Question 4.3.7: Using the Factor Theorem Given that 2 is a zero of the funct...

Because 2 is a zero of f(x), we have f(2)=0. The Factor Theorem tells us that (x-2) is a factor of f(x). Next, we use synthetic division to divide f(x) by (x-2).

$\begin{array}{r c}\begin{matrix} \underline{2|}  \\ \\ \\ \end{matrix} & \begin{matrix} 3 & 2& -19 & 6 \\ & 6&16& -6\\ \hline\underset{ \uparrow }{3} &\underset{ \uparrow }{8}&\underset{ \uparrow }{-3} & | 0 \end{matrix} &\begin{matrix} \\ \leftarrow \text{Remainder} \end{matrix} \begin{matrix} & & \end{matrix} \end{array} \\ \begin{matrix} \overbrace{\text{Coefficients of the quotient}}^{} \end{matrix}\qquad$

We now have the coefficients of the quotient Q(x), with f(x)=(x-2) Q(x).

$f(x)=3 x^{3}+2 x^{2}-19 x+6=(x-2)(3 x^{2}+\underset{\overset{\uparrow }{\text{Quotient}} }{8x} -3) .$

Any solution of the depressed equation 3 x²+8 x-3=0 is a zero of f. Because this equation is of degree 2, any method of solving a quadratic equation may be used to solve it.

$\begin{aligned}3 x^{2}+8 x-3=0 & \text { Depressed equation } \\(3 x-1)(x+3)=0 & \text { Factor } \\3 x-1=0 \text { or } x+3=0 & \text { Zero-product property } \\x=\frac{1}{3} \text { or } \quad x=-3 & \text { Solve each equation. }\end{aligned}$

The solution set is  $\left\{-3, \frac{1}{3}, 2\right\}.$