Factoring Polynomials - Revision history

Lila: /* Licensing */

2021-10-21T17:27:16Z

Licensing

Lila: /* The quadratic formula */

2021-10-21T17:25:37Z

The quadratic formula

Lila: /* Remainder and Factor Theorem */

2021-10-21T17:23:31Z

Remainder and Factor Theorem

Lila: /* Remainder and Factor Theorem */

2021-10-21T17:22:35Z

Remainder and Factor Theorem

Lila: /* Resources */

2021-10-21T17:20:30Z

Resources

Lila: /* Factor Theorem */

2021-10-04T17:23:52Z

Factor Theorem

Lila at 17:23, 4 October 2021

2021-10-04T17:23:01Z

Lila at 16:12, 30 September 2021

2021-09-30T16:12:21Z

Lila: /* Resources and Examples */

2021-09-15T15:22:17Z

Resources and Examples

Lila: /* Resources and Examples */

2021-09-14T23:47:50Z

Resources and Examples

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 == Licensing ==
-Content obtained from:
+Content obtained and/or adapted from:
-* [] under a CC BY-SA license
+* [https://en.wikibooks.org/wiki/Calculus/Algebra Algebra, Wikibooks: Calculus] under a CC BY-SA license

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 === The quadratic formula ===
 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>
-:<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.
+:<math>x=\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}</math>
 {| class="wikitable" style="margin: 0 auto;"
 !If

@@ Line 76: / Line 76: @@
 : 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 Problme====
+====Factor Theorem Example Problem====
 : Determine if x + 2 is a factor of <math>2x^2 + 3x -2</math>.

@@ Line 69: / Line 69: @@
 ===Remainder Theorem===
-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 d===ivided by <math> (x - c) </math> the remainder is <math> p(c) </math>.
+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>.

← Older revision		Revision as of 17:20, 21 October 2021
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	* [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 from:
		+	* [] under a CC BY-SA license

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 |There are no real solutions for the equation
 |}
 ==Example Problems==

← Older revision		Revision as of 16:12, 30 September 2021
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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>
		+
		+	<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>
		+
		+	=== 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
		+
		+	<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 ===
		+	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>
		+	:<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;"
		+	!If
		+	!Then
		+	\|-
		+	\|<math>b^2-4ac>0</math>
		+	\|There are two real solutions for the equation
		+	\|-
		+	\|<math>b^2-4ac=0</math>
		+	\|There are only one real solutions for the equation
		+	\|-
		+	\|<math>b^2-4ac<0</math>
		+	\|There are no real solutions for the equation
		+	\|}
		+
		+	==Example Problems==
		+	EXAMPLE 1: Find all the roots of <math>4x^2+7x-2</math>
		+	: 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>
		+	:: <math>x=\frac{-7\pm\sqrt{7^2-4(4)(-2)}}{2(4)}</math>
		+
		+	:: <math>x=\frac{-7\pm\sqrt{49+32}}{8}</math>
		+
		+	:: <math>x=\frac{-7\pm\sqrt{81}}{8}</math>
		+
		+	:: <math>x=\frac{-7\pm9}{8}</math>
		+
		+	:: <math>x=\frac{2}{8}\ ,\ x=\frac{-16}{8}</math>
		+
		+	:: <math>x=\frac{1}{4}\ ,\ x=-2</math>
		+
		+	: The quadratic formula can also help with factoring, as the next example demonstrates.
		+
		+	EXAMPLE 2: Factor the polynomial <math>4x^2+7x-2</math>
		+	: 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==		==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

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-==Resources and Examples==
+==Resources==
 * [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

← Older revision		Revision as of 23:47, 14 September 2021
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	* [https://www.youtube.com/watch?v=shHS-ERFVrc Difference of Squares]. Produced by TA Catherine Sporer, UTSA		* [https://www.youtube.com/watch?v=shHS-ERFVrc Difference of Squares]. Produced by TA Catherine Sporer, UTSA
	* [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