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• Calculators
• ::
• Polynomial Calculators
• ::
• Polynomial Roots Calculator

This free math tool finds the roots (zeros) of a given polynomial. The calculator computes exact solutions for quadratic, cubic, and quartic equations.
It also displays the step-by-step solution with a detailed explanation.

Enter polynomial:

= 0

Examples:

x^2 - 4x + 3

2x^2 - 3x + 1

x^3 – 2x^2 – x + 2

EXAMPLES

find roots of the polynomial $4x^2 - 10x + 4$

find polynomial roots $-2x^4 - x^3 + 189$

solve equation $6x^3 - 25x^2 + 2x + 8 = 0$

find polynomial roots $2x^3-x^2-x-3$

find roots $2x^5-x^4-14x^3-6x^2+24x+40$

Search our database of more than 200 calculators

TUTORIAL

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 $.

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

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