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Polynomial roots calculator

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This free math tool finds the roots (zeros) of a given polynomial. The calculator computes exact solutions for quadratic, cubic, and quartic equations. Calculator shows all the work and provides step-by-step on how to find zeros and their multiplicities.

Find roots of polynomial $$ p(x) = 16x+5 $$

solution

The root of polynomial $ p(x) $ is:

$$ \begin{aligned}x &= -\frac{ 5 }{ 16 } \end{aligned} $$

explanation

$$ \begin{aligned} 16x+5 & = 0 \\[1 em] 16 \cdot x & = -5 \\[1 em] x & = - \frac{ 5 }{ 16 } \end{aligned} $$

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Polynomial roots calculator
Find real and complex zeros for any polynomial.
help ↓↓ examples ↓↓ tutorial ↓↓
x^2-4x+3
2x^2-3x+1
x^3–2x^2–x+2
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Examples
ex 1:
find roots of the polynomial 4x2-10x+4
ex 2:
find polynomial roots -2x4-x^3+189
ex 3:
solve equation 6x3-25x3+2x+8=0
ex 4:
find polynomial roots 2x3-x2-x-3
ex 5:
find roots 2x4-x4-14x3-6x2+24x+40
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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)=8x2+3x-1 is 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 is

ax2+bx+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 2x2+3x-14.

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 2x2+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 2x2-18=0

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

2x2 - 18 = 0
      2x2 = 18
      x2 = 9

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 2x3-4x2-3x+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 x=2into 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, x=2 is the root of the equation. Now we have to divide polynomial by x-ROOT.

In this case we divide 2x3-x2-3x+6 by x-2.

(2x3-x2-3x+6)/(x-2) = 2x2-3

Now we use 2x2-3 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 factoring method.

Example 05: Solve equation 2x3-4x2-3x+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 smaller equations, which are much easier to solve. The first one is x-2=0 with a solution x=2, and the second one is 2x2-3=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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