Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
Chapter Review
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Exercise 12 Page 348

Recall the Factor Theorem.

Example Solution: y=x^3-4x^2-11x-6

Practice makes perfect
We want to write a polynomial function in standard form with the given zeros. To do so, we will use the Factor Theorem to write the factored form. We will then simplify it by applying the Distributive Property. Let's first recall the Factor Theorem. Factor Theorem The expressionx-a is a factor of a polynomial if and only if the valuea is a zero of the related polynomial function.We know that - 1, - 1, and 6 are zeros of our function. Therefore, we can write our polynomial function as the product of three factors. y = ( x-( - 1) ) ( x-( - 1) ) ( x-6) ⇕ y=(x+1)(x+1)(x-6) Finally, we can apply the Distributive Property to express the function in standard form.
y=(x+1)(x+1)(x-6)
y=( x(x+1)+1(x+1) ) (x-6)
â–Ľ
Distribute x & 1
y=( x^2+x+1(x+1) ) (x-6)
y=( x^2+x+x+1 ) (x-6)
y=( x^2+2x+1 ) (x-6)
y=x ( x^2+2x+1 ) -6 ( x^2+2x+1 )
â–Ľ
Distribute x & 6
y=x^3+2x^2+x -6( x^2+2x+1 )
y=x^3+2x^2+x-6x^2-12x-6
y=x^3-4x^2-11x-6

Checking Our Answer

Checking Our Answer
We can check our answer by substituting the given zeros for x. If the result is y=0, it means that the given numbers are actually zeros of the function and our answer is correct. Let's start by checking - 1.
y=x^3-4x^2-11x-6
y=( - 1)^3-4( - 1)^2-11( - 1)-6
â–Ľ
Simplify right-hand side
y=- 1-4(1)-11(- 1)-6
y=- 1-4-11(- 1)-6
y=- 1-4+11-6
y=0 âś“
We proved that - 1 is a zero of the function. Let's now check 6.
y=x^3-4x^2-11x-6
y= 6^3-4( 6)^2-11( 6)-6
â–Ľ
Simplify right-hand side
y=216-4(36)-11(6)-6
y=216-144-66-6
y=0 âś“
We found that 6 is also a zero. Since - 1 and 6 are zeros of the polynomial function, our answer is correct.