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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.

**solution**

**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:

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.

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Examples

ex 1:

Factor x^{2}-10x+25

ex 2:

Factor x^{2}-27

ex 3:

Factor 4x^{3}-21x^{2}+29x-6

ex 4:

Factor 3uv-12u+6v-24

ex 5:

Factor a^{2}+2a+1-b^{2}

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TUTORIAL

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.

**Example 01:** Factor 3ab^{3}-6a^{2}b

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

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 = 2a-4b + a^{2}-2ab

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

2a-4b+a^{2}-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)

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.

4x^{2}-y^{2}=(2x)^{2}-y^{2}

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

(2x)^{2}-y^{2}=(2x-y)(2x+y)

**Example 06:** Factor 9a^{2}b^{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^{2}b^{4}-4c^{2}=(3ab^{2})^{2}-4c^{2}= =(3ab^{2}-2c)(3ab^{2}+2c)

The second special case of factoring is the **Perfect Square Trinomial**

a^{2} ± 2ab + b^{2} = (a ± 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.

100+20x+x^{2}= 10^{2}+2·10·x+x^{2}= = (10+x)^{2}

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 1 and 4. As a result, the solution is::

x^2+5x+4 = (x+1)(x+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 is not hard to see that two numbers with such properties are -3 and -5, so the solution is.

x^{2}+5x+4 = (x+-3)(x+-5) = = (x-3)(x-5)

RESOURCES

1. Polynomial factoring tutorial — factoring out greatest common factor, factoring by grouping, quadratics and polynomials with degree greater than 2.

2. Youtube tutorial demonstrates some of the most important polynomial factoring techniques.

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