<PRE><B>Question 367308</B>


{{{r^3+3r^2-54r}}} Start with the given expression.



{{{r(r^2+3r-54)}}} Factor out the GCF {{{r}}}.



Now let&#39;s try to factor the inner expression {{{r^2+3r-54}}}



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Looking at the expression {{{r^2+3r-54}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{3}}}, and the last term is {{{-54}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{-54}}} to get {{{(1)(-54)=-54}}}.



Now the question is: what two whole numbers multiply to {{{-54}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{3}}}?



To find these two numbers, we need to list &lt;font size=4&gt;&lt;b&gt;all&lt;/b&gt;&lt;/font&gt; of the factors of {{{-54}}} (the previous product).



Factors of {{{-54}}}:

1,2,3,6,9,18,27,54

-1,-2,-3,-6,-9,-18,-27,-54



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-54}}}.

1*(-54) = -54
2*(-27) = -54
3*(-18) = -54
6*(-9) = -54
(-1)*(54) = -54
(-2)*(27) = -54
(-3)*(18) = -54
(-6)*(9) = -54


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{3}}}:



&lt;table border=&quot;1&quot;&gt;&lt;th&gt;First Number&lt;/th&gt;&lt;th&gt;Second Number&lt;/th&gt;&lt;th&gt;Sum&lt;/th&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-54&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+(-54)=-53&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;2&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-27&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;2+(-27)=-25&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;3&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-18&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;3+(-18)=-15&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;6&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-9&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;6+(-9)=-3&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;54&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+54=53&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;27&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+27=25&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-3&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;18&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-3+18=15&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-6&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;9&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-6+9=3&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{-6}}} and {{{9}}} add to {{{3}}} (the middle coefficient).



So the two numbers {{{-6}}} and {{{9}}} both multiply to {{{-54}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{3}}}



Now replace the middle term {{{3r}}} with {{{-6r+9r}}}. Remember, {{{-6}}} and {{{9}}} add to {{{3}}}. So this shows us that {{{-6r+9r=3r}}}.



{{{r^2+highlight(-6r+9r)-54}}} Replace the second term {{{3r}}} with {{{-6r+9r}}}.



{{{(r^2-6r)+(9r-54)}}} Group the terms into two pairs.



{{{r(r-6)+(9r-54)}}} Factor out the GCF {{{r}}} from the first group.



{{{r(r-6)+9(r-6)}}} Factor out {{{9}}} from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.



{{{(r+9)(r-6)}}} Combine like terms. Or factor out the common term {{{r-6}}}



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So {{{r(r^2+3r-54)}}} then factors further to {{{r(r+9)(r-6)}}}



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Answer:



So {{{r^3+3r^2-54r}}} completely factors to {{{r(r+9)(r-6)}}}.



In other words, {{{r^3+3r^2-54r=r(r+9)(r-6)}}}.



Note: you can check the answer by expanding {{{r(r+9)(r-6)}}} to get {{{r^3+3r^2-54r}}} or by graphing the original expression and the answer (the two graphs should be identical).



If you need more help, email me at &lt;a href=&quot;mailto:jim_thompson5910@hotmail.com?Subject=Algebra%20Help&quot;&gt;jim_thompson5910@hotmail.com&lt;/a&gt;


Also, feel free to check out my &lt;a href=&quot;http://www.freewebs.com/jimthompson5910/home.html&quot;&gt;tutoring website&lt;/a&gt;


Jim</PRE>