How to factor polynomials: Difference between revisions

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#REDIRECT [[Factorization of polynomials]]

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The methods that are used to factor a polynomial depend on how many terms the polynomial has.<br /> Note: This page assumes that the polynomials are being factored on the real field (that the answers are real numbers).

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== Any Polynomial ==

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<br />

The first step to factor '''any''' polynomial is '''always''' to factor out the '''Greatest Common Factor''', commonly referred to as the '''GCF'''.<br /><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>

<br /><br />

== 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===

<math>x^2-y^2=(x+y)(x-y)</math><br /><br />

'''For example:'''<br />

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

<br />or<br /><br />

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

<br /><br />

===Sum of Cubes===

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

'''For example:'''<br />

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

<br />or<br /><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===

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

'''For example:'''<br />

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

<br />or<br /><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.<br />

===Monic Trinomials===

A monic trinomial has 1 as the leading coefficient.<br />

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

where <math>(e)(d)=c</math> and <math>d+e=b</math>.<br /><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 />

<br />or<br /><br />

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

===Non-Monic Trinomials===

A non-monic trinomial has a constant other than 1 as the leading coefficient.<br /><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 /><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 /><br />

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

'''Step 1'''&nbsp;&nbsp;

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><br /><br />

'''Step 2'''&nbsp;&nbsp;

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

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

'''Step 3'''&nbsp;&nbsp;

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 /><br />

'''Step 4'''&nbsp;&nbsp;

Rewrite the middle term of '''bx''' using the factors found in step 3.<br />

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

'''Step 5'''&nbsp;&nbsp;

Factor by grouping.<br />

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

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

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

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

===Perfect Square Trinomials===

Perfect square trinomials are of the form <math>a^2+2ab+c^2</math> or <math>a^2-2ab+c^2</math><br />

<math>a^2+2ab+c^2=(a+c)^2</math> and<br />

<math>a^2-2ab+c^2=(a-c)^2</math><br /><br />

'''For example:'''<br />

<math>4h^2-20h+25=(2h)^2-2(2h)(5)+(5)^2=(2h-5)^2</math><br />

<br />or<br /><br />

<math>100x^2+180xy+81y^2=(10x)^2+2(10x)(9y)+(9y)^2=(10x+9y)^2</math><br />

<br /><br />

==Polynomials with 4 terms==

Polynomials with 4 terms can be factored by some form of grouping. There are special groupings but the most common form is referred to as '''factoring by grouping''' and is described step by step below.<br /><br />

'''Example:''' <br /><math>a^2-3ab+4ac-12bc</math><br /><br />

'''Step 1''' &nbsp;&nbsp;Split the polynomial into groups of 2 terms.<br />

<math>(a^2-3ab)+(4ac-12bc)</math><br /><br />

'''Step 2''' &nbsp;&nbsp;Find the GCF (greatest common factor) of each group.<br />

<math>a(a-3b)+4c(a-3b)</math><br /><br />

'''Step 3''' &nbsp;&nbsp;If the 'leftovers match' factor them out.<br />

Since there is a (a-3b) in each term, factor out (a-3b) from each term.<br />

<math>a\mathbf{(a-3b)}+4c\mathbf{(a-3b)}=(a-3b)(a+4c)</math><br /><br />

Therefore <math>a^2-3ab+4ac-12bc=(a-3b)(a+4c)</math><br /><br /><br />

==When nothing works==

If the polynomial can't be factored, then it is considered '''prime'''.

<br /><br />

==References==

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

{{cite book

| last1 = Lial | first1 = Margaret L.

| first2 =John | last2 = Hornsby

| first3 =Terry | last3 = McGinnis

| title = [[Intermediate Algebra 10th edition]]

| ___location = Boston

| publisher = [[Addison Wesley]]

| year= 2008

| pages = 346-388

| isbn = 978-0321443625 }}

<br />

{{cite book

| last1 = Tussy | first1 = Alan S.

| first2 =R. David | last2 = Gustafson

| title = [[Intermediate Algebra 4th edition]]

| publisher = [[Brooks Cole]]

| year= 2008

|month = March

| isbn = 978-0495389736 }}

<references/>

==External links==

* [http://www.purplemath.com/modules/factquad.htm Factoring Quadratics: The Simple Case]

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