Question 4.4.2: Solving a Polynomial Equation Solve 3x³ - 8x² - 8x + 8 = 0.

\begin{aligned}\text { Possible rational roots } &=\frac{\text { Factors of the constant term, } 8}{\text { Factors of the leading coefficient, } 3} \\&=\frac{\pm 1, \pm 2, \pm 4, \pm 8}{\pm 1, \pm 3} \\&=\pm 1, \pm 2, \pm 4, \pm 8, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}, \pm \frac{8}{3}\end{aligned}

The graph of y=3 x^{3}-8 x^{2}-8 x+8 is shown in Figure 16. The calculator graph is shown in the margin.

From the graph in Figure 16, it appears that an x-intercept is between 0.5 and 1.

We check whether x=\frac{2}{3} is a root by synthetic division.

\begin{matrix} \underline{\frac{2}{3} |}& \\ \\ \\ \end{matrix} \begin{matrix} 3 & -8& -8 & 8 \\ & 2 & -4 & -8\\ \hline 23 &7-6& 3-12& |0 \checkmark \text{Yes}\end{matrix}

Because the remainder in the synthetic division is 0,\left(x-\frac{2}{3}\right) is a factor of the original equation with the depressed equation 3 x^{2}-6 x-12=0.

So

\begin{aligned}3 x^{3}-8 x^{2}-8 x+8 &=0 & & \text { Original equation } \\\left(x-\frac{2}{3}\right)\left(3 x^{2}-6 x-12\right) &=0 & & \text { Synthetic division } \\\left(x-\frac{2}{3}\right)=0 & \text { or } 3 x^{2}-6 x-12=0 & & \text { Zero-product property }\end{aligned}

\begin {array}{rl|rlrl} x=\frac{2}{3} && x=\frac{-(-6) \pm \sqrt{(-6)^{2}-4(-12)(3)}}{2(3)} & \text{Solve for x using the quadratic formula} \\ && =\frac{6 \pm \sqrt{36+144}}{6} & \\ && =\frac{6 \pm \sqrt{180}}{6}=\frac{6 \pm \sqrt{(36)(5)}}{6} & \\ && =\frac{6 \pm 6 \sqrt{5}}{6}=1 \pm \sqrt{5}& \end {array}

You can see from the graph in Figure 16 that  1 \pm \sqrt{5} are also the x-intercepts of the graph.

The three real roots of the given equation are

\frac{2}{3}, 1+\sqrt{5}, \text { and } 1-\sqrt{5}.