Form a polynomial with given zeros and degree multiplicity calculator

For a polynomial, if #x=a# is a zero of the function, then #(x-a)# is a factor of the function.

We have two unique zeros: #-2# and #4#. However, #-2# has a multiplicity of #2#, which means that the factor that correlates to a zero of #-2# is represented in the polynomial twice.

Follow the colors to see how the polynomial is constructed:

#"zero at "color(red)(-2)", multiplicity "color(blue)2#
#"zero at "color(green)4", multiplicity "color(purple)1#

#p(x)=(x-(color(red)(-2)))^color(blue)2(x-color(green)4)^color(purple)1#

Thus,

#p(x)=(x+2)^2(x-4)#

Expand:

#p(x)=(x^2+4x+4)(x-4)#

#p(x)=x^3-12x-16#

We can graph the function to understand multiplicities and zeros visually:

graph{x^3-12x-16 [-6, 6, -43.83, 14.7]}

The zero at #x=-2# "bounces off" the #x#-axis. This behavior occurs when a zero's multiplicity is even.

The zero at #x=4# continues through the #x#-axis, as is the case with odd multiplicities.

Note that the function does have three zeros, which it is guaranteed by the Fundamental Theorem of Algebra, but one of such zeros is represented twice.

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Number Line

Factor using the AC method.

Consider the form . Find a pair of integers whose product is and whose sum is . In this case, whose product is and whose sum is .

Write the factored form using these integers.

If any individual factor on the left side of the equation is equal to , the entire expression will be equal to .

Set equal to and solve for .

Add to both sides of the equation.

Set equal to and solve for .

Subtract from both sides of the equation.

The final solution is all the values that make true. The multiplicity of a root is the number of times the root appears.

(Multiplicity of )

(Multiplicity of )

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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.
It also displays the step-by-step solution with a detailed explanation.

Enter polynomial:

= 0

Examples:

x^2 - 4x + 3

2x^2 - 3x + 1

x^3 – 2x^2 – x + 2

EXAMPLES

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

find polynomial roots $-2x^4 - x^3 + 189$

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

find polynomial roots $2x^3-x^2-x-3$

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.

Roots of quadratic polynomial

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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