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- 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 $2x^5-x^4-14x^3-6x^2+24x+40$
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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) = 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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