17Calculus Precalculus - Complex Roots of Polynomials

17Calculus

Polynomials With Complex Zeroes/Roots

For most of calculus, you will be working with real numbers. However, for precalculus it is good to make sure you know how to work with complex numbers since some other branches of mathematics use complex numbers. Additionally, you will definitely need to know about complex numbers in many areas of engineering.

This page discusses polynomials with complex roots. On this page, we will limit our discussion to polynomials with real coefficients. Surprisingly, polynomials with real coefficients can have complex roots. However, since the coefficients are all real, the complex roots are in a special form. First, let's have a short review of complex numbers.

Quick Review of Complex Numbers

 For a more indepth tutorial about complex numbers, see our complex numbers section, starting with this page. Note: In electrical engineering you may see $$j$$ instead of $$i$$ to represent $$\sqrt{-1}$$ since $$i$$ is the variable associated with current. In this section, we will stick with the mathematical form $$i$$.

1. Complex numbers are written in the form $$a+bi$$ where $$a$$ and $$b$$ are real numbers. $$a$$ represents the real part and $$bi$$ represents the imaginary part where $$i=\sqrt{-1}$$.

2. The complex conjugate of $$a+bi$$ is $$a-bi$$. To get the complex conjugate, you change the sign in front of the complex term.

3. When adding two complex numbers, just add the real parts together and add the complex parts together. For example, $$(1+5i) + (2+4i) = (1+2) + (5i+4i) = 3 + (5+4)i = 3+9i$$

4. When multiplying two complex numbers, just distribute like we do with polynomials. For example, $$(1+5i)(2+4i) = (1)(2) + (1)(4i) + (5i)(2) + (5i)(4i) = 3 + 4i + 10i + 20i^2 = 3 +14i + 20(-1) = -17 +14i$$
Notice that we used the idea that since $$i=\sqrt{-1}$$, $$i^2 = (\sqrt{-1})^2 = -1$$

5. The importance of the complex conjugate comes from the result that, when you multiply a complex number and it's conjugate together, you end up with a real number. For example, $$(1+5i)(1-5i) = (1)(1) + (1)(-5i) + (5i)(1) + (5i)(-5i) = 1 - 5i + 5i -25(-1) = 26$$

Okay, that should be enough to get you started for the next section on polynomial zeroes.

Polynomial Zeroes

1. Polynomials with real coefficients have the same number of roots as the highest power.
2. If the polynomial has complex zeroes, they will appear in complex conjugate pairs.

Let's expand a bit on these two important facts about polynomial zeros.

1. Polynomials with real coefficients have the same number of roots as the highest power. However, not all the roots are guaranteed to be real. They may be all real, all complex or a combination of real and complex numbers.

2. If the polynomial has complex zeroes, they will appear in complex conjugate pairs, i.e. if you have one complex root $$a+bi$$ then you are guaranteed to have a second complex root $$a-bi$$. This is true only for polynomials with real coefficients.

So, why do the complex roots appear in complex conjugate pairs? Here is a video clip that shows graphically what is going on in third degree polynomials. Once you understand this, you will get able to extend your understanding to higher order polynomials.

ExamSolutions - Complex Numbers: Roots of a cubic equation [3mins-15secs]

video by ExamSolutions

Finding complex roots is very similar to finding real roots except that, if a complex root exists, you will have to solve the square root of a negative number. So in that case, you just simplify the form using $$i$$. For example, say you have $$x^2+4=0$$. Solving this we get $$x^2+4=0 \to x^2=-4 \to x = \pm\sqrt{-4} \to x = \pm \sqrt{(4)(-1)} \to x = \pm 2\sqrt{-1} \to x= \pm 2i$$.

FAQ: When Do I Use $$\pm$$?

This is a very common question I get a lot as an instructor. It can be confusing to know when to use $$\pm$$. The short answer that I usually give is, use $$\pm$$ whenever you introduce a square root (or an even root) into an equation. Otherwise, do not add it to the equation. Here is an example.

If you are given the equation $$y=\sqrt{x}$$. This already has a square root in it, so you do not add $$\pm$$. In fact, if you did add it, you would change the equation to something different.

However, if you are asked to solve $$x^2=4$$, then $$x = \pm \sqrt{4} = \pm 2$$ is the correct answer, since $$2$$ and $$-2$$ both solve the original equation. Notice that you added a square root to the equation by taking the square root of both sides. The implied (missing) step is $$x^2=4 \to \sqrt{x^2} = \pm \sqrt{4}$$. Since we have an equation (with an equal sign), not an inequality (with $$\lt$$ or $$\geq$$, for example), the $$\pm$$ can go on either side and you only need one. Using two is not necessarily incorrect but it is redundant.

Okay, time for the practice problems.

Practice

Unless otherwise instructed, solve the equations or find the roots of the polynomial functions.

Write a polynomial given the roots $$3, 2+i, 2-i$$

Problem Statement

Write a polynomial given the roots $$3, 2+i, 2-i$$

$$y=x^3-7x^2+17x-15$$

Problem Statement

Write a polynomial given the roots $$3, 2+i, 2-i$$

Solution

Brian McLogan - 4204 video solution

video by Brian McLogan

$$y=x^3-7x^2+17x-15$$

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Find the zeroes of $$P(x) = x^3-4x^2+x-4$$ given $$i$$ is a zero.

Problem Statement

Find all the zeroes of $$P(x) = x^3-4x^2+x-4$$ given that $$i$$ is a zero.

$$4, i, -i$$

Problem Statement

Find all the zeroes of $$P(x) = x^3-4x^2+x-4$$ given that $$i$$ is a zero.

Solution

PatrickJMT - 4206 video solution

video by PatrickJMT

$$4, i, -i$$

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Find all the roots of $$z^3 + z^2 - 7z -15$$ given that $$z = -2+i$$ is one of the roots.

Problem Statement

Find all the roots of $$z^3 + z^2 - 7z -15$$ given that $$z = -2+i$$ is one of the roots.

$$-2+i, -2-i, 3$$

Problem Statement

Find all the roots of $$z^3 + z^2 - 7z -15$$ given that $$z = -2+i$$ is one of the roots.

Solution

Mrs O'Gram's Math - 4209 video solution

$$-2+i, -2-i, 3$$

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Find all the roots of $$g(x) = x^3 - 7x^2 - x + 87$$ given that $$x = 5 + 2i$$ is one of the roots.

Problem Statement

Find all the roots of $$g(x) = x^3 - 7x^2 - x + 87$$ given that $$x = 5 + 2i$$ is one of the roots.

Solution

Brian McLogan - 4245 video solution

video by Brian McLogan

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

Problem Statement

Find the zeroes of $$f(x) = x^3 + 2x$$

Solution

Brian McLogan - 4244 video solution

video by Brian McLogan

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$$p(x) = x^3 - x^2 +x - 1$$

Problem Statement

Find the zeroes of $$p(x) = x^3 - x^2 +x - 1$$

$$1, i, -i$$

Problem Statement

Find the zeroes of $$p(x) = x^3 - x^2 +x - 1$$

Solution

Mrs O'Gram's Math - 4208 video solution

$$1, i, -i$$

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$$x^3 - 4x^2 + x + 26 = 0$$

Problem Statement

Find the zeroes of $$x^3 - 4x^2 + x + 26 = 0$$

Solution

ExamSolutions - 4246 video solution

video by ExamSolutions

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$$x^3 - 7x^2 + 17x -15 = 0$$

Problem Statement

Find the zeroes of $$x^3 - 7x^2 + 17x -15 = 0$$

Solution

ExamSolutions - 4247 video solution

video by ExamSolutions

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$$f(x) = x^3 - 5x^2 + 4x - 20$$

Problem Statement

Find the zeroes of $$f(x) = x^3 - 5x^2 + 4x - 20$$

Solution

Brian McLogan - 4249 video solution

video by Brian McLogan

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$$5x^3 - 10x^2 +4x - 8 = 0$$

Problem Statement

Solve $$5x^3 - 10x^2 +4x - 8 = 0$$

$$2, \pm 2i/\sqrt{5}$$

Problem Statement

Solve $$5x^3 - 10x^2 +4x - 8 = 0$$

Solution

Factored form with all real numbers $$(x-2)(5x^2+4) = 0$$

The Organic Chemistry Tutor - 4216 video solution

$$2, \pm 2i/\sqrt{5}$$

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$$x^4 + 8x^2 -9$$

Problem Statement

Factor $$x^4 + 8x^2 -9$$

$$(x+1)(x-1)(x+3i)(x-3i)$$

Problem Statement

Factor $$x^4 + 8x^2 -9$$

Solution

The zeroes are $$1, -1, 3i, -3i$$

The Organic Chemistry Tutor - 4213 video solution

$$(x+1)(x-1)(x+3i)(x-3i)$$

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$$x^4 - 5x^2 - 36$$

Problem Statement

Factor $$x^4 - 5x^2 - 36$$

$$(x+3)(x-3)(x+2i)(x-2i)$$

Problem Statement

Factor $$x^4 - 5x^2 - 36$$

Solution

The zeroes are $$-3, 3, 2i, -2i$$

The Organic Chemistry Tutor - 4214 video solution

$$(x+3)(x-3)(x+2i)(x-2i)$$

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$$f(x) = x^4 + x^3 + 5x^2 - x - 6$$

Problem Statement

Find the zeroes of $$f(x) = x^4 + x^3 + 5x^2 - x - 6$$

$$1, -1, (-1 \pm i \sqrt{23})/2$$

Problem Statement

Find the zeroes of $$f(x) = x^4 + x^3 + 5x^2 - x - 6$$

Solution

Brian McLogan - 4205 video solution

video by Brian McLogan

$$1, -1, (-1 \pm i \sqrt{23})/2$$

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$$f(x) = 3x^4 - 8x^3 - 37x^2 + 2x + 40$$

Problem Statement

Find the zeroes of $$f(x) = 3x^4 - 8x^3 - 37x^2 + 2x + 40$$

$$1, -2, -4/3, 5$$

Problem Statement

Find the zeroes of $$f(x) = 3x^4 - 8x^3 - 37x^2 + 2x + 40$$

Solution

PatrickJMT - 4207 video solution

video by PatrickJMT

$$1, -2, -4/3, 5$$

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$$f(x) = 3x^4 + 5x^3 + 25x^2 + 45x -18$$

Problem Statement

Find the zeroes of $$f(x) = 3x^4 + 5x^3 + 25x^2 + 45x -18$$

$$-2, 1/3, 3i, -3i$$

Problem Statement

Find the zeroes of $$f(x) = 3x^4 + 5x^3 + 25x^2 + 45x -18$$

Solution

The instructor in the video also writes the polynomial in two factored forms, $$f(x) = 3(x+2)(x-1/3)(x-3i)(x+3i)$$ and $$f(x) = (x+2)(3x-1)(x-3i)(x+3i)$$. From the first form, it is easier to see the roots/zeroes. The second form looks cleaner and simpler. As usual, check with your instructor to see what they require.

SuperMathPrincess - 4210 video solution

$$-2, 1/3, 3i, -3i$$

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$$4x^4 - 3x^3 - 33x^2 + 16x + 6 = 0$$

Problem Statement

Solve $$4x^4 - 3x^3 - 33x^2 + 16x + 6 = 0$$

$$3, -1/4, -1 \pm \sqrt{3}$$

Problem Statement

Solve $$4x^4 - 3x^3 - 33x^2 + 16x + 6 = 0$$

Solution

Thinkwell - 4212 video solution

video by Thinkwell

$$3, -1/4, -1 \pm \sqrt{3}$$

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$$x^5 - 5x^3 + 4x = 0$$

Problem Statement

Find the zeroes of $$x^5 - 5x^3 + 4x = 0$$

Solution

Brian McLogan - 4248 video solution

video by Brian McLogan

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$$2x^5 - 2x^3 = 144x$$

Problem Statement

Solve $$2x^5 - 2x^3 = 144x$$

$$0, 3, -3, 2i\sqrt{2}, -2i\sqrt{2}$$

Problem Statement

Solve $$2x^5 - 2x^3 = 144x$$

Solution

Thinkwell - 4211 video solution

video by Thinkwell

$$0, 3, -3, 2i\sqrt{2}, -2i\sqrt{2}$$

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$$x^6 - 2x^4 - 4x^2 + 8$$

Problem Statement

Factor $$x^6 - 2x^4 - 4x^2 + 8$$

$$\pm \sqrt{2}, \pm i\sqrt{2}$$

Problem Statement

Factor $$x^6 - 2x^4 - 4x^2 + 8$$

Solution

Factored form with all real numbers $$(x^2-2)^2(x^2+2)$$

The Organic Chemistry Tutor - 4217 video solution

$$\pm \sqrt{2}, \pm i\sqrt{2}$$

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