The Special Role of Zero in Factoring

Lesson 11: The Special Role of Zero in Factoring

Classwork Opening Exercise

Find all solutions to the equation (𝑎𝑎2+ 5𝑎𝑎+ 6)(𝑎𝑎2−3𝑎𝑎 −4)= 0.

Exercise 1

1. Find the solutions of (𝑎𝑎2−9)(𝑎𝑎2−16) = 0.

Example 1

Suppose we know that the polynomial equation 4𝑎𝑎3−12𝑎𝑎2+ 3𝑎𝑎+ 5 = 0 has three real solutions and that one of the factors of 4𝑎𝑎3−12𝑎𝑎2+ 3𝑎𝑎+ 5 is (𝑎𝑎 −1). How can we find all three solutions to the given equation?

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Exercises 2–5

2. Find the zeros of the following polynomial functions, with their multiplicities.

a. (𝑎𝑎) = (𝑎𝑎+ 1)(𝑎𝑎 −1)(𝑎𝑎2+ 1)

b. (𝑎𝑎) = (𝑎𝑎 −4)3(𝑎𝑎 −2)

c. (𝑎𝑎) = (2𝑎𝑎 −3)

d. (𝑎𝑎) = (3𝑎𝑎+ 4)100(𝑎𝑎 −17)

3. Find a polynomial function that has the following zeros and multiplicities. What is the degree of your polynomial?

Zero Multiplicity

2 3

−4 1

6 6

−8 10

4. Is there more than one polynomial function that has the same zeros and multiplicities as the one you found in Exercise 3?

5. Can you find a rule that relates the multiplicities of the zeros to the degree of the polynomial function?

Lesson 11: The Special Role of Zero in Factoring

Relevant Vocabulary Terms

In the definitions below, the symbol stands for the set of real numbers.

FUNCTION: A function is a correspondence between two sets, and , in which each element of is assigned to one and only one element of .

The set in the definition above is called the domain of the function. The range (or image) of the function is the subset of , denoted ( ), that is defined by the following property: 𝑦𝑦 is an element of ( ) if and only if there is an 𝑎𝑎 in such that (𝑎𝑎) =𝑦𝑦.

If (𝑎𝑎) =𝑎𝑎2 where 𝑎𝑎 can be any real number, then the domain is all real numbers (denoted ), and the range is the set of nonnegative real numbers.

POLYNOMIAL FUNCTION: Given a polynomial expression in one variable, a polynomial function in one variable is a function : such that for each real number 𝑎𝑎 in the domain, (𝑎𝑎) is the value found by substituting the number 𝑎𝑎 into all instances of the variable symbol in the polynomial expression and evaluating.

It can be shown that if a function : is a polynomial function, then there is some non-negative integer 𝑛𝑛 and collection of real numbers 𝑎𝑎0, 𝑎𝑎1, 𝑎𝑎2,… , 𝑎𝑎 with 𝑎𝑎 0 such that the function satisfies the equation

(𝑎𝑎) =𝑎𝑎 𝑎𝑎 +𝑎𝑎 1𝑎𝑎 1+ +𝑎𝑎1𝑎𝑎+𝑎𝑎0,

for every real number 𝑎𝑎 in the domain, which is called the standard form of the polynomial function. The function (𝑎𝑎) = 3𝑎𝑎3+ 4𝑎𝑎2+ 4𝑎𝑎+ 7, where 𝑎𝑎 can be any real number, is an example of a function written in standard form.

DEGREE OF A POLYNOMIAL FUNCTION: The degree of a polynomial function is the degree of the polynomial expression used to define the polynomial function.

The degree of (𝑎𝑎) = 8𝑎𝑎3+ 4𝑎𝑎2+ 7𝑎𝑎+ 6 is 3, but the degree of (𝑎𝑎) = (𝑎𝑎+ 1)2−(𝑎𝑎 −1)2 is 1 because when is put into standard form, it is (𝑎𝑎) = 4𝑎𝑎.

CONSTANT FUNCTION: A constant function is a polynomial function of degree 0. A constant function is of the form (𝑎𝑎) =𝑐𝑐, for a constant 𝑐𝑐.

LINEAR FUNCTION: A linear function is a polynomial function of degree 1. A linear function is of the form (𝑎𝑎) =𝑎𝑎𝑎𝑎+𝑏𝑏, for constants 𝑎𝑎 and 𝑏𝑏 with 𝑎𝑎 0.

QUADRATIC FUNCTION: A quadratic function is a polynomial function of degree 2. A quadratic function is in standard form if it is written in the form (𝑎𝑎) =𝑎𝑎𝑎𝑎2+𝑏𝑏𝑎𝑎+𝑐𝑐, for constants 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 with 𝑎𝑎 0 and any real number 𝑎𝑎.

CUBIC FUNCTION: A cubic function is a polynomial function of degree 3. A cubic function is of the form (𝑎𝑎) =𝑎𝑎𝑎𝑎3+𝑏𝑏𝑎𝑎2+𝑐𝑐𝑎𝑎+𝑑𝑑, for constants 𝑎𝑎, 𝑏𝑏, 𝑐𝑐,𝑑𝑑 with 𝑎𝑎 0.

ZEROS OR ROOTS OF A FUNCTION: A zero (or root) of a function : is a number 𝑎𝑎 of the domain such that (𝑎𝑎) = 0. A zero of a function is an element in the solution set of the equation (𝑎𝑎) = 0.

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Problem Set

For Problems 1–4, find all solutions to the given equations.

1. (𝑎𝑎 −3)(𝑎𝑎+ 2) = 0

2. (𝑎𝑎 −5)(𝑎𝑎+ 2)(𝑎𝑎+ 3) = 0

3. (2𝑎𝑎 −4)(𝑎𝑎+ 5) = 0

4. (2𝑎𝑎 −2)(3𝑎𝑎+ 1)(𝑎𝑎 −1) = 0

5. Find four solutions to the equation (𝑎𝑎2−9)(𝑎𝑎 −16) = 0.

6. Find the zeros with multiplicity for the function (𝑎𝑎) = (𝑎𝑎3−8)(𝑎𝑎 −4𝑎𝑎3).

7. Find two different polynomial functions that have zeros at 1, 3, and 5 of multiplicity 1.

8. Find two different polynomial functions that have a zero at 2 of multiplicity 5 and a zero at −4 of multiplicity 3.

9. Find three solutions to the equation (𝑎𝑎2−9)(𝑎𝑎3−8) = 0.

10. Find two solutions to the equation (𝑎𝑎3−64)(𝑎𝑎 −1) = 0.

11. If , , , 𝑠𝑠 are nonzero numbers, find the solutions to the equation ( 𝑎𝑎+ )( 𝑎𝑎+𝑠𝑠) = 0 in terms of , , , 𝑠𝑠.

Use the identity 𝑎𝑎2− 𝑏𝑏2= (𝑎𝑎 − 𝑏𝑏)(𝑎𝑎+𝑏𝑏) to solve the equations given in Problems 12–13.

12. (3𝑎𝑎 −2)2= (5𝑎𝑎+ 1)2

13. (𝑎𝑎+ 7)2= (2𝑎𝑎+ 4)2 Lesson Summary

Given any two polynomial functions and , the solution set of the equation (𝑎𝑎) (𝑎𝑎) = 0 can be quickly found by solving the two equations (𝑎𝑎) = 0 and (𝑎𝑎) = 0 and combining the solutions into one set.

The number 𝑎𝑎 is a zero of a polynomial function with multiplicity if the factored form of contains (𝑎𝑎 − 𝑎𝑎) .

Lesson 11: The Special Role of Zero in Factoring

14. Consider the polynomial function (𝑎𝑎) =𝑎𝑎3+ 2𝑎𝑎2+ 2𝑎𝑎 −5.

a. Divide by the divisor (𝑎𝑎 −1) and rewrite in the form (𝑎𝑎) = (divisor)(quotient) + remainder.

b. Evaluate (1).

15. Consider the polynomial function (𝑎𝑎) =𝑎𝑎 −3𝑎𝑎 + 4𝑎𝑎3−12𝑎𝑎2+𝑎𝑎 −3.

a. Divide by the divisor (𝑎𝑎 −3) and rewrite in the form (𝑎𝑎) = (divisor)(quotient) + remainder. b. Evaluate (3).

16. Consider the polynomial function (𝑎𝑎) =𝑎𝑎 + 2𝑎𝑎3−2𝑎𝑎2−3𝑎𝑎+ 2.

a. Divide by the divisor (𝑎𝑎+ 2) and rewrite in the form (𝑎𝑎) = (divisor)(quotient) + remainder. b. Evaluate (−2).

17. Consider the polynomial function (𝑎𝑎) =𝑎𝑎 +𝑎𝑎 − 𝑎𝑎 − 𝑎𝑎 +𝑎𝑎3+𝑎𝑎2− 𝑎𝑎 −1.

a. Divide by the divisor (𝑎𝑎+ 1) and rewrite in the form (𝑎𝑎) = (divisor)(quotient) + remainder. b. Evaluate (−1).

18. Make a conjecture based on the results of Problems 14—17.

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