Chapter 7.3, Problem 117E

a

To determine

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

Answer to Problem 117E

the real zeros are $x=3,x=-4,x=\frac{-3}{2}$

Explanation of Solution

Given information: Polynomial function $f(x)=2x^3+5x^2-21x-36$

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 $2$ and constant term $-36$

Possible rational zeros $=\frac{\text{factor of}-36}{\text{factor of }2}$

By Synthetic division $x=3$ is a rational zero

So, $f(x)$ factors as

  $\begin{array}{l}f(x)=(x-3)(2x^2+11x+12)\\ f(x)=(x-3)(x+4)(2x+3)\end{array}$

So, the real zeros are $x=3,x=-4,x=\frac{-3}{2}$

b

To determine

To graph: Sketch the graph of $f(x)=2x^3+5x^2-21x-36$

Explanation of Solution

Given information: $f(x)=2x^3+5x^2-21x-36$

Graph:

  

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

  $x=3,x=-4,x=\frac{-3}{2}$