Chapter 2.5, Problem 39E

(a)

To determine

Find the values of x for which $f\left(x\right)=0$ using the Rational Zero Test

Answer to Problem 39E

Possible rational zeros:

  $=\pm 1,\pm 3,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{1}{4},\pm \frac{3}{4},\pm \frac{1}{8},\pm \frac{3}{8},\pm \frac{1}{16},\pm \frac{3}{16},\pm \frac{1}{32},\pm \frac{3}{32}$

Explanation of Solution

Given:

  $f\left(x\right)=32x^3–52x^2+17x+3$

To find the values of x for which $f\left(x\right)=0$ using the Rational Zero Test.

The leading coefficient is 32 and the constant term is 3.

Possible rational zeros:

  $\frac{\text{factor of}3}{\text{factor of}32}=\frac{\pm 1,\pm 3}{\pm 1,\pm 2,\pm 4,\pm 8,\pm 16,\pm 32}$

  $=\pm 1,\pm 3,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{1}{4},\pm \frac{3}{4},\pm \frac{1}{8},\pm \frac{3}{8},\pm \frac{1}{16},\pm \frac{3}{16},\pm \frac{1}{32},\pm \frac{3}{32}$

(b)

To determine

Sketch the graph of the polynomial function.

Answer to Problem 39E

Graph is shown below.

Explanation of Solution

Given:

  $f\left(x\right)=32x^3–52x^2+17x+3$

Now draw the graph of $f\left(x\right),$ it shown below:

  

(b)

To determine

Determine all real zeros of f.

Answer to Problem 39E

All real zeros of fare $-\frac{1}{8},\frac{3}{4},1$ .

Explanation of Solution

Given:

  $f\left(x\right)=32x^3–52x^2+17x+3$

It is clear the polynomial will have only three solutions because of its degree is three, so the rational solution from graph of $f\left(x\right)$ are $-\frac{1}{8},\frac{3}{4},1$