Big Ideas Math Algebra 2, 2014
BI
Big Ideas Math Algebra 2, 2014 View details
8. Analyzing Graphs of Polynomial Functions
Continue to next subchapter

Exercise 13 Page 216

Use the Zero Product Property to find the zeros of the polynomial function.

Practice makes perfect

We want to graph the given polynomial function. h(x)=(x-3)(x^2+x+1) Let's start by finding the zeros of the function.

Zeros of the Function

To do so, we need to find the values of x for which h(x)=0. h(x)=0 ⇔ (x-3)(x^2+x+1)=0 Since the function is already written in factored form, we will use the Zero Product Property. Note that the second part does not factor out anymore and has no real zeros, so we will omit it.
x-3=0
x=3
We found that the zero of the function is 3.

Graph

To draw the graph of the function, we will find some additional points and consider the end behavior. Let's use a table to find additional points.

x (x-3)(x^2+x+1) h(x)=(x-3)(x^2+x+1)
- 1 ( - 1-3)(( - 1)^2+( - 1)+1) - 4
0 ( 0-3)( 0^2+ 0+1) - 3
1 ( 1-3)( 1^2+ 1+1) - 6
2 ( 2-3)( 2^2+ 2+1) -7
The points ( - 1, -4), ( 0, -3), ( 1, -6), and ( 2, -7) are on the graph of the function. Finally, let's apply the Distributive Property. This will simplify the equation and determine the leading coefficient and degree of the polynomial function.
h(x)=(x-3)(x^2+x+1)
â–Ľ
Multiply
h(x)=x^2(x-3)+x(x-3)+1(x-3)
h(x)=x^3-3x^2+x(x-3)+1(x-3)
h(x)=x^3-3x^2+x^2-3x+1(x-3)
h(x)=x^3-3x^2+x^2-3x+x-3
h(x)=x^3-2x^2-2x-3
We can see now that the leading coefficient is 1, which is a positive number. Also, the degree is 3, which is an odd number. Therefore, the end behavior is down and up. Now, let's draw the graph!