 2\), \(a = 1\), \(b=2\), \(c = 2\). Then \(x=\frac{-2\pm\sqrt{2^{2}-4\times1\times2}}{2\times1}=\frac{-2\pm\sqrt{4 - 8}}{2}=\frac{-2\pm2i}{2}=-1\pm i\).
The factor \(x + 9 = 0\) gives \(x=-9\).

Answer:

\(x=-9,x=-1 + i,x=-1 - i\)