How to factor polynomials: Difference between revisions

Revision as of 01:22, 8 October 2009

Latest revision as of 22:17, 23 August 2025

(78 intermediate revisions by 23 users not shown)

Line 1:

{{<includeonly>safesubst:</includeonly>#invoke:RfD||2=How-to redirects|month = August

{{Userspace draft|date=October 2009}}

|day = 14

|year = 2025

|time = 19:25

|timestamp = 20250814192547

<!-- The above content is generated by {{subst:Rfd}}. -->

<!-- End of RFD message. Don't edit anything above here. Feel free to edit below here, but do NOT change the redirect's target. -->|content=

#REDIRECT [[Factorization of polynomials]]

{{Redirect category shell|

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

{{R with history}}

{{R from how-to name}}

}}

== Any Polynomial ==

<!-- Don't add anything after this line unless you're drafting a disambiguation page or article to replace the redirect. -->

<br />

}}

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===

A monic trinomial has 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===

A non-monic trinomial has 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 />

Example: Factor <math>6x^2+7x-3</math>

====Step 1====

Multiply '''a''' and '''c'''. (Multiply the number in front of <math>x^2</math> and the [[constant term|constant]])<br />

Multipy 6 and -3, <math>6(-3)=-18</math>

====Step 2====

Find factors of '''ac'''.<br />

Find factors of -18: -1(18), 1(-18), -2(9), 2(-9), -3(6), and 3(-6).<br />

====Step 3====

Decide which factors of '''ac''' that when added together will give '''b'''.<br />

The combination of -2 and 9 is the one needed since -2+9=7.<br />

====Step 4====

Rewrite the middle term of '''bx''' using the factors found in step 3.

Instead of <math>7x</math>, write <math>6x^2-2x+9x-18</math><br />

====Step 5====

Factor by grouping.

<math>(6x^2-2x)+(9x-3)</math><br />

<math>2x(3x-1)+3(3x-1)</math>

<math>(3x-1)(2x+3)</math>

Therefore <math>6x^2+7x-3=(3x-1)(2x+3)</math>

==References==

<!--- See [[Wikipedia:Footnotes]] on how to create references using <ref></ref> tags which will then appear here automatically -->

<references/>

==External links==

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

<!--- Categories --->

[[Category:Articles created via the Article Wizard]]