How to factor polynomials: Difference between revisions

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#REDIRECT [[Factorization of polynomials]]

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The methods that you will use to factor any polynomial depend on how many terms the polynomial has.<br />

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== Any Polynomial ==

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The first step, no matter what you are factoring, is always to factor out the '''Greatest Common Factor''', commonly referred to as the '''GCF'''.<br />

For example:

:<math>2a^2b^3+4a^4b^4c^3-2a^3b^3=2a^2b^3(1+2a^2bc^3-a),\,\!</math>

<br />or<br />

: <math>3x^3-6x+9x^4=3x(x^2-2+3x^3),\,\!</math>

<br />or<br />

: <math>4x(x+2)+3x^2(x+2)=(x+2)(4x+3x^2),\,\!</math>

== Binomial--2 Terms ==

<br />

Again, the first step is to factor out the GCF. If there is no GCF, then there are only 3 possibilities:

'''Difference of Squares''', '''Sum of Cubes''', or '''Difference of Cubes'''.<br /><br/>

'''Difference of Squares:'''<br />

:<math>x^2-y^2=(x+y)(x-y),\,\!</math><br />

For example:

:<math>y^2-9=(y+3)(y-3),\,\!</math><br />

or<br />

:<math>16a^2-49b^2=(4a+7b)(4a-7b).\,\!</math>

<br /><br />

'''Sum of Cubes:'''<br />

:<math>x^3+y^3=(x+y)(x^2-xy+y^2),\,\!</math><br />

For example:

:<math>z^3+27=(z+3)(z^2-3z+9),\,\!</math><br />

or<br />

:<math>8x^3+125=(2x)^3+(5)^3=(2x+5)[(2x)^2-(5)(2x)+(5)^2]=(2x+5)(4x^2-10x+5).\,\!</math>

<br /><br />

'''Difference of Cubes:'''<br />

:<math>x^3-y^3=(x-y)(x^2+xy+y^2),\,\!</math><br />

For example:

:<math>z^3-27=(z-3)(z^2+3z+9),\,\!</math><br />

or<br />

:<math>8x^3-125=(2x)^3-(5)^3=(2x-5)[(2x)^2+(5)(2x)+(5)^2]=(2x-5)(4x^2+10x+5).\,\!</math>

<br /><br />

==Trinomial--3 Terms==

There are three possibilities for factoring a trinomial depending on which type of trinomial it is.

===Monic Trinomials===--There is a 1 as the leading coefficient.<br />

<math>x^2+bx+c=(x+d)(x+e),</math>

where

<br />

<math> ed=c </math>

and<br />

<math> d+e=b</math><br />

For example:<br />

<math>x^2-x-6=(x-3)(x+2)</math> because <math>(-3)(2)=-6</math> and <math>-3+2=-1</math><br />

<math>x^2-11x+24=(x-3)(x-8)</math> because <math>(-3)(-8)=24</math> and <math>-3+-8=-11</math>

<br /><br />

===Non-Monic Trinomials===--There is a constant other than 1 as the leading coefficient.<br />

<math>ax^2+bx+c=(mx+p)(nx+q)</math><br />

where <math>mn=a</math>, <math>pq=c</math>, and <math>mq+pn=b</math><br />

Many times students are taught that to factor a non-monic trinomials, they must guess different combinations of m,n,p,and q and then [[FOIL]] the factors to see if they had guessed correctly. There is a method of factoring that, while not often taught, will work.<br />

====Step 2==== Multiply '''a''' and '''c'''. (Multiply the number in front of <math>x^2</math> and the [[constant]]

==References==

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<references/>

==External links==

* [http://www.example.com example.com]

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