Rules For Factoring Polynomials

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Factoring a polynomial is the opposed procedure of multiply polynomials. When we factor a number, we are appear for prime factors to multiplying together to provide the number; for example 16 = 2 8, otherwise 16 = 2 x 2x 4. When we factor a polynomial, we are generally just concerned within breaking it along into polynomials to contain integer coefficients also constants.

Rules for factoring polynomials

Factoring polynomials having the following rules,

a2 – b2 = (a + b) (a – b)

a3 – b3 = (a – b) (a2 + ab + b2)

a3 + b3 = (a + b) (a2 – ab + b2)

These rules are most commonly used in factoring polynomials

The easiest kind of factoring is when there is a factor general to all term. In that case, we will be able to factor out general factor.

Using distributive law,

a(b + c) = ab + ac.

It can be written in reverse means, if you observe ab + ac, you can inscribed as a(b + c).

Consider the trinomial approach from multiplying two first-degree binomials.

Consider, (x + 2)(x + 4)

Using the FOIL technique, we obtain

(x + 2)(x + 4) = x2 + 4x + 2x +8

Then, grouping similar terms we obtain,

(x + 2)(x + 4) = x2 + 6x + 8


3×2 + 6x

In this example all term contain a factor of 3x, therefore we can rewrite it as:

3×2 + 6x = 3x(x + 2)

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x2 – 9 = (x – 3)(x + 3)

This only contain for a difference of two squares. There is no method toward factor a total of two squares such as a2 + b2 into factors through real numbers.

Examples for rules for factoring polynomials

Example 1 for rules for factoring polynomials.

Solve 25×2 – 256 = 0


Solve this through factoring.

Apply rule then we can write,

a2 – b2 = (a + b) (a – b)

25×2 – 256 =0

(5x)2 = 162

And we can apply the rule with a = 5x and b = 16 to obtain

25×2 – 256 = (5x + 16) (5x -16)

And we place this equivalent to zero:

(5x + 16) (5x – 16) = 0

Therefore, 5x + 16 = 0


5x – 16 = 0

Therefore that x = -16/5 and x = -16/5 are the solutions.

Example 2 for rules for factoring polynomials

Simplify x24 + 1.


Exponent 24 is a multiple of 3.

Using rules of exponents we get the following.

(x8)3 = x(8×3) = x24

So we can write x24= (x8)3 .

x8can fit within the position of factoring rule.

Since 1 also can be written as 13

=> So we can written the given problem, x24+ 1 = (x8)3 + 13

Using the factoring polynomial rule a3 + b3 = (a + b) (a2 – ab + b2)

Therefore, (x8)3 + 13= (x8 + 1) ((x8)2 – x8 + 1)

This can be simplified we get,

(x8 + 1) (x16 – x8 + 1)

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Updated: March 19, 2018 — 5:03 am
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