<PRE><B>Question 335982</B>
Please try to avoid writing in all caps. Thanks.





{{{18x^2-78x-60}}} Start with the given expression.



{{{6(3x^2-13x-10)}}} Factor out the GCF {{{6}}}.



Now let&#39;s try to factor the inner expression {{{3x^2-13x-10}}}



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Looking at the expression {{{3x^2-13x-10}}}, we can see that the first coefficient is {{{3}}}, the second coefficient is {{{-13}}}, and the last term is {{{-10}}}.



Now multiply the first coefficient {{{3}}} by the last term {{{-10}}} to get {{{(3)(-10)=-30}}}.



Now the question is: what two whole numbers multiply to {{{-30}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{-13}}}?



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 {{{-30}}} (the previous product).



Factors of {{{-30}}}:

1,2,3,5,6,10,15,30

-1,-2,-3,-5,-6,-10,-15,-30



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-30}}}.

1*(-30) = -30
2*(-15) = -30
3*(-10) = -30
5*(-6) = -30
(-1)*(30) = -30
(-2)*(15) = -30
(-3)*(10) = -30
(-5)*(6) = -30


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{-13}}}:



&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;-30&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+(-30)=-29&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;2&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-15&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;2+(-15)=-13&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;-10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;3+(-10)=-7&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;5&lt;/font&gt;&lt;/td&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;5+(-6)=-1&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;30&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+30=29&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;15&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+15=13&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;10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-3+10=7&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-5&lt;/font&gt;&lt;/td&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;-5+6=1&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{2}}} and {{{-15}}} add to {{{-13}}} (the middle coefficient).



So the two numbers {{{2}}} and {{{-15}}} both multiply to {{{-30}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{-13}}}



Now replace the middle term {{{-13x}}} with {{{2x-15x}}}. Remember, {{{2}}} and {{{-15}}} add to {{{-13}}}. So this shows us that {{{2x-15x=-13x}}}.



{{{3x^2+highlight(2x-15x)-10}}} Replace the second term {{{-13x}}} with {{{2x-15x}}}.



{{{(3x^2+2x)+(-15x-10)}}} Group the terms into two pairs.



{{{x(3x+2)+(-15x-10)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(3x+2)-5(3x+2)}}} Factor out {{{5}}} 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.



{{{(x-5)(3x+2)}}} Combine like terms. Or factor out the common term {{{3x+2}}}



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So {{{6(3x^2-13x-10)}}} then factors further to {{{6(x-5)(3x+2)}}}



===============================================================



Answer:



So {{{18x^2-78x-60}}} completely factors to {{{6(x-5)(3x+2)}}}.



In other words, {{{18x^2-78x-60=6(x-5)(3x+2)}}}.



So the answer is C (since the order of the factors doesn&#39;t matter).



Note: you can check the answer by expanding {{{6(x-5)(3x+2)}}} to get {{{18x^2-78x-60}}} 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=I%20Need%20Algebra%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;website&lt;/a&gt;. 


Jim</PRE>