<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Skip Navigation
Our Terms of Use (click here to view) have changed. By continuing to use this site, you are agreeing to our new Terms of Use.

Fundamental Theorem of Algebra

Find all zeroes of a polynomial, including complex solutions

Atoms Practice
Estimated6 minsto complete
Practice Fundamental Theorem of Algebra
Estimated6 minsto complete
Practice Now
Turn In
Finding Imaginary Solutions

Louis calculates that the area of a rectangle is represented by the equation \begin{align*}3x^4 + 7x^2 = 2\end{align*}. Did Louis calculate it right? Explain based on the degree and zeros of the function.

Imaginary Solutions

Remember, imaginary solutions always come in pairs. To find the imaginary solutions to a function, use the Quadratic Formula.



Let's solve \begin{align*}f(x)=3x^4-x^2-14\end{align*}

First, this quartic function can be factored just like a quadratic equation.


Now, because neither factor can be factored further and there is no \begin{align*}x-\end{align*}term, we can set each equal to zero and solve.

\begin{align*}& && 3x^2-7=0\\ & x^2+2=0 && 3x^2=7\\ & x^2=-2 \qquad \qquad \qquad \quad and && x^2=\frac{7}{3}\\ & x=\pm \sqrt{-2} \ or \ \pm i \sqrt{2} && x=\pm \sqrt{\frac{7}{3}} \ or \ \pm \frac{\sqrt{21}}{3}\end{align*}

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

Now, let's find all the solutions of the function \begin{align*}g(x)=x^4+21x^2+90\end{align*}.

When graphed, this function does not touch the \begin{align*}x-\end{align*}axis. 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.

\begin{align}g(x) &= x^4+21x^2+90\\ 0 &= (x^2+6)(x^2+15)\end{align}

Now, set each factor equal to zero and solve.

\begin{align*}& x^2+6=0 && x^2+15=0\\ & x^2=-6 \qquad \qquad \qquad and && x^2=-15\\ & x=\pm i \sqrt{6} && x=\pm i \sqrt{15}\end{align*}

Finally, let's find the function that has the solution 3, -2, and \begin{align*}4 + i\end{align*}.

Notice that one of the given solutions involves an imaginary number. Imaginary and complex solutions always come in pairs, so \begin{align*}4 - i\end{align*} is also a factor. The two factors are complex conjugates. Translate each solution into a factor and multiply them all together.

Any multiple of this function would also have these roots. For example, \begin{align*}2x^4-18x^3+38x^2+62x-204\end{align*} would have these roots as well.


Example 1

Earlier, you were asked to determine if Louis calculated his work correctly. 

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

\begin{align}3x^4 + 7x^2 &= 2\\ 3x^4 + 7x^2 - 2 &= 0\\ (3x^2 + 1)(x^2 + 2) &= 0 \end{align}

Solving for x we get

\begin{align*} 3x^2+1=0 && x^2+2=0\\ & x^2=\frac{-1}{3} \qquad \qquad \qquad and && x^2=-2\\ & x=\pm i \sqrt{\frac{1}{3}} && x=\pm i \sqrt{2}\end{align*}

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

Example 2

Find all the solutions to the following function: \begin{align*}f(x)=25x^3-120x^2+81x-4\end{align*}.

First, graph the function.

Using the Rational Root Theorem, the possible realistic zeros could be \begin{align*}\frac{1}{25}\end{align*}, 1, or 4. Let’s try these three possibilities using synthetic division.

Of these three possibilities, only 4 is a zero. The leftover polynomial, \begin{align*}25x^2-20x+1\end{align*} is not factorable, so we need to use the Quadratic Formula to find the last two zeros.

\begin{align}x &= \frac{20 \pm \sqrt{20^2-4(25)(1)}}{2(25)}\\ &= \frac{20 \pm \sqrt{400-100}}{50}\\ &=\frac{20 \pm 10 \sqrt{3}}{50} \ or \ \frac{2 \pm \sqrt{3}}{5} \approx 0.746 \ and \ 0.054 \end{align}

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

Example 3

Find all the solutions to the following function: \begin{align*}f(x)=4x^4+35x^2-9\end{align*}.

\begin{align*}f(x)=4x^4+35x^2-9\end{align*} is factorable. \begin{align*}ac = -36\end{align*}.

\begin{align*} 4x^4+35x^2-9\\ 4x^4+36x^2-x^2-9\\ 4x^2(x^2+9)-1(x^2+9)\\ (x^2+9)(4x^2-1)\end{align*}

Setting each factor equal to zero, we have:

\begin{align*}& && 4x^2-1=0\\ & x^2+9=0 && 4x^2=1\\ & x^2=-9 \quad \qquad \qquad or && x^2=\frac{1}{4}\\ & x=\pm 3i && x=\pm \frac{1}{2}\end{align*}

Example 4

Find the equation of a function with roots 4, \begin{align*}\sqrt{2}\end{align*} and \begin{align*}1-i\end{align*}.

Recall that irrational and imaginary roots come in pairs. Therefore, all the roots are 4, \begin{align*}\sqrt{2}, {\color{red}-\sqrt{2}},1+i,{\color{red}1-i}\end{align*}. Multiply all 5 roots together.

\begin{align*} (x-4)(x-\sqrt{2})(x+\sqrt{2})(x-(1+i))(x-(1-i))\\ (x-4)(x^2-2)(x^2-2x+2)\\ (x^3-4x^2-2x+8)(x^2-2x+2)\\ x^5-6x^4+8x^3-4x^2-20x+16\end{align*}


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

  1. \begin{align*}f(x)=x^4+x^3-12x^2-10x+20\end{align*}
  2. \begin{align*}f(x)=4x^3-20x^2-3x+15\end{align*}
  3. \begin{align*}f(x)=2x^4-7x^2-30\end{align*}
  4. \begin{align*}f(x)=x^3+5x^2+12x+18\end{align*}
  5. \begin{align*}f(x)=4x^4+4x^3-22x^2-8x+40\end{align*}
  6. \begin{align*}f(x)=3x^4+4x^2-15\end{align*}
  7. \begin{align*}f(x)=2x^3-6x^2+9x-27\end{align*}
  8. \begin{align*}f(x)=6x^4-7x^3-280x^2-419x+280\end{align*}
  9. \begin{align*}f(x)=9x^4+6x^3-28x^2+2x+11\end{align*}
  10. \begin{align*}f(x)=2x^5-19x^4+30x^3+97x^2-20x+150\end{align*}

Find a function with the following roots.

  1. \begin{align*}4, i\end{align*}
  2. \begin{align*}-3, -2i\end{align*}
  3. \begin{align*}\sqrt{5}, -1 + i\end{align*}
  4. \begin{align*}2, \frac{1}{3}, 4-\sqrt{2}\end{align*}
  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 \begin{align*}f(x)=x^5+x^3+8x^2+8\end{align*}.

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 6.12. 

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Please to create your own Highlights / Notes
Show More


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

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

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

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

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

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.

Image Attributions

Explore More

Sign in to explore more, including practice questions and solutions for Fundamental Theorem of Algebra.
Please wait...
Please wait...