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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**

Both the first and third terms are perfect squares.

$$ x^2 = \left( \color{blue}{ x } \right)^2 ~~ \text{and} ~~ 100 = \left( \color{red}{ 10 } \right)^2 $$The middle term ( $ 20x $ ) is two times the product of the terms that are squared.

$$ 20x = 2 \cdot \color{blue}{x} \cdot \color{red}{10} $$We can conclude that the polynomial $ x^{2}+20x+100 $ is a **perfect square trinomial**, so we will use the formula below.

In this example we have $ \color{blue}{ A = x } $ and $ \color{red}{ B = 10 } $ so,

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

working...

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.

445 987 255 solved problems

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