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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.

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Examples

ex 1:

find roots of the polynomial 4x^{2}-10x+4

ex 2:

find polynomial roots -2x^{4}-x^3+189

ex 3:

solve equation 6x^{3}-25x^{3}+2x+8=0

ex 4:

find polynomial roots 2x^{3}-x^{2}-x-3

ex 5:

find roots 2x^{4}-x^{4}-14x^{3}-6x^{2}+24x+40

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TUTORIAL

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^{2}+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).

This is the standard form of a quadratic equation is

ax^{2}+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 2x^{2}+3x-14.

In this case we have a=2, b=3, c=-14, so the roots are:

$$ \begin{aligned} x_1, x_2 &= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\ x_1, x_2 &= \dfrac{-3 \pm \sqrt{3^2-4 \cdot 2 \cdot (-14)}}{2\cdot2} \\ x_1, x_2 &= \dfrac{-3 \pm \sqrt{9 + 4 \cdot 2 \cdot 14}}{4} \\ x_1, x_2 &= \dfrac{-3 \pm \sqrt{121}}{4} \\ x_1, x_2 &= \dfrac{-3 \pm 11}{4} \\ x_1 &= \dfrac{-3 + 11}{4} = \dfrac{8}{4} = 2 \\ x_2 &= \dfrac{-3 - 11}{4} = \dfrac{-14}{4} = -\dfrac{7}{2} \end{aligned} $$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.

**Example 03:** Solve equation 2x^{2}-18=0

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

2x^{2}- 18 = 0 2x^{2}= 18 x^{2}= 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

$$ \color{blue}{x_2 = -\sqrt{9} = -3 }$$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 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 2x^{3}-x^{2}-3x+6 by
x-2.

(2x^{3}-x^{2}-3x+6)/(x-2) = 2x^{2}-3

Now we use 2x^{2}-3 to find remaining roots

To solve cubic equations, we usually use the factoring 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 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
2x^{2}-3=0.

RESOURCES

1. Roots of Polynomials — find roots of linear, quadratic and cubic polynomials.

2. Rational Root Theorem with examples and explanations.

443 639 807 solved problems

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