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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 $$ p(x) = x^8+11x^7-18x^6-472x^5-320x^4+6576x^3+7776x^2-27648x-41472 $$
solution
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 3\\[1 em]x_2 &= 4\\[1 em]x_3 &= -2\\[1 em]x_4 &= -6\\[1 em]x_5 &= 4\\[1 em]x_6 &= -2\\[1 em]x_7 &= -6\\[1 em]x_8 &= -6 \end{aligned} $$explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 3 } $ is a root of polynomial $ x^8+11x^7-18x^6-472x^5-320x^4+6576x^3+7776x^2-27648x-41472 $.
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}{ 41472 } $, with a single factor of 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 81, 96, 108, 128, 144, 162, 192, 216, 256, 288, 324, 384, 432, 512, 576, 648, 768, 864, 1152, 1296, 1536, 1728, 2304, 2592, 3456, 4608, 5184, 6912, 10368, 13824, 20736 and 41472.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 41472 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 81, 96, 108, 128, 144, 162, 192, 216, 256, 288, 324, 384, 432, 512, 576, 648, 768, 864, 1152, 1296, 1536, 1728, 2304, 2592, 3456, 4608, 5184, 6912, 10368, 13824, 20736, 41472 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 16}{ 1} \pm \frac{ 18}{ 1} \pm \frac{ 24}{ 1} \pm \frac{ 27}{ 1} \pm \frac{ 32}{ 1} \pm \frac{ 36}{ 1} \pm \frac{ 48}{ 1} \pm \frac{ 54}{ 1} \pm \frac{ 64}{ 1} \pm \frac{ 72}{ 1} \pm \frac{ 81}{ 1} \pm \frac{ 96}{ 1} \pm \frac{ 108}{ 1} \pm \frac{ 128}{ 1} \pm \frac{ 144}{ 1} \pm \frac{ 162}{ 1} \pm \frac{ 192}{ 1} \pm \frac{ 216}{ 1} \pm \frac{ 256}{ 1} \pm \frac{ 288}{ 1} \pm \frac{ 324}{ 1} \pm \frac{ 384}{ 1} \pm \frac{ 432}{ 1} \pm \frac{ 512}{ 1} \pm \frac{ 576}{ 1} \pm \frac{ 648}{ 1} \pm \frac{ 768}{ 1} \pm \frac{ 864}{ 1} \pm \frac{ 1152}{ 1} \pm \frac{ 1296}{ 1} \pm \frac{ 1536}{ 1} \pm \frac{ 1728}{ 1} \pm \frac{ 2304}{ 1} \pm \frac{ 2592}{ 1} \pm \frac{ 3456}{ 1} \pm \frac{ 4608}{ 1} \pm \frac{ 5184}{ 1} \pm \frac{ 6912}{ 1} \pm \frac{ 10368}{ 1} \pm \frac{ 13824}{ 1} \pm \frac{ 20736}{ 1} \pm \frac{ 41472}{ 1} ~~ \end{aligned} $$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( 3 \right) = 0 $ so $ x = 3 $ 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-3 }$
$$ \frac{ x^8+11x^7-18x^6-472x^5-320x^4+6576x^3+7776x^2-27648x-41472}{ x-3} = x^7+14x^6+24x^5-400x^4-1520x^3+2016x^2+13824x+13824 $$Step 2:
The next rational root is $ x = 3 $
$$ \frac{ x^8+11x^7-18x^6-472x^5-320x^4+6576x^3+7776x^2-27648x-41472}{ x-3} = x^7+14x^6+24x^5-400x^4-1520x^3+2016x^2+13824x+13824 $$Step 3:
The next rational root is $ x = 4 $
$$ \frac{ x^7+14x^6+24x^5-400x^4-1520x^3+2016x^2+13824x+13824}{ x-4} = x^6+18x^5+96x^4-16x^3-1584x^2-4320x-3456 $$Step 4:
The next rational root is $ x = -2 $
$$ \frac{ x^6+18x^5+96x^4-16x^3-1584x^2-4320x-3456}{ x+2} = x^5+16x^4+64x^3-144x^2-1296x-1728 $$Step 5:
The next rational root is $ x = -6 $
$$ \frac{ x^5+16x^4+64x^3-144x^2-1296x-1728}{ x+6} = x^4+10x^3+4x^2-168x-288 $$Step 6:
The next rational root is $ x = 4 $
$$ \frac{ x^4+10x^3+4x^2-168x-288}{ x-4} = x^3+14x^2+60x+72 $$Step 7:
The next rational root is $ x = -2 $
$$ \frac{ x^3+14x^2+60x+72}{ x+2} = x^2+12x+36 $$Step 8:
The next rational root is $ x = -6 $
$$ \frac{ x^2+12x+36}{ x+6} = x+6 $$Step 9:
To find the last zero, solve equation $ x+6 = 0 $
$$ \begin{aligned} x+6 & = 0 \\[1 em] x & = -6 \end{aligned} $$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.