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6.12: Finding Imaginary Solutions

Difficulty Level: At Grade Created by: CK-12
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Louis calculates that the area of a rectangle is represented by the equation 3x4+7x2=2. Did is calculate right? Explain.

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James Sousa: Ex 4: Find the Zeros of a Polynomial Function with Imaginary Zeros


In #12 from the previous problem set, there are two imaginary solutions. Imaginary solutions always come in pairs. To find the imaginary solutions to a function, use the Quadratic Formula. If you need a little review on imaginary numbers and how to solve a quadratic equation with complex solutions see the Quadratic Equations chapter.

Example A

Solve f(x)=3x4x214. (#12 from the previous problem set.)

Solution: First, this quartic function can be factored just like a quadratic equation. See the Factoring Polynomials in Quadratic Form concept from this chapter for review.


Now, because neither factor can be factored further and there is no xterm, we can set each equal to zero and solve.

x2+2=0x2=2andx=±2 or ±i23x27=03x2=7x2=73x=±73 or ±213

Including the imaginary solutions, there are four, which is what we would expect because the degree of this function is four.

Example B

Find all the solutions of the function g(x)=x4+21x2+90.

Solution: When graphed, this function does not touch the xaxis. Therefore, all the solutions are imaginary. To solve, this function can be factored like a quadratic equation. The factors of 90 that add up to 21 are 6 and 15.


Now, set each factor equal to zero and solve.


Example C

Find the function that has the solution 3, -2, and 4+i.

Solution: Notice that one of the given solutions is imaginary. Imaginary solutions always come in pairs, so 4i is also a factor, they are complex conjugates. Now, translate each solution into a factor and multiply them all together.

Any multiple of this function would also have these roots. For example, 2x418x3+38x2+62x204 would have these roots as well.

Intro Problem Revisit First we need to change the equation to standard form. Then we can factor it.


Solving for x we get


All of the solutions are imaginary and the area of a rectangle must have real solutions. Therefore Louis did not calculate correctly.

Guided Practice

Find all the solutions to the following functions.

1. f(x)=25x3120x2+81x4

2. f(x)=4x4+35x29

3. Find the equation of a function with roots 4, 2 and 1i.


1. First, graph the function.

Using the Rational Root Theorem, the possible realistic zeros could be 125, 1 or 4. Let’s try these three possibilities using synthetic division.

Of these three possibilities, only 4 is a zero. The leftover polynomial, 25x220x+1 is not factorable, so we need to do the Quadratic Formula to find the last two zeros.

x=20±2024(25)(1)2(25)=20±40010050=20±10350 or 2±350.746 and 0.054

Helpful Hint: Always find the decimal values of each zero to make sure they match up with the graph.

2. f(x)=4x4+35x29 is factorable. ac=36.


Setting each factor equal to zero, we have:


This problem could have also been done by using the same method from #1.

3. Recall that irrational and imaginary roots come in pairs. Therefore, all the roots are 4, 2,2,1+i,1i. Multiply all 5 roots together.



Find all solutions to the following functions. Use any method.

  1. f(x)=x4+x312x210x+20
  2. f(x)=4x320x23x+15
  3. f(x)=2x47x230
  4. f(x)=x3+5x2+12x+18
  5. f(x)=4x4+4x322x28x+40
  6. f(x)=3x4+4x215
  7. f(x)=2x36x2+9x27
  8. f(x)=6x47x3280x2419x+280
  9. f(x)=9x4+6x328x2+2x+11
  10. f(x)=2x519x4+30x3+97x220x+150

Find a function with the following roots.

  1. 4,i
  2. 3,2i
  3. 5,1+i
  4. 2,13,42
  5. Writing Write down the steps you use to find all the zeros of a polynomial function.
  6. Writing Why do imaginary and irrational roots always come in pairs?
  7. Challenge Find all the solutions to f(x)=x5+x3+8x2+8.


Complex Conjugate

Complex Conjugate

Complex conjugates are pairs of complex binomials. The complex conjugate of a+bi is a-bi. When complex conjugates are multiplied, the result is a single real number.
complex number

complex number

A complex number is the sum of a real number and an imaginary number, written in the form a + bi.
conjugate pairs theorem

conjugate pairs theorem

The conjugate pairs theorem states that if f(z) is a polynomial of degree n, with n\ne0 and with real coefficients, and if f(z_{0})=0, where z_{0}=a+bi, then f(z_{0}^{*})=0. Where z_{0}^{*} is the complex conjugate of z_{0}.
fundamental theorem of algebra

fundamental theorem of algebra

The fundamental theorem of algebra states that if f(x) is a polynomial of degree n\ge 1, then f(x) has at least one zero in the complex number domain. In other words, there is at least one complex number c such that f(c)=0. The theorem can also be stated as follows: an n^{th} degree polynomial with real or complex coefficients has, with multiplicity, exactly n complex roots.
Imaginary Number

Imaginary Number

An imaginary number is a number that can be written as the product of a real number and i.
Imaginary Numbers

Imaginary Numbers

An imaginary number is a number that can be written as the product of a real number and i.


A polynomial is an expression with at least one algebraic term, but which does not indicate division by a variable or contain variables with fractional exponents.


The roots of a function are the values of x that make y equal to zero.


The zeros of a function f(x) are the values of x that cause f(x) to be equal to zero.


The zeroes of a function f(x) are the values of x that cause f(x) to be equal to zero.

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Difficulty Level:
At Grade
Date Created:
Mar 12, 2013
Last Modified:
Mar 23, 2016
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