Chapter 2.6, Problem 19E

Solution

To find: all real zeros of the function.

The real zeros of the function $1,\frac{-1\pm \sqrt{17}}{2}$ .

Given information: 

The function $f\left(x\right)=4x^3-20x+16$ .

  

Calculation:

Consider the function

  $f\left(x\right)=4x^3-20x+16$

Take the GCF (Greatest Common Factor) out.

  $\begin{array}{c}f\left(x\right)=4x^3-20x+16\\ =4\left(x^3-5x+4\right)\end{array}$

Every rational zero of $f\left(x\right)$ has to divide the last coefficient, $4$ .

Possible zeros: $\pm 1,\pm 2,\pm 4$

Try out the $6$ values, and find that $f\left(1\right)=0$ so $x=1$ is a zero.

Divide $x^3-5x+4$ by $\left(x-1\right)$ . You can use synthetic division.

The quotient is ${\text{x}}^2+\text{x}-4$ and the remainder, as expected, is $0.$

  $\begin{array}{ccccc}1& 1& 0& -5& 4\\ & & 1& 1& -4\\ & 1& 1& -4& 0\end{array}$

Write the factoring of $f\left(x\right)$ so far

  $\begin{array}{c}f\left(x\right)=4\left(x^3-5x+4\right)\\ =4\left(x-1\right)\left(x^2+x-4\right)\end{array}$

Then $f\left(x\right)=0$ when $x=0$ or $x^2+x-4=0$ .

For the roots of the quadratic equation, use the quadratic formula.

Use $a=1,b=1,c=-4$

  $\begin{array}{c}x=\frac{-1\pm \sqrt{1^2-4\cdot 1\cdot (-4)}}{2\cdot 1}\\ x=\frac{-1\pm \sqrt{17}}{2}\end{array}$

Hence, the real zeros of the function $1,\frac{-1\pm \sqrt{17}}{2}$ .