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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. Calculator shows all the work and provides step-by-step on how to find zeros and their multiplicities.
Find roots of polynomial $$ p(x) = 16x^7-21x^6+6x^5+25x^4+12x^3+6x^2+6x-5 $$
solution
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0.4099\\[1 em]x_2 &= -0.0005+0.6881i\\[1 em]x_3 &= -0.0005-0.6881i\\[1 em]x_4 &= -0.7427+0.3397i\\[1 em]x_5 &= -0.7427-0.3397i\\[1 em]x_6 &= 1.1946+0.9937i\\[1 em]x_7 &= 1.1946-0.9937i \end{aligned} $$explanation
Polynomial $ 16x^7-21x^6+6x^5+25x^4+12x^3+6x^2+6x-5 $ has no rational roots that can be found using Rational Root Test, so the roots were found using Newton method.
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.
$$ \begin{aligned} 2x^2 - 3 &= 0 \\ x^2 = \frac{3}{2} \\ x_1x_2 = \pm \sqrt{\frac{3}{2}} \end{aligned} $$1. Roots of Polynomials — find roots of linear, quadratic and cubic polynomials.
2. Rational Root Theorem with examples and explanations.