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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 $$p(x) = 2x^2+3x$$

solution

The roots of polynomial $p(x)$ are:

\begin{aligned}x_1 &= 0\\[1 em]x_2 &= -\frac{ 3 }{ 2 } \end{aligned}

explanation

Step 1:

Factor out $\color{blue}{ x }$ from $2x^2+3x$ and solve two separate equations:

\begin{aligned} 2x^2+3x & = 0\\[1 em] \color{blue}{ x }\cdot ( 2x+3 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ 2x+3 & = 0 \end{aligned}

One solution is $\color{blue}{ x = 0 }$. Use second equation to find the remaining roots.

Step 2:

To find the second zero, solve equation $2x+3 = 0$

\begin{aligned} 2x+3 & = 0 \\[1 em] 2 \cdot x & = -3 \\[1 em] x & = - \frac{ 3 }{ 2 } \end{aligned}

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Form values: 2x^2+3x , g , Find roots of 2x^2+3x ,

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Polynomial Roots Calculator
find real and complex zeros of a polynomial
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x^2-4x+3
2x^2-3x+1
x^3–2x^2–x+2
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EXAMPLES
example 1:ex 1:
find roots of the polynomial $4x^2 - 10x + 4$
example 2:ex 2:
find polynomial roots $-2x^4 - x^3 + 189$
example 3:ex 3:
solve equation $6x^3 - 25x^2 + 2x + 8 = 0$
example 4:ex 4:
find polynomial roots $2x^3-x^2-x-3$
example 5:ex 5:
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.

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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