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# Polynomial factoring calculator

Polynomial factoring calculator converts polynomials to factored form. The calculator can be used to factor polynomials having one or more variables. Calculator shows all the work and provides detailed explanation how to factor the expression.

## Factor polynomial $$x^{2}+5x+4$$

solution

$$x^{2}+5x+4 = ( x+1 )( x+4 )$$

explanation

Step 1: Identify constants $\color{blue}{ b }$ and $\color{red}{ c }$. ( $\color{blue}{ b }$ is a number in front of the $x$ term and $\color{red}{ c }$ is a constant). In our case:

$$\color{blue}{ b = 5 } ~ \text{ and } ~ \color{red}{ c = 4 }$$

Now we must discover two numbers that sum up to $\color{blue}{ 5 }$ and multiply to $\color{red}{ 4 }$.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = 4 }$.

 PRODUCT = 4 1     4 -1     -4 2     2 -2     -2

Step 3: Find out which pair sums up to $\color{blue}{ b = 5 }$

 PRODUCT = 4 and SUM = 5 1     4 -1     -4 2     2 -2     -2

Step 4: Put 1 and 4 into placeholders to get factored form.

\begin{aligned} x^{2}+5x+4 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}+5x+4 & = (x + 1)(x + 4) \end{aligned}

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Script name : polynomial-factoring-calculator

Form values: x^2+5x+4 , g , Factor x^2+5x+4 , ,

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Polynomial factoring calculator
Factor polynomials with single or multiple variables.
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2x^2+3x-5
10ab+15b-2a-3
9u^2v^4+6uv^2+1

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## Polynomial Factoring Techniques

To find the factored form of a polynomial, this calculator employs the following methods:

1. Factoring GCF, 2 Factoring by grouping, 3 Using the difference of squares, and 4 Factoring quadratic polynomials.

### Method 1 : Factoring GCF

Example 01: Factor 3ab3-6a2b

3ab3-6a2b=3·a·b·b·b-2·3·a·a·b=
=3ab(b2-2a)


### Method 2 : Factoring By Grouping

The method is very useful for finding the factored form of the four term polynomials.

Example 03: Factor 2a-4b+a2-2ab

We usually group the first two, and the last two terms.

2a-4b+a2-2ab = 2a-4b + a2-2ab

We now factor 2 out of the blue terms and a out of from red ones.

2a-4b+a2-2ab = 2·(a-2b)+a·(a-2b)

At the end we factor out common factor of (a-2b)

2·(a-2b)+a·(a-2b)=(a-2b)(2+a)

Example 04: Factor 5ab+2b+5ac+2c

This is a rare situation where the first two terms of a polynomial do not have a common factor, so we have to group the first and third terms together.

5ab+2b+5ac+2c = 5ab+5ac+2b+2c=
= 5a(b+c) + 2(b+c)
= (b+c)(5a + 2)


### Method 3 : Special Form – A

The most common special case is the difference of two squares.

a2-b2=(a+b)(a-b)

We usually use this method when the polynomial has only two terms.

Example 05: Factor 4x2-y2.

First, we need to notice that the polynomial can be written as the difference of two perfect squares.

4x2-y2=(2x)2-y2

Now we can apply above formula with a=2x and b=y

(2x)2-y2=(2x-y)(2x+y)

Example 06: Factor 9a2b4-4c2

The binomial we have here is the difference of two perfect squares, thus the calculation will be similar to the last one.

9a2b4-4c2=(3ab2)2-4c2=
=(3ab2-2c)(3ab2+2c)


### Method 4 : Special Form – B

The second special case of factoring is the Perfect Square Trinomial

a2 ± 2ab + b2 = (a ± b)2

Example 07: Factor 100+20x+x2

In this case, the first and third terms are perfect squares. So we can use the above formula.

100+20x+x2 = 102+2·10·x+x2 =
= (10+x)2


### Method 5: Factoring Quadratic Polynomials

The best way to explain this method is by using an example.

Example 08: Factor x2 + 3x + 4

We know that the factored form has the following pattern

x2+5x+4 = (x + _ )(x + _ )

All we have to do now is fill in the blanks with the two numbers. To do this, notice that the product of these two numbers has to be 4 and their sum has to be 5. We conclude, after some trial and error, that the missing numbers are 1 and 4. As a result, the solution is::

x^2+5x+4 = (x+1)(x+4)

Example 09: Factor x2 - 8x + 15

Like in the previous example, we look again for the solution in the form

x2-8x+15 = (x + _)(x + _)

The sum of missing numbers is -8 so we need to find two negative numbers such that the product is 15 and the sum is -8. It is not hard to see that two numbers with such properties are -3 and -5, so the solution is.

x2+5x+4 = (x+-3)(x+-5) =
= (x-3)(x-5)

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