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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 $$ p(x) = x^4-3x^3-13x^2+9x+30 $$
solution
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 5\\[1 em]x_2 &= -2\\[1 em]x_3 &= \sqrt{ 3 }\\[1 em]x_4 &= -\sqrt{ 3 } \end{aligned} $$explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 5 } $ is a root of polynomial $ x^4-3x^3-13x^2+9x+30 $.
The Rational Root Theorem tells us that if the polynomial has a rational zero then it must be a fraction $ \dfrac{ \color{blue}{p}}{ \color{red}{q} } $, where $ p $ is a factor of the constant term and $ q $ is a factor of the leading coefficient.
The constant term is $ \color{blue}{ 30 } $, with a single factor of 1, 2, 3, 5, 6, 10, 15 and 30.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 30 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 2, 3, 5, 6, 10, 15, 30 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 10}{ 1} \pm \frac{ 15}{ 1} \pm \frac{ 30}{ 1} ~~ \end{aligned} $$Substitute the possible roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.
We can see that $ p\left( 5 \right) = 0 $ so $ x = 5 $ is a root of a polynomial $ p(x) $.
To find remaining zeros we use Factor Theorem. This theorem states that if $ \dfrac{p}{q} $ is root of the polynomial then the polynomial can be divided by $ \color{blue}{qx − p} $. In this example we divide polynomial $ p $ by $ \color{blue}{ x-5 }$
$$ \frac{ x^4-3x^3-13x^2+9x+30}{ x-5} = x^3+2x^2-3x-6 $$Step 2:
The next rational root is $ x = 5 $
$$ \frac{ x^4-3x^3-13x^2+9x+30}{ x-5} = x^3+2x^2-3x-6 $$Step 3:
The next rational root is $ x = -2 $
$$ \frac{ x^3+2x^2-3x-6}{ x+2} = x^2-3 $$Step 4:
The solutions of $ x^2-3 = 0 $ are: $ x = -\sqrt{ 3 } ~ \text{and} ~ x = \sqrt{ 3 }$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.
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.