Lila: /* Resources */

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Resources

Lila: /* Example 5 */

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Example 5

Lila: /* Example Problem 4 */

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Example Problem 4

Lila: /* Example Problem 4 */

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Example Problem 4

Lila: /* Example Problem 2 */

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Example Problem 2

Lila: /* Example 5 */

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Example 5

Lila: /* Trinomials */

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Trinomials

Lila: /* =Example Problem 3 */

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=Example Problem 3

Lila: /* =Example Problem 3 */

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=Example Problem 3

Lila: /* =Example Problem 3 */

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=Example Problem 3

← Older revision		Revision as of 19:19, 21 October 2021
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	* [https://courses.lumenlearning.com/ivytech-collegealgebra/chapter/use-arrow-notation/ Arrow Notation for Polynomials]		* [https://courses.lumenlearning.com/ivytech-collegealgebra/chapter/use-arrow-notation/ Arrow Notation for Polynomials]
	* [https://mathresearch.utsa.edu/wikiFiles/MAT1053/More_on_Polynomial_and_Power_Functions/MAT1053_M4.1More_on_Power_Functions_and_Polynomial_FunctionsGN.pdf, Guided Notes]		* [https://mathresearch.utsa.edu/wikiFiles/MAT1053/More_on_Polynomial_and_Power_Functions/MAT1053_M4.1More_on_Power_Functions_and_Polynomial_FunctionsGN.pdf, Guided Notes]
		+
		+	== Licensing ==
		+	Content obtained and/or adapted from:
		+	* [https://en.wikibooks.org/wiki/CLEP_College_Algebra/Polynomials Polynomials, Wikibooks: CLEP College Algebra] under a CC BY-SA license

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 Therefore, <math>c=f_{2}</math>. This tells us a very important truth about factoring the above expression:<br>
-{{Equation|<math>f_{1}f_{2}=-2</math>|3.2.3.2|left}}
+<math>f_{1}f_{2}=-2</math>
 This is a result of applying our algorithm! Using what we know, let us rewrite the above problem in terms of what we know: <math>x^{2}+x-2=x^{2}+(f_{1}+f_{2})x-f_{1}f{2}</math>, where <math>-2=f_{1}f_{2}</math> and <math>1=f_{1}+f_{2}</math>. LOOK AT THIS! We can use a systems of equations!

@@ Line 99: / Line 99: @@
 :<math>xy\left(xy^{2}+2x+4y^{2}+8\right)</math>.
-We are not done however. Notice that you could still group the polynomial inside the parentheses with common terms. The <math>xy</math> may seem to complicate things; however, you can simply ignore it and say you are working on the inside of the parentheses. If you do get easily confused when too many things are on the paper or screen, it may be helpful to ignore them but always be sure to come back to them. Let us rewrite the expression inside of the parentheses in the same way we did in '''''Example 2.2.a''''':
+We are not done however. Notice that you could still group the polynomial inside the parentheses with common terms. The <math>xy</math> may seem to complicate things; however, you can simply ignore it and say you are working on the inside of the parentheses. If you do get easily confused when too many things are on the paper or screen, it may be helpful to ignore them but always be sure to come back to them. Let us rewrite the expression inside of the parentheses in the same way we did in '''''Example 3''''':
 :<math>{\color{Red}xy^{2}+2x}+{\color{Green}4y^{2}+8}</math>.

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 Keep in mind, this trick doesn't always work because the middle terms often end up having the same common variables and get simplified. It's helpful to start with the first terms and then go to the last terms. If you can get these two things you should be able to find the middle terms. Another thing to note is that we had to factor <math>xy</math> to make this work. While it could be possible to do the original problem like this, it is much easier to use FOIL when the last term does not have any variables. Before we move on to the next section, we must answer one final pressing question that we asked at the beginning of this section.
-'''Factoring by grouping: a trinomial then a binomial?''' This question is fundamental to the curious student. Let us try it out using '''''Example 2.2.a''''':
+'''Factoring by grouping: a trinomial then a binomial?''' This question is fundamental to the curious student. Let us try it out using '''''Example 3''''':
 :<math>x^3+x^2+2x+4</math>.

@@ Line 49: / Line 49: @@
 The coefficients should not matter if you factor because when multiplying the expression <math>\left(ca^{p}\right)a^{q}</math>, where <math>c</math> is some constant, as per the associative property, <math>\left(ca^{p}\right)a^{q}=c\left(a^{p}a^{q}\right)</math>. From there, multiplying one term by a constant does not matter: <math>c\left(a^{p}a^{q}\right)=c\left(a^{p+q}\right)=ca^{p+q}</math>.<sup>*</sup>
-Therefore, we can apply the same general rule we discovered for '''Example 2.1.a''':
+Therefore, we can apply the same general rule we discovered for '''Example 1''':
 <math display="block">42x^3+12x^2+24x=6x(7x^2+2x+4)\blacksquare</math>

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 Notice how our algorithm is confused at steps 3 and 4. Rightly, it should be because our algorithm assumes that we are factoring by grouping, which is not always applicable to all situations, such as this one. However, we can "force" the algorithm to fit into this situation. For example, we can try making the middle term split up into two terms. That is, we set <math>x=(f_{1}+f_{2})x=f_{1}x+f_{2}x</math>, where<br>
-{{Equation|<math>f_{1}+f_{2}=1</math>|3.2.3.1|left}}
+<math>f_{1}+f_{2}=1</math>
 However, if we are to get anywhere with this, we want the middle term to have common factors. This is an interesting dilemma!

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 As before, we will introduce the concept by exploring the consequences and behaviors that a mathematician may have first discovered techniques to solve. First, let us apply our algorithm and see if it applies to problems like these.
-{{ExampleRobox|theme=3|title='''Example 2.3.a''': Factor the expression <math>x^{2}+x-2</math>.}}
+====Example 5====
 # <math>\text{GCF}(1,1,-2)=1</math>.
 # We can only factor <math>x^{0}=1</math> for the following expression.
@@ Line 205: / Line 206: @@
   &=x(x+1)-2(x+1)\\
   &=(x-2)(x+1)\blacksquare
-\end{align}</math>{{Robox/Close}}
+\end{align}</math>
-'''Example 2.3.a''' showed us a very important algorithm that is taught all around the world. It is called the '''ac-method''' or the '''diamond method''', the steps for which we will outline below.
+'''Example 5''' showed us a very important algorithm that is taught all around the world. It is called the '''ac-method''' or the '''diamond method''', the steps for which we will outline below.
-{{Calculus/Def | title = ac-method of factoring trinomials |text = Given a trinomial of the form <math>ax^{2}+bx+c</math>, with <math>a\ne0</math>, the following procedure should be followed.
+'''AC-method of factoring trinomials'''
-# Find the greatest common factor for each of the coefficients.
+:Given a trinomial of the form <math>ax^{2}+bx+c</math>, with <math>a\ne0</math>, the following procedure should be followed.
-# If there is no constant (a value that has no variable attached to it), factor the variable to the lowest power value in the expression. If there is a constant, simply keep the greatest common factor outside the parentheses.
+:# Find the greatest common factor for each of the coefficients.
-# Find two values <math>f_{1}</math> and <math>f_{2}</math> whereby <math>ac=f_{1}f_{2}</math> and <math>b=f_{1}+f_{2}</math>.
+:# If there is no constant (a value that has no variable attached to it), factor the variable to the lowest power value in the expression. If there is a constant, simply keep the greatest common factor outside the parentheses.
-# Rewrite the trinomial into the form <math>ax^{2}+f_{1}x+f_{2}x+c</math>; then, group the polynomial into two binomials, and apply steps 1 and 2 to the different groupings. Do this separately for the two groupings.
+:# Find two values <math>f_{1}</math> and <math>f_{2}</math> whereby <math>ac=f_{1}f_{2}</math> and <math>b=f_{1}+f_{2}</math>.
-# After completely factoring the two binomials. Put them together. If there is a common factor with the two combined binomials, factor it.}}
+:# Rewrite the trinomial into the form <math>ax^{2}+f_{1}x+f_{2}x+c</math>; then, group the polynomial into two binomials, and apply steps 1 and 2 to the different groupings. Do this separately for the two groupings.
 You should have realized by now that the ac-method is a modified version of the factoring by grouping method. This is why we decided to show factoring trinomials later, because the work involved is actually more than what you would expect from a carefully designed polynomial.
 === Difference of Two Squares ===

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 When there is no common term that travels throughout the ''entire'' polynomial you can factor by grouping. When factoring by grouping you need to split up the polynomial into two binomials.
-=====Example Problem 3====
+====Example Problem 3====
 Factor the expression <math>x^3+x^2+2x+4</math>.
 Before we group, we need to do a quick check to see if we can factor the entire expression with just one term outside the parenthesis. If "4" was not a constant, we could factor the expression using the second step of the procedure. Another thing is that <math>\text{GCF}\left(1,1,2,4\right)=1</math>, which if we were to factor, that would mean that we are multiplying the same thing by 1, which is not very helpful.

More on Polynomial Functions - Revision history

Lila: /* Resources */

Lila: /* Example 5 */

Lila: /* Example Problem 4 */

Lila: /* Example Problem 4 */

Lila: /* Example Problem 2 */

Lila: /* Example 5 */

Lila: /* Trinomials */

Lila: /* =Example Problem 3 */

Lila: /* =Example Problem 3 */

Lila: /* =Example Problem 3 */