Chapter 2.5, Problem 93E

To determine

Find all zeros of the given function.

Answer to Problem 93E

The real solution of $f\left(x\right)$ is $x=–\frac{3}{4}$

Explanation of Solution

Given:

The polynomial function

  $f\left(y\right)=4y^3+3y^2+8y+6$

Let consider the following polynomial function

  $f\left(y\right)=4y^3+3y^2+8y+6$

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

The leading coefficient is 4 and the constant term is 6.

Possible rational Zeros:

  $\frac{\text{factor of 6}}{\text{factor of 4}}=\frac{\pm 1,\pm 2,\pm 3,\pm 6}{\pm 1,\pm 2,\pm 4}$

There are lot of possibility of rational solution, so it is better to draw a graph, that some of possibility discarded.

  

Now from the graph, it is clear $x=–\frac{3}{4}$ is an only real solution of $f\left(x\right)$ .Since $f\left(x\right)$ is three degree polynomial, so other two solutions will be complex solutions.

Therefore the real solution of $f\left(x\right)$ is $x=–\frac{3}{4}$