Lesson Objectives

- Learn about the Fundamental Theorem of Algebra
- Learn about the Complete Factorization Theorem
- Learn about the Number of Zeros Theorem
- Learn how to write a polynomial function given zeros and a point

## How to Write a Polynomial Function Given Real Zeros and a Point

In this lesson, we will learn about the Fundamental Theorem of Algebra, the Complete Factorization Theorem, and the Number of Zeros Theorem. We will then learn how to write a polynomial function when we are given the real zeros and a point.

As another example, consider the function g(x). $$g(x) = (x + 1)(x - 3)^2$$ $$= (x - (-1))(x - 3)(x - 3)$$ Here, we have the following zeros: $$-1, 3 (\text{multiplicity }2)$$ Again, we can see that the zero of 3 occurs twice. We can use the theorems presented to write a polynomial function given the real zeros and a point. Let's look at some examples.

Example #1: Find a function f defined by a polynomial of degree 3 that satisfies the given conditions. $$\text{Zeros of: }2, 5, 1$$ $$f(0) = -20$$ Let's set up a polynomial function: $$f(x)=a(x - k_1)(x - k_2)(x - k_3)$$ Now, let's plug in for the zeros. The order does not matter: $$f(x)=a(x - 2)(x - 5)(x - 1)$$ $$f(x)=a(x^3 - 8x^2 + 17x - 10)$$ To find a, use the fact that f(0) is -20. $$-20 = a(0^3 - 8(0)^2 + 17(0) - 10)$$ $$-20=a(-10)$$ $$a=\frac{-20}{-10}=2$$ We know that a is 2: $$f(x)=2(x^3 - 8x^2 + 17x - 10)$$ $$f(x)=2x^3 - 16x^2 + 34x - 20$$ Example #2: Find a function f defined by a polynomial of degree 3 that satisfies the given conditions. $$\text{Zeros of: }{-1}, 2, 4$$ $$f(0) = 12$$ Let's set up a polynomial function: $$f(x)=a(x - k_1)(x - k_2)(x - k_3)$$ Now, let's plug in for the zeros. The order does not matter: $$f(x)=a(x - (-1))(x - 2)(x - 4)$$ $$f(x)=a(x + 1)(x - 2)(x - 4)$$ $$f(x)=a(x^3 - 5x^2 + 2x + 8)$$ To find a, use the fact that f(0) is 12. $$12 = a(0^3 - 5(0)^2 + 2(0) + 8)$$ $$12=a(8)$$ $$a=\frac{12}{8}=\frac{3}{2}$$ We know that a is 3/2: $$f(x)=\frac{3}{2}(x^3 - 5x^2 + 2x + 8)$$ $$f(x)=\frac{3}{2}x^3 - \frac{15}{2}x^2 + 3x + 12$$ Example #3: Find a function f defined by a polynomial of degree 4 that satisfies the given conditions. $$\text{Zeros of: }{-1}, 2, 8$$ Where -1 is a zero of multiplicity 2. $$f(1) = 140$$ Let's set up a polynomial function: $$f(x)=a(x - k_1)(x - k_2)(x - k_3)(x - k_4)$$ Now, let's plug in for the zeros. The order does not matter: $$f(x)=a(x - (-1))(x - (-1))(x - 2)(x - 8)$$ $$f(x)=a(x + 1)(x + 1)(x - 2)(x - 8)$$ $$f(x)=a(x + 1)^2(x - 2)(x - 8)$$ $$f(x)=a(x^4 - 8x^3 - 3x^2 + 22x + 16)$$ To find a, use the fact that f(1) is 140. $$140 = a(1^4 - 8(1)^3 - 3(1)^2 + 22(1) + 16)$$ $$140 = a(1 - 8 - 3 + 22 + 16)$$ $$140 = a(28)$$ $$a=\frac{140}{28}=5$$ We know that a is 5: $$f(x)=5(x^4 - 8x^3 - 3x^2 + 22x + 16)$$ $$f(x)=5x^4 - 40x^3 - 15x^2 + 110x + 80$$

### Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra tells us that if f(x) is a polynomial function of degree n with complex coefficients: $$f(x) = a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0$$ Where: $$n ≥ 1, a_n ≠ 0$$ Then the equation f(x) = 0 has at least one complex zero. Since all real numbers are also complex numbers, this theorem applies to any polynomial function with real coefficients.### Complete Factorization Theorem

We can use the Fundamental Theorem of Algebra with the factor theorem to show that a polynomial function can be factored completely into linear factors. If f(x) is a polynomial function of degree n, where n is 1 or more, then there exist complex numbers: $$a, k_1, k_2, ... , k_n$$ Where: $$a ≠ 0$$ Such that: $$f(x) = a(x - k_1)(x - k_2) \cdots (x - k_n)$$#### Proof

From the Fundamental Theorem of Algebra, if f(x) is of degree n, where n is 1 or more, then there is some number k_{1}such that: $$f(k_1) = 0$$ Using the Factor Theorem: $$f(x) = (x - k_1)q_1(x)$$ If q_{1}(x) is of degree 1 or more, then we can apply the same thought process. There will be some number k_{2}such that: $$q_1(k_2) = 0$$ Using the Factor Theorem: $$f(x) = (x - k_1)(x - k_2)q_2(x)$$ Since f(x) is of degree n, we repeat this process n times to obtain our model below: $$f(x) = a(x - k_1)(x - k_2) \cdots (x - k_n)$$ Each of these factors gives us a zero for f(x), so f(x) of degree n, where n is 1 or more, has n zeros, given by: $$k_1, k_2, ..., k_n$$ Theses zeros need not all be different. The number of times a factor appears in the factored form of the equation is known as the multiplicity. This result leads us to the Number of Zeros theorem.### Number of Zeros Theorem

A polynomial function of degree n ≥ 1 has at most n distinct zeros. Additionally, this polynomial function of degree n ≥ 1 will have exactly n zeros, provided that a zero of multiplicity k is counted k times. For example, suppose we have the following function. $$f(x) = (x + 2)^2$$ This function is of degree 2, so how many zeros are we going to have? $$f(x) = (x + 2)^2$$ $$f(-2) = 0$$ It would appear at first glance that this function only has one zero but notice that it is a zero that occurs twice. $$f(x) = (x + 2)^2 = (x + 2)(x + 2)$$ $$= (x - (-2))(x - (-2))$$ This is what we mean by a zero of multiplicity 2.As another example, consider the function g(x). $$g(x) = (x + 1)(x - 3)^2$$ $$= (x - (-1))(x - 3)(x - 3)$$ Here, we have the following zeros: $$-1, 3 (\text{multiplicity }2)$$ Again, we can see that the zero of 3 occurs twice. We can use the theorems presented to write a polynomial function given the real zeros and a point. Let's look at some examples.

Example #1: Find a function f defined by a polynomial of degree 3 that satisfies the given conditions. $$\text{Zeros of: }2, 5, 1$$ $$f(0) = -20$$ Let's set up a polynomial function: $$f(x)=a(x - k_1)(x - k_2)(x - k_3)$$ Now, let's plug in for the zeros. The order does not matter: $$f(x)=a(x - 2)(x - 5)(x - 1)$$ $$f(x)=a(x^3 - 8x^2 + 17x - 10)$$ To find a, use the fact that f(0) is -20. $$-20 = a(0^3 - 8(0)^2 + 17(0) - 10)$$ $$-20=a(-10)$$ $$a=\frac{-20}{-10}=2$$ We know that a is 2: $$f(x)=2(x^3 - 8x^2 + 17x - 10)$$ $$f(x)=2x^3 - 16x^2 + 34x - 20$$ Example #2: Find a function f defined by a polynomial of degree 3 that satisfies the given conditions. $$\text{Zeros of: }{-1}, 2, 4$$ $$f(0) = 12$$ Let's set up a polynomial function: $$f(x)=a(x - k_1)(x - k_2)(x - k_3)$$ Now, let's plug in for the zeros. The order does not matter: $$f(x)=a(x - (-1))(x - 2)(x - 4)$$ $$f(x)=a(x + 1)(x - 2)(x - 4)$$ $$f(x)=a(x^3 - 5x^2 + 2x + 8)$$ To find a, use the fact that f(0) is 12. $$12 = a(0^3 - 5(0)^2 + 2(0) + 8)$$ $$12=a(8)$$ $$a=\frac{12}{8}=\frac{3}{2}$$ We know that a is 3/2: $$f(x)=\frac{3}{2}(x^3 - 5x^2 + 2x + 8)$$ $$f(x)=\frac{3}{2}x^3 - \frac{15}{2}x^2 + 3x + 12$$ Example #3: Find a function f defined by a polynomial of degree 4 that satisfies the given conditions. $$\text{Zeros of: }{-1}, 2, 8$$ Where -1 is a zero of multiplicity 2. $$f(1) = 140$$ Let's set up a polynomial function: $$f(x)=a(x - k_1)(x - k_2)(x - k_3)(x - k_4)$$ Now, let's plug in for the zeros. The order does not matter: $$f(x)=a(x - (-1))(x - (-1))(x - 2)(x - 8)$$ $$f(x)=a(x + 1)(x + 1)(x - 2)(x - 8)$$ $$f(x)=a(x + 1)^2(x - 2)(x - 8)$$ $$f(x)=a(x^4 - 8x^3 - 3x^2 + 22x + 16)$$ To find a, use the fact that f(1) is 140. $$140 = a(1^4 - 8(1)^3 - 3(1)^2 + 22(1) + 16)$$ $$140 = a(1 - 8 - 3 + 22 + 16)$$ $$140 = a(28)$$ $$a=\frac{140}{28}=5$$ We know that a is 5: $$f(x)=5(x^4 - 8x^3 - 3x^2 + 22x + 16)$$ $$f(x)=5x^4 - 40x^3 - 15x^2 + 110x + 80$$

#### Skills Check:

Example #1

Write a polynomial function of degree 3 that satisfies the given conditions. $$\text{Zeros}: 4, 2, -5$$ $$f(0)=40$$

Please choose the best answer.

A

$$f(x)=-7x^3 + 5x^2 - 8x + 1$$

B

$$f(x)=x^3 - 3x^2 + 9x + 1$$

C

$$f(x)=x^3 - x^2 - 22x + 40$$

D

$$f(x)=-2x^3 - 3x^2 - 11x + 6$$

E

$$f(x)=2x^3 - 3x^2 - 11x + 6$$

Example #2

Write a polynomial function of degree 3 that satisfies the given conditions.

Please choose the best answer. $$\text{Zeros}: 3 \hspace{.2em}\text{mult.}\hspace{.2em}2, 1$$ $$f(0)=-9$$

A

$$f(x)=-5x^3 + 2x^2 + 3x - 4$$

B

$$f(x)=-x^3 + 5x^2 + 2x + 1$$

C

$$f(x)=x^3 + 7x^2 + 5$$

D

$$f(x)=x^3 - 7x^2 + 15x - 9$$

E

$$f(x)=2x^3 - 3x^2 + 5x + 1$$

Example #3

Write a polynomial function of degree 3 that satisfies the given conditions. $$\text{Zeros}: -1, 2, 4$$ $$f(0)=8$$

Please choose the best answer.

A

$$f(x)=3x^3 - 8x^2 + 16x + 80$$

B

$$f(x)=\frac{1}{2}x^3 - \frac{5}{2}x^2 + x + 4$$

C

$$f(x)=2x^3 - 10x^2 + 20x + 16$$

D

$$f(x)=x^3 - 5x^2 + 2x + 8$$

E

$$f(x)=2x^3 + 10x^2 - 20x - 40$$

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