<PRE><B>Question 790717</B>


{{{4a^3+12a^2+5a}}} Start with the given expression.



{{{a(4a^2+12a+5)}}} Factor out the GCF {{{a}}}.



Now let&#39;s try to factor the inner expression {{{4a^2+12a+5}}}



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Looking at the expression {{{4a^2+12a+5}}}, we can see that the first coefficient is {{{4}}}, the second coefficient is {{{12}}}, and the last term is {{{5}}}.



Now multiply the first coefficient {{{4}}} by the last term {{{5}}} to get {{{(4)(5)=20}}}.



Now the question is: what two whole numbers multiply to {{{20}}} (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 {{{20}}} (the previous product).



Factors of {{{20}}}:

1,2,4,5,10,20

-1,-2,-4,-5,-10,-20



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{20}}}.

1*20 = 20
2*10 = 20
4*5 = 20
(-1)*(-20) = 20
(-2)*(-10) = 20
(-4)*(-5) = 20


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;20&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+20=21&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;10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;2+10=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;5&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;4+5=9&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;-20&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+(-20)=-21&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;-10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+(-10)=-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;-5&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-4+(-5)=-9&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{2}}} and {{{10}}} add to {{{12}}} (the middle coefficient).



So the two numbers {{{2}}} and {{{10}}} both multiply to {{{20}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{12}}}



Now replace the middle term {{{12a}}} with {{{2a+10a}}}. Remember, {{{2}}} and {{{10}}} add to {{{12}}}. So this shows us that {{{2a+10a=12a}}}.



{{{4a^2+highlight(2a+10a)+5}}} Replace the second term {{{12a}}} with {{{2a+10a}}}.



{{{(4a^2+2a)+(10a+5)}}} Group the terms into two pairs.



{{{2a(2a+1)+(10a+5)}}} Factor out the GCF {{{2a}}} from the first group.



{{{2a(2a+1)+5(2a+1)}}} 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.



{{{(2a+5)(2a+1)}}} Combine like terms. Or factor out the common term {{{2a+1}}}



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So {{{a(4a^2+12a+5)}}} then factors further to {{{a(2a+5)(2a+1)}}}



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Answer:



So {{{4a^3+12a^2+5a}}} completely factors to {{{a(2a+5)(2a+1)}}}.



In other words, {{{4a^3+12a^2+5a=a(2a+5)(2a+1)}}}.



Note: you can check the answer by expanding {{{a(2a+5)(2a+1)}}} to get {{{4a^3+12a^2+5a}}} or by graphing the original expression and the answer (the two graphs should be identical).
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