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

Solving a Polynomial Equation

Solve 3x³ – 8x² – 8x + 8 = 0.

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 Possible rational roots = Factors of the constant term, 8 Factors of the leading coefficient, 3=±1,±2,±4,±8±1,±3=±1,±2,±4,±8,±13,±23,±43,±83\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=3x38x28x+8 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=23 x=\frac{2}{3} is a root by synthetic division.

23388824823763120Yes\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,(x23) 0,\left(x-\frac{2}{3}\right) is a factor of the original equation with the depressed equation 3x26x12=0. 3 x^{2}-6 x-12=0.

So

3x38x28x+8=0 Original equation (x23)(3x26x12)=0 Synthetic division (x23)=0 or 3x26x12=0 Zero-product property \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}

 

x=23x=(6)±(6)24(12)(3)2(3)Solve for x using the quadratic formula=6±36+1446=6±1806=6±(36)(5)6=6±656=1±5\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±5 1 \pm \sqrt{5} are also the x-intercepts of the graph.

The three real roots of the given equation are

23,1+5, and 15. \frac{2}{3}, 1+\sqrt{5}, \text { and } 1-\sqrt{5}.

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