Pearson Algebra 2 Common Core, 2011
PA
Pearson Algebra 2 Common Core, 2011 View details
End-of-Course Assessment

Exercise 11 Page 965

Use the zeros of the function to write the equation in factored form.

B

Practice makes perfect

We are given the graph of a polynomial function with a degree of 4. Let's take a look at it by considering the points that the graph intersects the x-axis.

We want to write an equation for this graph. To do so, we will use the zeros of the graph. As we can see from the graph, our zeros lie at ( -2,0), ( 0,0), ( 1,0), and ( 3,0). By using these points we can write our equation in factored form. y= a(x-( -2)) * (x- 0) * (x- 1) * (x- 3) Since all the leading coefficients are 1 in the given options, we can set a equal to 1. We will now simplify this product. Let's begin with multiplying the first two binomials together, (x-( -2)) and (x- 0).

y=1*(x-(-2)) * (x-0) * (x-1) * (x-3)
y=1*(x-(-2)) * (x-0) * (x-1) * (x-3)
y=(x+2) * (x-0) * (x-1) * (x-3)
y=(x+2) * x * (x-1) * (x-3)
y=(x^2+2x) * (x-1) * (x-3)

Next we will multiply the second pair of binomials, (x- 1) and (x- 3).

y=(x^2+2x) * (x-1) * (x-3)
y=(x^2+2x) * (x^2-3x-x+3)
y=(x^2+2x) * (x^2-4x+3)

Finally, we can multiply those two expressions by distributing (x^2+2x) to every term inside the second parentheses.

y=(x^2+2x) * (x^2-4x+3)
y=(x^2+2x) * x^2 + (x^2+2x) * (- 4x) + (x^2+2x)* 3
y=x^2 * x^2 +2x * x^2 + (x^2+2x) * (- 4x) + (x^2+2x)* 3
y=x^2 * x^2 +2x * x^2 + x^2 * (- 4x) +2x * (- 4x) + (x^2+2x)* 3
y=x^2 * x^2 +2x * x^2 + x^2 * (- 4x) +2x * (- 4x) + x^2* 3 +2x * 3
y=x^4+2x^3 -4x^3-8x^2+3x^2+6x
y=x^4-2x^3-5x^2+6x

Note that the equation that we found corresponds to option B.