Difference between revisions of "Factoring Polynomials"

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==Resources and Examples==
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==Useful Formulas==
 +
Computing factors of polynomials requires knowledge of different formulas
 +
and some experience to find out which formula to be applied. Below, we give
 +
some important formulas:
 +
 
 +
<math>{x^2-y^2=(x+y)(x-y)}</math>
 +
 
 +
<math>{x^2+2xy+y^2=(x+y)^2}</math>
 +
 
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<math>{x^2-2xy+y^2=(x-y)^2}</math>
 +
 
 +
<math>{x^3-y^3=(x-y)(x^2+xy+y^2)}</math>
 +
 
 +
<math>{x^3+y^3=(x+y)(x^2-xy+y^2)}</math>
 +
 
 +
<math>{a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)}</math>
 +
 
 +
==Methods==
 +
Given the expression <math>x^2+3x+2</math> , one may ask "what are the values of <math>x</math> that make this expression 0?" If we factor we obtain
 +
<center><math>x^2+3x+2=(x+2)(x+1)</math></center> .
 +
If <math>x=-1,-2</math> , then one of the factors on the right becomes zero. Therefore, the whole must be zero. So, by factoring we have discovered the values of <math>x</math> that render the expression zero. These values are termed "roots." In general, given a quadratic polynomial <math>px^2+qx+r</math> that factors as
 +
<center><math>px^2+qx+r=(ax+c)(bx+d)</math></center>
 +
then we have that <math>x=-\frac{c}{a}</math> and <math>x=-\frac{d}{b}</math> are roots of the original polynomial.
 +
 
 +
A special case to be on the look out for is the difference of two squares, <math>a^2-b^2</math> . In this case, we are always able to factor as
 +
<center><math>a^2-b^2=(a+b)(a-b)</math></center>For example, consider <math>4x^2-9</math> . On initial inspection we would see that both <math>4x^2</math> and <math>9</math> are squares of <math>2x</math> and <math>3</math>, respectively. Applying the previous rule we have
 +
<center><math>4x^2-9=(2x+3)(2x-3)</math></center>
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 +
=== The AC method ===
 +
 
 +
There is a way of simplifying the process of factoring using the AC method. Suppose that a quadratic polynomial has a formula of
 +
 
 +
<center><math>ax^2+bx+c</math></center>
 +
 
 +
If there are numbers <math>r</math> and <math>q</math> that satisfy both
 +
 
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<center><math>r\cdot q=ac</math> and <math>r+q=b</math></center>
 +
 
 +
Then, we can rewrite the polynomial as
 +
 
 +
<center><math>ax^2+bx+c=ax^2 + rx + qx + c</math></center>
 +
 
 +
and factor out a common term from <math>(ax^2 + rx)</math> and <math>(qx + c)</math> to factor the polynomial.
 +
 
 +
 
 +
For example, take the polynomial <math> 2x^2 + 7x - 15 </math>.
 +
 
 +
<math> ac = 2(-15) = -30</math>. <math> 10(-3) = -30 = ac </math> and <math> 10 + (-3) = 7 = b</math>, so we can rewrite the polynomial as <math> 2x^2 + 10x - 3x - 15 </math>. From here, we can factor <math> 2x </math> from the first two terms, and <math> -3 </math> from the last two, which gives us <math> 2x(x + 5) + (-3)(x + 5) = (2x - 3)(x + 5) </math>. Thus, the factored form of <math> 2x^2 + 7x - 15 </math> is <math> (2x - 3)(x + 5) </math>. Note that if we reverse the order of <math>10x</math> and <math>-3x</math>, we get <math> 2x^2 -3x + 10x - 15 = x(2x - 3) + 5(2x - 3) = (x + 5)(2x - 3)</math>, which is the same factored form that we got previously. Thus, the order of <math>rx</math> and <math>qx</math> should not matter when using the AC method to factor a quadratic polynomial.
 +
 
 +
=== The quadratic formula ===
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Given any quadratic equation <math>ax^2+bx+c=0\ ,\ a\ne0</math>, all solutions of the equation are given by the quadratic formula:</br>
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:<math>x=\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}</math>
 +
 
 +
Note that the value of <math>b^2-4ac</math> will affect the number of ''real'' solutions of the equation.
 +
{| class="wikitable" style="margin: 0 auto;"
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!If
 +
!Then
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|-
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|<math>b^2-4ac>0</math>
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|There are two real solutions for the equation
 +
|-
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|<math>b^2-4ac=0</math>
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|There are only one real solutions for the equation
 +
|-
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|<math>b^2-4ac<0</math>
 +
|There are no real solutions for the equation
 +
|}
 +
 
 +
==Remainder and Factor Theorem==
 +
The polynomial division algorithm is as follows: suppose <math> d(x) </math> and <math> p(x) </math> are nonzero polynomials where the degree of <math> p(x) </math> is greater than or equal to the degree of <math> d(x) </math>. Then there exist two unique polynomials, <math> q(x) </math> and <math> r(x) </math>, such that  <math> p(x) = d(x)q(x) + r(x) </math>, where either <math> r(x) = 0 </math> or the degree of <math> r(x) </math> is strictly less than the degree of <math> d(x) </math>.
 +
 
 +
===Remainder Theorem===
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Suppose <math> p(x) </math> is a polynomial of degree at least 1 and c is a real number. When <math> p(x) </math> is divided by <math> (x - c) </math> the remainder is <math> p(c) </math>.
 +
: Proof: By the division algorithm, <math> p(x) = (x - c)q(x) + r </math>, where r must be a constant since <math> d(x) = x - c </math> has a degree of 1. <math> p(x) = (x - c)q(x) + r </math> must hold for all values of <math> x </math>, so we can set <math> x = c </math> and get that <math> p(c) = (c - c)q(x) + r = r </math>. Thus the remainder <math> r = p(c) </math>.
 +
 
 +
===Factor Theorem===
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Suppose <math> p(x) </math> is a nonzero polynomial. The real number <math> c </math> is a zero of <math> p(x) </math> if and only if <math> (x - c) </math> is a factor of <math> p(x) </math>.
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: By the division algorithm, <math> x - c </math> is a factor of <math> p(x) </math> if and only if <math> r = 0 </math>. So, since <math> p(c) = r </math> when <math> p(x) </math> is divided by <math> x - c </math>, <math> x - c </math> is a factor of <math> p(x) </math> if and only if <math> p(c) = 0 </math>; that is, if <math> c </math> is a zero of <math> p(x) </math>.
 +
 
 +
====Factor Theorem Example Problem====
 +
 
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: Determine if x + 2 is a factor of <math>2x^2 + 3x -2</math>.
 +
 
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Since c is positive instead of negative we need to use this basic identity:
 +
 
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<math>x + 2 = x - \left ( - 2 \right )</math>
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Now we can use the factor theorem.
 +
 
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<math>2 \left (-2 \right )^2 + 3 \left (-2 \right ) -2 = 8 - 6 - 2 = 0</math>.
 +
 
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Since the resultant is 0, <math>(x+2)</math> is a factor of <math>2x^2 + 3x -2</math>.
 +
 
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This means it is possible to re-state the polynomial in the form (x+2)( some linear expression of x).
 +
 +
So <math>2x^2 + 3x -2 = (x+2)(ax+b)</math>
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Expanding the right hand side we get :
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<math>2x^2 + 3x -2 = ax^2 + x( 2a+b) +2b</math>
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Equating like terms we get :
 +
 
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<math>2= a</math>
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<math> 2a+b = 3 </math>, and
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<math> 2b = -2 </math>
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Giving <math>a = 2</math>, <math> b= -1 </math> from the first and third equations and this works in the second, so
 +
 
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<math>2x^2 + 3x -2 = (x+2)(2x-1)</math>
 +
 
 +
==Example Problems==
 +
EXAMPLE 1: Find all the roots of <math>4x^2+7x-2</math>
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: Finding the roots is equivalent to solving the equation <math>4x^2+7x-2=0</math> . Applying the quadratic formula with <math>a=4\ ,\ b=7\ ,\ c=-2</math> , we have:</br>
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:: <math>x=\frac{-7\pm\sqrt{7^2-4(4)(-2)}}{2(4)}</math>
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 +
:: <math>x=\frac{-7\pm\sqrt{49+32}}{8}</math>
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 +
:: <math>x=\frac{-7\pm\sqrt{81}}{8}</math>
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:: <math>x=\frac{-7\pm9}{8}</math>
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:: <math>x=\frac{2}{8}\ ,\ x=\frac{-16}{8}</math>
 +
 
 +
:: <math>x=\frac{1}{4}\ ,\ x=-2</math>
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 +
: The quadratic formula can also help with factoring, as the next example demonstrates.
 +
 
 +
EXAMPLE 2: Factor the polynomial <math>4x^2+7x-2</math>
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: We already know from the previous example that the polynomial has roots <math>x=\frac{1}{4}</math> and <math>x=-2</math> . Our factorization will take the form <br/> <math>C(x+2)\left(x-\tfrac{1}{4}\right)</math><br/>.
 +
: All we have to do is set this expression equal to our polynomial and solve for the unknown constant C:</br>
 +
:: <math>C(x+2)\left(x-\tfrac{1}{4}\right)=4x^2+7x-2</math>
 +
 
 +
:: <math>C\left(x^2+\left(-\tfrac{1}{4}+2\right)x-\tfrac{2}{4}\right)=4x^2+7x-2</math>
 +
 
 +
:: <math>C\left(x^2+\tfrac{7}{4}x-\tfrac{1}{2}\right)=4x^2+7x-2</math>
 +
 
 +
: You can see that <math>C=4</math> solves the equation. So the factorization is</br>
 +
:: <math>4x^2+7x-2=4(x+2)\left(x-\tfrac{1}{4}\right)=(x+2)(4x-1)</math>
 +
 
 +
EXAMPLE 3: Factor the polynomial <math> 6x^2 + 7x + 2 </math>.
 +
: For this problem, let's use the AC method. <math> a = 6 </math>, <math> b = 7 </math>, and <math> c = 2 </math>. So, <math> ac = 6(2) = 12 </math>, and we need to find two numbers whose product equals 12, and whose sum equals 7. Both ac and b are positive, so we are only concerned with the positive factors of 12, which are <math> 1, 2, 3, 4, 6, </math>and <math> 12 </math>. <math> 3(4) = 12 = ac </math> and <math> 3 + 4 = 7 = b </math>, so we can rewrite the polynomial as <math> 6x^2 + 3x + 4x + 2 = (3x)(2x + 1) = 2(2x + 1) = (3x + 2)(2x+1)</math>.
 +
 
 +
 
 +
==Resources==
 
* [https://tutorial.math.lamar.edu/classes/alg/factoring.aspx Factoring Polynomials with Example Problems], Paul's Online Notes
 
* [https://tutorial.math.lamar.edu/classes/alg/factoring.aspx Factoring Polynomials with Example Problems], Paul's Online Notes
 
* [https://courses.lumenlearning.com/suny-beginalgebra/chapter/5-2-1-factor-trinomials/ Factoring Polynomials], Lumen Learning
 
* [https://courses.lumenlearning.com/suny-beginalgebra/chapter/5-2-1-factor-trinomials/ Factoring Polynomials], Lumen Learning
Line 5: Line 151:
 
* [http://www.algebralab.org/lessons/lesson.aspx?file=algebra_factoring.xml 6 Methods of Factoring with Practive Problems], AlgebraLAB
 
* [http://www.algebralab.org/lessons/lesson.aspx?file=algebra_factoring.xml 6 Methods of Factoring with Practive Problems], AlgebraLAB
 
* [https://txwes.edu/media/twu/content-assets/images/academics/academic-success-center/A-C-Method.pdf AC Factoring Method], Texas Wesleyan University
 
* [https://txwes.edu/media/twu/content-assets/images/academics/academic-success-center/A-C-Method.pdf AC Factoring Method], Texas Wesleyan University
 +
 +
== Licensing ==
 +
Content obtained and/or adapted from:
 +
* [https://en.wikibooks.org/wiki/Calculus/Algebra Algebra, Wikibooks: Calculus] under a CC BY-SA license
 +
* [https://en.wikibooks.org/wiki/Algebra/Factoring_Polynomials Factoring Polynomials, Wikibooks: Algebra] under a CC BY-SA license

Latest revision as of 11:27, 21 October 2021

Useful Formulas

Computing factors of polynomials requires knowledge of different formulas and some experience to find out which formula to be applied. Below, we give some important formulas:

Methods

Given the expression , one may ask "what are the values of that make this expression 0?" If we factor we obtain

.

If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=-1,-2} , then one of the factors on the right becomes zero. Therefore, the whole must be zero. So, by factoring we have discovered the values of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} that render the expression zero. These values are termed "roots." In general, given a quadratic polynomial Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle px^2+qx+r} that factors as

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle px^2+qx+r=(ax+c)(bx+d)}

then we have that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=-\frac{c}{a}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=-\frac{d}{b}} are roots of the original polynomial.

A special case to be on the look out for is the difference of two squares, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^2-b^2} . In this case, we are always able to factor as

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^2-b^2=(a+b)(a-b)}

For example, consider Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 4x^2-9} . On initial inspection we would see that both Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 4x^2} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 9} are squares of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2x} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3} , respectively. Applying the previous rule we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 4x^2-9=(2x+3)(2x-3)}

The AC method

There is a way of simplifying the process of factoring using the AC method. Suppose that a quadratic polynomial has a formula of

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ax^2+bx+c}

If there are numbers Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q} that satisfy both

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r\cdot q=ac} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r+q=b}

Then, we can rewrite the polynomial as

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ax^2+bx+c=ax^2 + rx + qx + c}

and factor out a common term from Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (ax^2 + rx)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (qx + c)} to factor the polynomial.


For example, take the polynomial Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2x^2 + 7x - 15 } .

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ac = 2(-15) = -30} . Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 10(-3) = -30 = ac } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 10 + (-3) = 7 = b} , so we can rewrite the polynomial as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2x^2 + 10x - 3x - 15 } . From here, we can factor Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2x } from the first two terms, and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -3 } from the last two, which gives us Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2x(x + 5) + (-3)(x + 5) = (2x - 3)(x + 5) } . Thus, the factored form of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2x^2 + 7x - 15 } is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (2x - 3)(x + 5) } . Note that if we reverse the order of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 10x} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -3x} , we get Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2x^2 -3x + 10x - 15 = x(2x - 3) + 5(2x - 3) = (x + 5)(2x - 3)} , which is the same factored form that we got previously. Thus, the order of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle rx} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle qx} should not matter when using the AC method to factor a quadratic polynomial.

The quadratic formula

Given any quadratic equation Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ax^2+bx+c=0\ ,\ a\ne0} , all solutions of the equation are given by the quadratic formula:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}}

Note that the value of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b^2-4ac} will affect the number of real solutions of the equation.

If Then
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b^2-4ac>0} There are two real solutions for the equation
There are only one real solutions for the equation
There are no real solutions for the equation

Remainder and Factor Theorem

The polynomial division algorithm is as follows: suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d(x) } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) } are nonzero polynomials where the degree of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) } is greater than or equal to the degree of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d(x) } . Then there exist two unique polynomials, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q(x) } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r(x) } , such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) = d(x)q(x) + r(x) } , where either Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r(x) = 0 } or the degree of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r(x) } is strictly less than the degree of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d(x) } .

Remainder Theorem

Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) } is a polynomial of degree at least 1 and c is a real number. When Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) } is divided by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x - c) } the remainder is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(c) } .

Proof: By the division algorithm, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) = (x - c)q(x) + r } , where r must be a constant since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d(x) = x - c } has a degree of 1. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) = (x - c)q(x) + r } must hold for all values of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x } , so we can set Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = c } and get that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(c) = (c - c)q(x) + r = r } . Thus the remainder Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r = p(c) } .

Factor Theorem

Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) } is a nonzero polynomial. The real number Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c } is a zero of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) } if and only if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x - c) } is a factor of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) } .

By the division algorithm, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x - c } is a factor of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) } if and only if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r = 0 } . So, since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(c) = r } when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) } is divided by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x - c } , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x - c } is a factor of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) } if and only if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(c) = 0 } ; that is, if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c } is a zero of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) } .

Factor Theorem Example Problem

Determine if x + 2 is a factor of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2x^2 + 3x -2} .

Since c is positive instead of negative we need to use this basic identity:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x + 2 = x - \left ( - 2 \right )}

Now we can use the factor theorem.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2 \left (-2 \right )^2 + 3 \left (-2 \right ) -2 = 8 - 6 - 2 = 0} .

Since the resultant is 0, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x+2)} is a factor of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2x^2 + 3x -2} .

This means it is possible to re-state the polynomial in the form (x+2)( some linear expression of x).

So Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2x^2 + 3x -2 = (x+2)(ax+b)}

Expanding the right hand side we get :

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2x^2 + 3x -2 = ax^2 + x( 2a+b) +2b}

Equating like terms we get :

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2= a}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2a+b = 3 } , and

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2b = -2 }

Giving Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a = 2} , from the first and third equations and this works in the second, so

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2x^2 + 3x -2 = (x+2)(2x-1)}

Example Problems

EXAMPLE 1: Find all the roots of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 4x^2+7x-2}

Finding the roots is equivalent to solving the equation Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 4x^2+7x-2=0} . Applying the quadratic formula with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a=4\ ,\ b=7\ ,\ c=-2} , we have:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=\frac{-7\pm\sqrt{7^2-4(4)(-2)}}{2(4)}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=\frac{-7\pm\sqrt{49+32}}{8}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=\frac{-7\pm\sqrt{81}}{8}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=\frac{-7\pm9}{8}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=\frac{2}{8}\ ,\ x=\frac{-16}{8}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=\frac{1}{4}\ ,\ x=-2}
The quadratic formula can also help with factoring, as the next example demonstrates.

EXAMPLE 2: Factor the polynomial Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 4x^2+7x-2}

We already know from the previous example that the polynomial has roots Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=\frac{1}{4}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=-2} . Our factorization will take the form
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C(x+2)\left(x-\tfrac{1}{4}\right)}
.
All we have to do is set this expression equal to our polynomial and solve for the unknown constant C:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C(x+2)\left(x-\tfrac{1}{4}\right)=4x^2+7x-2}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C\left(x^2+\left(-\tfrac{1}{4}+2\right)x-\tfrac{2}{4}\right)=4x^2+7x-2}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C\left(x^2+\tfrac{7}{4}x-\tfrac{1}{2}\right)=4x^2+7x-2}
You can see that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C=4} solves the equation. So the factorization is
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 4x^2+7x-2=4(x+2)\left(x-\tfrac{1}{4}\right)=(x+2)(4x-1)}

EXAMPLE 3: Factor the polynomial Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 6x^2 + 7x + 2 } .

For this problem, let's use the AC method. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a = 6 } , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b = 7 } , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c = 2 } . So, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ac = 6(2) = 12 } , and we need to find two numbers whose product equals 12, and whose sum equals 7. Both ac and b are positive, so we are only concerned with the positive factors of 12, which are Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1, 2, 3, 4, 6, } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 12 } . Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3(4) = 12 = ac } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3 + 4 = 7 = b } , so we can rewrite the polynomial as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 6x^2 + 3x + 4x + 2 = (3x)(2x + 1) = 2(2x + 1) = (3x + 2)(2x+1)} .


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