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.

### 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 k1, k2, and k3, 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 k1, k2, and k3, 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 a2 + b2 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 = x2 and then factor: $$u^2 + 5u - 24 = (u + 8)(u - 3)$$ Replace u with x2: $$f(x) = (x^2 + 8)(x^2 - 3)$$ $$x^2 - 3 = (x + \sqrt{3})(x - \sqrt{3})$$ x2 + 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:
1. $$\overline{z + w} = \overline{z} + \overline{w}$$
2. $$\overline{z \cdot w} = \overline{z} \cdot \overline{w}$$
3. $$\overline{z^n} = (\overline{z})^n$$
4. if z is a real number, then:
• $$\overline{z} = z$$
• In other words, the complex conjugate of any real number is just the number
Now that we have the properties that we need, let's begin our proof for the Conjugate Zeros Theorem. $$f(x) = a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0$$ Where all the coefficients are real numbers. If the complex number z is a zero of f(x): $$f(z) = 0$$ $$a_nz^n + a_{n - 1}z^{n - 1} + \cdots + a_1z + a_0 = 0$$ If we take the conjugate of both sides we obtain the following: $$\overline{a_nz^n + a_{n - 1}z^{n - 1} + \cdots + a_1z + a_0} = \overline{0}$$ Using property #1 from the properties of complex conjugates above: $$\overline{a_nz^n} + \overline{a_{n - 1}z^{n - 1}} + \cdots + \overline{a_1z} + \overline{a_0} = \overline{0}$$ Using property #2 from the properties of complex conjugates above: $$\overline{a_n} \cdot \overline{z^n} + \overline{a_{n - 1}} \cdot \overline{z^{n - 1}} + \cdots + \overline{a_1} \cdot \overline{z} + \overline{a_0} = \overline{0}$$ Using property #3 from the properties of complex conjugates above: $$\overline{a_n} \cdot (\overline{z})^n + \overline{a_{n - 1}} \cdot (\overline{z})^{n - 1} + \cdots + \overline{a_1} \cdot (\overline{z}) + \overline{a_0} = \overline{0}$$ Using property #4 from the properties of complex conjugates above: $$a_n \cdot (\overline{z})^n + a_{n - 1} \cdot (\overline{z})^{n - 1} + \cdots + a_1 \cdot (\overline{z}) + a_0 = 0$$ From this result, we can state that: $$f(\overline{z}) = 0$$

### 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

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

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.

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