Lesson Objectives

- Demonstrate an understanding of the Fundamental Theorem of Algebra
- Demonstrate an understanding of the Complete Factorization Theorem
- Demonstrate an understanding of the Number of Zeros Theorem
- Learn about the Conjugate Zeros Theorem
- Learn how to write a polynomial function given certain conditions

## How to Use the Conjugate Zeros Theorem to Write a Polynomial Function

In the last lesson, we learned about the Fundamental Theorem of Algebra, the Complete Factorization Theorem, and the Number of Zeros Theorem. We learned that a polynomial function of degree n ≥ 1 has exactly n zeros, provided that a zero of multiplicity k is counted k times. In this lesson, we will learn about the Conjugate Zeros Theorem, which tells us that non-real complex zeros come in conjugate pairs.

Example #1: Write a polynomial function of least degree, having only real coefficients.

Zeros: -1, 1 + 4i

From the conjugate zeros theorem, we know that if (1 + 4i) is a zero, then its conjugate (1 - 4i) is also a zero. This tells us that we will have three zeros.

Set up the polynomial function: $$f(x) = a(x - k_1)(x - k_2)(x - k_3)$$ Plug in for k

Zeros: 5 + i, 1/2

From the conjugate zeros theorem, we know that if (5 + i) is a zero, then its conjugate (5 - i) is also a zero. This tells us that we will have three zeros.

Set up the polynomial function: $$f(x) = a(x - k_1)(x - k_2)(x - k_3)$$ Plug in for k

Example #3: Factor f(x) into linear and irreducible quadratic factors with real coefficients. After this is done, continue the process to factor f(x) completely using linear factors with complex coefficients. $$f(x) = x^4 + 5x^2 - 24$$ Notice that f(x) is quadratic in form. $$f(x) = (x^2)^2 + 5(x^2) - 24$$ let u = x

Property #1: $$\overline{z + w} = \overline{z} + \overline{w}$$ $$\overline{z} = \overline{a + bi} = a - bi$$ $$\overline{w} = \overline{c + di} = c - di$$ $$\overline{z} + \overline{w} = (a - bi) + (c - di)$$ $$\overline{z + w} = \overline{(a + bi) + (c + di)}$$ $$= \overline{(a + c) + (bi + di)}$$ $$= \overline{(a + c) + (b + d)i}$$ $$= (a + c) - (b + d)i$$ $$= (a + c) - [bi + di]$$ $$= a + c -bi - di$$ $$= (a - bi) + (c - di)$$ Property #2: $$\overline{z \cdot w} = \overline{z} \cdot \overline{w}$$ $$\overline{z} \cdot \overline{w}$$ $$=\overline{a + bi} \cdot \overline{c + di}$$ $$=(a - bi)(c - di) = ac - adi - bci + bdi^2$$ $$=(ac + bdi^2) + (-adi - bci)$$ $$=(ac - bd) - (ad + bc)i$$ $$\overline{z \cdot w} = \overline{(a + bi)(c + di)}$$ $$=\overline{ac + adi + bci + bdi^2}$$ $$=\overline{ac + adi + bci - bd}$$ $$=\overline{(ac - bd) + (ad + bc)i}$$ $$=(ac - bd) - (ad + bc)i$$ Property #3: $$\overline{z^n} = \left(\overline{z}\right)^n$$ $$\overline{z^n} = \overline{z \cdot z \cdots z}$$ $$= \overline{z} \cdot \overline{z} \cdots \overline{z}$$ $$=(\overline{z})^n$$ Property #4: If z is a real number, then: $$z = a + bi, b = 0$$ $$z = a + 0i = a$$ $$\overline{z} = z$$ $$\overline{z} = \overline{a + 0i} = a - 0i = a$$

### Conjugate Zeros Theorem

If f(x) is a polynomial function with only real coefficients, then if (a + bi) is a zero (where a and b are both real numbers), its conjugate (a - bi) is also a zero. Since the proof for this theorem is quite involved, we will show it below. For now, we can use this theorem to write a polynomial function given certain conditions. Let's look at some examples.Example #1: Write a polynomial function of least degree, having only real coefficients.

Zeros: -1, 1 + 4i

From the conjugate zeros theorem, we know that if (1 + 4i) is a zero, then its conjugate (1 - 4i) is also a zero. This tells us that we will have three zeros.

Set up the polynomial function: $$f(x) = a(x - k_1)(x - k_2)(x - k_3)$$ Plug in for k

_{1}, k_{2}, and k_{3}, the order does not matter. $$f(x) = a(x - (-1))(x - (1 + 4i))(x - (1 - 4i))$$ $$f(x) = a(x + 1)(x - 1 - 4i)(x - 1 + 4i)$$ $$f(x) = a(x + 1)(x^2 - 2x + 17)$$ $$f(x)=a(x^3 - x^2 + 15x + 17)$$ Here, we were not given any specific information about a, the leading coefficient. In this case, we typically just use 1. $$f(x)=1(x^3 - x^2 + 15x + 17)$$ $$f(x)=x^3 - x^2 + 15x + 17$$ Notice that any nonzero multiple would also satisfy what we were given for zeros. $$f(x)=a(x^3 - x^2 + 15x + 17)$$ For example, we could have used an a-value of 3, which still satisfies the given zeros. $$f(x) = 3(x^3 - x^2 + 15x + 17)$$ $$f(x) = 3x^3 - 3x^2 + 45x + 51$$ Example #2: Write a polynomial function of least degree that has integer coefficients.Zeros: 5 + i, 1/2

From the conjugate zeros theorem, we know that if (5 + i) is a zero, then its conjugate (5 - i) is also a zero. This tells us that we will have three zeros.

Set up the polynomial function: $$f(x) = a(x - k_1)(x - k_2)(x - k_3)$$ Plug in for k

_{1}, k_{2}, and k_{3}, the order does not matter. $$f(x) = a\left(x -\frac{1}{2}\right)(x - (5 + i))(x - (5 - i))$$ $$f(x) = a\left(x -\frac{1}{2}\right)(x - 5 - i)(x - 5 + i)$$ $$f(x) = a\left(x -\frac{1}{2}\right)(x^2 - 10x + 26)$$ $$f(x) = a\left(x^3 - \frac{21}{2}x^2 + 31x - 13\right)$$ Since we want integer coefficients, we can set a = 2, which is the LCD of the denominators. $$f(x) = 2\left(x^3 - \frac{21}{2}x^2 + 31x - 13\right)$$ $$f(x) = 2x^3 - 21x^2 + 62x - 26$$ Other polynomial functions that are integer multiples would also work.### Linear and Quadratic Factors Theorem

At this point, we have learned by the Complete Factorization Theorem that a polynomial function can be factored completely into linear factors when we use complex numbers. If we are not using complex numbers, then we need to make a slight adjustment. We will say that a polynomial with real coefficients can be factored into a product of linear factors corresponding to the real zeros and irreducible quadratic factors that correspond to the non-real zeros. When we say "irreducible" quadratic, it just means it does not have any real zeros. One such example would be: $$f(x) = x^2 +4$$ The zeros here are ± 2i. This means we can only factor this polynomial using complex numbers. $$f(x) = x^2 + 4 = (x - 2i)(x + 2i)$$ Here is a quick proof of this theorem: $$(x - (a + bi))(x - (a - bi))$$ $$=(x - a - bi)(x - a + bi)$$ $$=((x - a) - bi)((x - a) + bi)$$ $$=(x - a)^2 - b^2i^2$$ $$=x^2 - 2ax + a^2 - b^2 \cdot -1$$ $$=x^2 - 2ax + a^2 + b^2$$ $$=x^2 - 2ax + (a^2 + b^2)$$ Notice that a^{2}+ b^{2}is a real number, this would be the constant term. Let's look at an example.Example #3: Factor f(x) into linear and irreducible quadratic factors with real coefficients. After this is done, continue the process to factor f(x) completely using linear factors with complex coefficients. $$f(x) = x^4 + 5x^2 - 24$$ Notice that f(x) is quadratic in form. $$f(x) = (x^2)^2 + 5(x^2) - 24$$ let u = x

^{2}and then factor: $$u^2 + 5u - 24 = (u + 8)(u - 3)$$ Replace u with x^{2}: $$f(x) = (x^2 + 8)(x^2 - 3)$$ $$x^2 - 3 = (x + \sqrt{3})(x - \sqrt{3})$$ x^{2}+ 8 is irreducible, it does not have any real zeros. $$f(x) = (x^2 + 8)(x^2 - 3)$$ $$= (x^2 + 8)(x + \sqrt{3})(x - \sqrt{3})$$ Now, we will complete the second part. First, let's find our complex zeros. We can do this with some basic factoring or use the quadratic formula. $$x^2 + 8 = x^2 - (-8)$$ $$= x^2 - (2i\sqrt{2})^2$$ $$= (x + 2i\sqrt{2})(x - 2i\sqrt{2})$$ Let's now return to our problem and give the complete factorization using complex numbers. $$f(x) = (x^2 + 8)(x + \sqrt{3})(x - \sqrt{3})$$ $$=(x + 2i\sqrt{2})(x - 2i\sqrt{2})(x + \sqrt{3})(x - \sqrt{3})$$### Proof for the Conjugate Zeros Theorem

In order to show a proof for the Conjugate Zeros Theorem, we will need a few properties of complex conjugates. In most math textbooks, the letters z and w are used for complex numbers, although this can vary based on the author. The notation that is normally used for the conjugate of a complex number is a bar that sits on top of the letter or number itself. For example: $$z = 5 + 7i$$ $$\overline{z} = \overline{5 + 7i} = 5 - 7i$$ For any complex numbers z and w:- $$\overline{z + w} = \overline{z} + \overline{w}$$
- $$\overline{z \cdot w} = \overline{z} \cdot \overline{w}$$
- $$\overline{z^n} = (\overline{z})^n$$
- if z is a real number, then:
- $$\overline{z} = z$$
- In other words, the complex conjugate of any real number is just the number

### Properties of Complex Conjugates Proofs

$$z = a + bi$$ $$w = c + di$$ $$i^2 = -1$$ Where a, b, c, and d are real numbers.Property #1: $$\overline{z + w} = \overline{z} + \overline{w}$$ $$\overline{z} = \overline{a + bi} = a - bi$$ $$\overline{w} = \overline{c + di} = c - di$$ $$\overline{z} + \overline{w} = (a - bi) + (c - di)$$ $$\overline{z + w} = \overline{(a + bi) + (c + di)}$$ $$= \overline{(a + c) + (bi + di)}$$ $$= \overline{(a + c) + (b + d)i}$$ $$= (a + c) - (b + d)i$$ $$= (a + c) - [bi + di]$$ $$= a + c -bi - di$$ $$= (a - bi) + (c - di)$$ Property #2: $$\overline{z \cdot w} = \overline{z} \cdot \overline{w}$$ $$\overline{z} \cdot \overline{w}$$ $$=\overline{a + bi} \cdot \overline{c + di}$$ $$=(a - bi)(c - di) = ac - adi - bci + bdi^2$$ $$=(ac + bdi^2) + (-adi - bci)$$ $$=(ac - bd) - (ad + bc)i$$ $$\overline{z \cdot w} = \overline{(a + bi)(c + di)}$$ $$=\overline{ac + adi + bci + bdi^2}$$ $$=\overline{ac + adi + bci - bd}$$ $$=\overline{(ac - bd) + (ad + bc)i}$$ $$=(ac - bd) - (ad + bc)i$$ Property #3: $$\overline{z^n} = \left(\overline{z}\right)^n$$ $$\overline{z^n} = \overline{z \cdot z \cdots z}$$ $$= \overline{z} \cdot \overline{z} \cdots \overline{z}$$ $$=(\overline{z})^n$$ Property #4: If z is a real number, then: $$z = a + bi, b = 0$$ $$z = a + 0i = a$$ $$\overline{z} = z$$ $$\overline{z} = \overline{a + 0i} = a - 0i = a$$

#### Skills Check:

Example #1

Write a polynomial function of least degree, having only real coefficients.

Zeros: 1 - 7i, 3

Please choose the best answer.

A

$$f(x)=4x^3 + 5x + 1$$

B

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

C

$$f(x)=x^3 + 9x^2 - 4x + 3$$

D

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

E

$$f(x)=x^3 - 5x^2 + 56x - 150$$

Example #2

Write a polynomial function of least degree, having only real coefficients.

Zeros: 2 + i, 9

Please choose the best answer.

A

$$f(x)=7x^3 + 4x^2 + 3x - 1$$

B

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

C

$$f(x)=2x^3 + 5x^2 + 4x - 1$$

D

$$f(x)=x^3 - 13x^2 + 41x - 45$$

E

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

Example #3

Factor f(x) into linear and irreducible quadratic factors with real coefficients. $$f(x) = x^3 - x^2 + 9x - 9$$ Hint: This polynomial can be factored using grouping.

Please choose the best answer.

A

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

B

$$f(x)=(x^2 - x - 2)(x - 9)$$

C

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

D

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

E

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

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