How to find complex zeros of a polynomial function calculator

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Table of Contents Show

Output Type
Zeros of a Polynomial
Use of the zeros Calculator
More References and Links
How to find polynomial roots ?
Roots of quadratic polynomial
Quadratic equation - special cases
Roots of cubic polynomial
Cubic polynomial - factoring method
How do you find all real zeros of a polynomial function?
Can polynomial have complex zeros?

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A calculator to calculate the real and complex zeros of a polynomial is presented.

Zeros of a Polynomial

$ a $ is a zero of a polynomial $ P(x) $ if and only if $ P(a) = 0 $
or
$ a $ is a zero of a polynomial $ P(x) $ if and only if $ x - a $ is a factor of $ P(x) $
Note that the zeros of the polynomial $ P(x) $ refer to the values of $ x $ that makes $ P(x) $ equal to zero. But both the zeros and the roots of a polynomial are found using factoring and the factor theorem [1 2].

Example
Find the zeros of the polynomial $ P(x) = x^2 + 5x - 14 $.
Solution
Factor $ P(x) $ as follows
$ P(x) = (x-2)(x+7) $
Set $ P(x) = 0 $ and solve
$ P(x) = (x-2)(x+7) = 0 $
Apply the factor theorem [1 2] and write that each factor is equal to zero.
$ x-2 = 0 $ or $ x+7 = 0 $
Solve to obtain
$ x = 2 $ and $ x = - 7 $
Hence the zeros of $ P(x) $ are $ x = 2 $ and $ x = - 7 $

Use of the zeros Calculator

1 - Enter and edit polynomial $ P(x) $ and click "Enter Polynomial" then check what you have entered and edit if needed.
Note that the five operators used are: + (plus) , - (minus), , ^ (power) and * (multiplication). (example: P(x) = -2*x^4+8*x^3+14*x^2-44*x-48).(more notes on editing functions are located below)
2 - Click "Calculate Zeros" to obain the zeros of the polynomial.
Note that the zeros of some polynomials take a large amount of time to be computated and their expressions may be quite complicated to understand.

Notes: In editing functions, use the following:
1 - The five operators used are: + (plus) , - (minus), / (division) , ^ (power) and * (multiplication). (example: P(x) = 2*x^2 - 2*x - 4 )
Here are some examples of polynomials that you may copy and paste to practice:
x^2 - 9 x^2 + 9 x^2 + 2*x + 7 x^3 + 2*x - 3 3*x^4 - 3
x^5+5*x^4+3*x^3+x^2-10*x-120 x^5+4x^4-7x^3-28x^2+6x+24
x^4 - 4*x^3 + 3 (this one has very complicated zeros and takes time to compute; try it to have an idea.)

How to find polynomial roots ?

The process of finding polynomial roots depends on its degree. The degree is the largest exponent in the polynomial. For example, the degree of polynomial $ p(x) = 8x^\color{red}{2} + 3x -1 $ is $\color{red}{2}$.

We name polynomials according to their degree. For us, the most interesting ones are: quadratic - degree 2, Cubic - degree 3, and Quartic - degree 4.

Roots of quadratic polynomial

This is the standard form of a quadratic equation

$$ a\,x^2 + b\,x + c = 0 $$

The formula for the roots is

$$ x_1, x_2 = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} $$

Example 01: Solve the equation $ 2x^2 + 3x - 14 = 0 $

In this case we have $ a = 2, b = 3 , c = -14 $, so the roots are:

$$ \begin{aligned} x_1, x_2 &= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\ x_1, x_2 &= \dfrac{-3 \pm \sqrt{3^2-4 \cdot 2 \cdot (-14)}}{2\cdot2} \\ x_1, x_2 &= \dfrac{-3 \pm \sqrt{9 + 4 \cdot 2 \cdot 14}}{4} \\ x_1, x_2 &= \dfrac{-3 \pm \sqrt{121}}{4} \\ x_1, x_2 &= \dfrac{-3 \pm 11}{4} \\ x_1 &= \dfrac{-3 + 11}{4} = \dfrac{8}{4} = 2 \\ x_2 &= \dfrac{-3 - 11}{4} = \dfrac{-14}{4} = -\dfrac{7}{2} \end{aligned} $$

Quadratic equation - special cases

Sometimes, it is much easier not to use a formula for finding the roots of a quadratic equation.

Example 02: Solve the equation $ 2x^2 + 3x = 0 $

Because our equation now only has two terms, we can apply factoring. Using factoring we can reduce an original equation to two simple equations.

$$ \begin{aligned} 2x^2 + 3x &= 0 \\ \color{red}{x} \cdot \left( \color{blue}{2x + 3} \right) &= 0 \\ \color{red}{x = 0} \,\,\, \color{blue}{2x + 3} & \color{blue}{= 0} \\ \color{blue}{2x } & \color{blue}{= -3} \\ \color{blue}{x} &\color{blue}{= -\frac{3}{2}} \end{aligned} $$

Example 03: Solve equation $ 2x^2 - 10 = 0 $

This is also a quadratic equation that can be solved without using a quadratic formula.

. $$ \begin{aligned} 2x^2 - 18 &= 0 \\ 2x^2 &= 18 \\ x^2 &= 9 \\ \end{aligned} $$

The last equation actually has two solutions. The first one is obvious

$$ \color{blue}{x_1 = \sqrt{9} = 3} $$

and the second one is

$$ \color{blue}{x_2 = -\sqrt{9} = -3 }$$

Roots of cubic polynomial

To solve a cubic equation, the best strategy is to guess one of three roots.

Example 04: Solve the equation $ 2x^3 - 4x^2 - 3x + 6 = 0 $.

Step 1: Guess one root.

The good candidates for solutions are factors of the last coefficient in the equation. In this example, the last number is -6 so our guesses are

1, 2, 3, 6, -1, -2, -3 and -6

if we plug in $ \color{blue}{x = 2} $ into the equation we get,

$$ 2 \cdot \color{blue}{2}^3 - 4 \cdot \color{blue}{2}^2 - 3 \cdot \color{blue}{2} + 6 = 2 \cdot 8 - 4 \cdot 4 - 6 - 6 = 0$$

So, $ \color{blue}{x = 2} $ is the root of the equation. Now we have to divide polynomial with $ \color{red}{x - \text{ROOT}} $

In this case we divide $ 2x^3 - x^2 - 3x - 6 $ by $ \color{red}{x - 2}$.

$$ ( 2x^3 - 4x^2 - 3x + 6 ) \div (x - 2) = 2x^2 - 3 $$

Now we use $ 2x^2 - 3 $ to find remaining roots

$$ \begin{aligned} 2x^2 - 3 &= 0 \\ 2x^2 &= 3 \\ x^2 &= \frac{3}{2} \\ x_1 & = \sqrt{ \frac{3}{2} } = \frac{\sqrt{6}}{2}\\ x_2 & = -\sqrt{ \frac{3}{2} } = - \frac{\sqrt{6}}{2} \end{aligned} $$

Cubic polynomial - factoring method

To solve cubic equations, we usually use the factoting method:

Example 05: Solve equation $ 2x^3 - 4x^2 - 3x + 6 = 0 $.

Notice that a cubic polynomial has four terms, and the most common factoring method for such polynomials is factoring by grouping.

$$ \begin{aligned} 2x^3 - 4x^2 - 3x + 6 &= \color{blue}{2x^3-4x^2} \color{red}{-3x + 6} = \\ &= \color{blue}{2x^2(x-2)} \color{red}{-3(x-2)} = \\ &= (x-2)(2x^2 - 3) \end{aligned} $$

Now we can split our equation into two, which are much easier to solve. The first one is $ x - 2 = 0 $ with a solution $ x = 2 $, and the second one is $ 2x^2 - 3 = 0 $.

$$ \begin{aligned} 2x^2 - 3 &= 0 \\ x^2 = \frac{3}{2} \\ x_1x_2 = \pm \sqrt{\frac{3}{2}} \end{aligned} $$

228 282 421 solved problems

How do you find all real zeros of a polynomial function?

To find the real zeros of a polynomial, first convert the polynomial to factored form. Once all factors are found, set each individual factor equal to zero to solve for the real zeros.

Can polynomial have complex zeros?

The Fundamental Theorem of Algebra states that every polynomial function of positive degree with complex coefficients has at least one complex zero. For example, the polynomial function P(x) = 4ix² + 3x - 2 has at least one complex zero.