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**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.

**solution**

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

$$ \begin{aligned}x_1 &= 2\\[1 em]x_2 &= \frac{\sqrt{ 6 }}{ 2 }\\[1 em]x_3 &= - \frac{\sqrt{ 6 }}{ 2 } \end{aligned} $$**explanation**

**Step 1:**

Use **rational root test** to find out that the $ \color{blue}{ x = 2 } $ is a root of polynomial $ 2x^3-4x^2-3x+6 $.

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}{ 6 } $, with factors of ** 1, 2, 3 and 6**.

The leading coefficient is $ \color{red}{ 2 }$, with factors of ** 1 and 2**.

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 6 }}{\text{ factors of 2 }} = \pm \dfrac{\text{ ( 1, 2, 3, 6 ) }}{\text{ ( 1, 2 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 6}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 6}{ 2} ~~ \end{aligned} $$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( 2 \right) = 0 $ so $ x = 2 $ 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-2 }$

**Step 2:**

The next rational root is $ x = 2 $

$$ \frac{ 2x^3-4x^2-3x+6}{ x-2} = 2x^2-3 $$**Step 3:**

The solutions of $ 2x^2-3 = 0 $ are: $ x = - \dfrac{\sqrt{ 6 }}{ 2 } ~ \text{and} ~ x = \dfrac{\sqrt{ 6 }}{ 2 }$.

You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.

working...

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

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:

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

$$ \color{blue}{x_2 = -\sqrt{9} = -3 }$$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

$$ \begin{aligned} 2x^2 - 3 &= 0 \\ 2x^2 &= 3 \\ x^2 &= \frac{3}{2} \\ x_1 & = \sqrt{ \frac{3}{2} } = \frac{\sqrt{6}}{2}\\ x_2 & = -\sqrt{ \frac{3}{2} } = - \frac{\sqrt{6}}{2} \end{aligned} $$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 $.

$$ \begin{aligned} 2x^2 - 3 &= 0 \\ x^2 = \frac{3}{2} \\ x_1x_2 = \pm \sqrt{\frac{3}{2}} \end{aligned} $$RESOURCES

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

2. Rational Root Theorem with examples and explanations.

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