Chapter 7.3, Problem 118E

a

To determine

To Find: The real zeros of $f(x)$

Answer to Problem 118E

the real zeros are the real zeros are $x=5,x=\frac{1}{3},x=\frac{-1}{2}$

Explanation of Solution

Given information: Polynomial function $f(x)=6x^3-29x^2-6x+5$

Concept Used:

Polynomial $f(x)$ has integer coefficients, every rational zero of $f(x)$ has the form

  $\text{Possible rational zero}=\frac{\text{factor of constant term}}{\text{factor of leading coefficient}}$

Synthetic division is used for long division of polynomials when dividing by the divisors of the form $x-k$

Calculation:

  $f(x)$ have leading coefficient $6$ and constant term $5$

Possible rational zeros $= \frac{\text{factor of }5}{\text{factor of }6}$

By Synthetic division $x=5$ is a rational zero

So, $f(x)$ factors as

  $\begin{array}{l}f(x)=(x-5)(6x^2+x-1)\\ f(x)=(x-5)(2x+1)(3x-1)\end{array}$

So, the real zeros are $x=5,x=\frac{1}{3},x=\frac{-1}{2}$

b

To determine

To graph: Sketch the graph of $f(x)=6x^3-29x^2-6x+5$ 5

Explanation of Solution

Given information: $f(x)=6x^3-29x^2-6x+5$ +5

Graph:

  

Interpretation: As indicated y the multiplicities of the zeros, the graph crosses the $x$ axis at

  $x=5,x=\frac{1}{3},x=\frac{-1}{2}$