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# Polynomial factoring calculator

This calculator is a free online math tool that transforms polynomial into 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 $$3ab^{3}-6a^{2}b$$

solution

$$3ab^{3}-6a^{2}b = 3ab(b^{2}-2a)$$

explanation

Factor out common factor $\color{blue}{ 3ab }$:

$$3ab^3-6a^2b = 3ab ( b^2-2a )$$

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Script name : polynomial-factoring-calculator

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Polynomial Factoring Calculator
factor polynomials with both single and multiple variables
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2x^2+3x-5
10ab+15b-2a-3
9u^2v^4+6uv^2+1

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EXAMPLES
example 1:ex 1:
factor $x^2 - 10x + 25$
example 2:ex 2:
factor $3x^2 - 27$
example 3:ex 3:
factor $4x^3 - 21x^2 + 29x -6$
example 4:ex 4:
factor $3uv-12u+6v-24$
example 5:ex 5:
factor $a^2 + 2a + 1 - b^2$
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TUTORIAL

## 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 $3ab^3 - 6a^2b$

\begin{aligned} 3ab^3 - 6a^2b &= \color{blue}{3 \cdot a \cdot b} \cdot b \cdot b - 2 \cdot \color{blue}{3 \cdot a } \cdot a \color{blue}{\cdot b} = \\[1 em] &= \color{blue}3ab(b^2-2a) \end{aligned}

### Method 2 : Factoring By Grouping

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

Example 03: Factor $2a - 4b + a^2 - 2ab$

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

$$2a - 4b + a^2 - 2ab = \color{blue}{2a - 4b} + \color{red}{a^2 - 2ab}$$

We now factor $\color{blue}{2}$ out of the blue terms and $\color{red}{a}$ out of from red ones.

$$\color{blue}{2a - 4b} + \color{red}{a^2 - 2ab} = \color{blue}{2 \cdot (a - 2b)} + \color{red}{a \cdot (a - 2b) }$$

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

$$2 \cdot \color{blue}{(a - 2b)} + a \cdot \color{blue}{(a - 2b)} = (a - 2b) (2 + a)$$

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.

\begin{aligned} 5ab + 2b + 5ac + 2c &= \color{blue}{5ab + 5ac} + \color{red}{2b + 2c} = \\ &=\color{blue}{5a (b + c)} + \color{red}{2 (b + c)} = \\ &= (b+c)(\color{blue}{5a} + \color{red}{2}) \end{aligned}

### Method 3 : Special Form - A

The most common special case is the difference of two squares

$$a^2 - b^2 = (a+b) (a-b)$$

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

Example 05: Factor $4x^2 - y^2$.

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

$$\color{blue}{4x^2} - \color{red}{y^2} = \color{blue}{ \left(2x\right)}^2 - \color{red}{y}^2$$

Now we can apply above formula with $\color{blue}{a = 2x}$ and $\color{red}{b = y}$

$$\left( \color{blue}{2x} \right)^2 - \color{red}{y}^2 = (2x - b) (2x + b)$$

Example 06: Factor $9a^2b^4 - 4c^2$.

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

$$9a^2b^4 - 4c^2 = \left( 3ab^2 \right)^2 - \left( 2c \right)^2 = (3ab^2-2c)(3ab^2+2c)$$

### Method 4 : Special Form - B

The second special case of factoring is the Perfect Square Trinomial

$$a^2 \pm 2ab + b^2 = (a \pm b)^2$$

Example 07: Factor $100 + 20x + x^2$

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

$$\color{blue}{100} + 20x + \color{red}{x^2} = \color{blue}{10}^2 + 2 \cdot \color{blue}{10} \cdot \color{red}{x} + \color{red}{x}^2 = (10 + x)^2$$

### Method 5: Factoring Quadratic Polynomials

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

Example 08: Factor $x^2 + 3x + 4$

We know that the factored form has the following pattern

$$x^2 + 5x + 4 = (x + \_ ) (x + \_ )$$

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 $\color{blue}{1}$ and $\color{red}{4}$. As a result, the solution is::

$$x^2 + 5x + 4 = (x + \color{blue}{1} ) (x + \color{red}{4} )$$

Example 09: Factor $x^2 - 8x + 15$

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

$$x^2 - 8x + 15 = (x + \_ ) (x + \_ )$$

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's not hard to see that two numbers with such properties are $-3$ and $-5$, so the solution is.

$$x^2 + 5x + 4 = (x + \color{blue}{-3} ) (x + \color{red}{-5} ) = (x-3)(x-5)$$
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