<PRE><B>Question 703574</B>


{{{21y^3-36y^2-12y}}} Start with the given expression.



{{{3y(7y^2-12y-4)}}} Factor out the GCF {{{3y}}}.



Now let&#39;s try to factor the inner expression {{{7y^2-12y-4}}}



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Looking at the expression {{{7y^2-12y-4}}}, we can see that the first coefficient is {{{7}}}, the second coefficient is {{{-12}}}, and the last term is {{{-4}}}.



Now multiply the first coefficient {{{7}}} by the last term {{{-4}}} to get {{{(7)(-4)=-28}}}.



Now the question is: what two whole numbers multiply to {{{-28}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{-12}}}?



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 {{{-28}}} (the previous product).



Factors of {{{-28}}}:

1,2,4,7,14,28

-1,-2,-4,-7,-14,-28



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-28}}}.

1*(-28) = -28
2*(-14) = -28
4*(-7) = -28
(-1)*(28) = -28
(-2)*(14) = -28
(-4)*(7) = -28


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{-12}}}:



&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;-28&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+(-28)=-27&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;-14&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;2+(-14)=-12&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;4&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-7&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;4+(-7)=-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;28&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+28=27&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;14&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+14=12&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-4&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;7&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-4+7=3&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{2}}} and {{{-14}}} add to {{{-12}}} (the middle coefficient).



So the two numbers {{{2}}} and {{{-14}}} both multiply to {{{-28}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{-12}}}



Now replace the middle term {{{-12y}}} with {{{2y-14y}}}. Remember, {{{2}}} and {{{-14}}} add to {{{-12}}}. So this shows us that {{{2y-14y=-12y}}}.



{{{7y^2+highlight(2y-14y)-4}}} Replace the second term {{{-12y}}} with {{{2y-14y}}}.



{{{(7y^2+2y)+(-14y-4)}}} Group the terms into two pairs.



{{{y(7y+2)+(-14y-4)}}} Factor out the GCF {{{y}}} from the first group.



{{{y(7y+2)-2(7y+2)}}} Factor out {{{2}}} 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.



{{{(y-2)(7y+2)}}} Combine like terms. Or factor out the common term {{{7y+2}}}



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So {{{3y(7y^2-12y-4)}}} then factors further to {{{3y(y-2)(7y+2)}}}



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Answer:



So {{{21y^3-36y^2-12y}}} completely factors to {{{3y(y-2)(7y+2)}}}.



In other words, {{{21y^3-36y^2-12y=3y(y-2)(7y+2)}}}.



Note: you can check the answer by expanding {{{3y(y-2)(7y+2)}}} to get {{{21y^3-36y^2-12y}}} or by graphing the original expression and the answer (the two graphs should be identical).
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