<PRE><B>Question 640411</B>


{{{12x^3+8x^2-20x}}} Start with the given expression.



{{{4x(3x^2+2x-5)}}} Factor out the GCF {{{4x}}}.



Now let&#39;s try to factor the inner expression {{{3x^2+2x-5}}}



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Looking at the expression {{{3x^2+2x-5}}}, we can see that the first coefficient is {{{3}}}, the second coefficient is {{{2}}}, and the last term is {{{-5}}}.



Now multiply the first coefficient {{{3}}} by the last term {{{-5}}} to get {{{(3)(-5)=-15}}}.



Now the question is: what two whole numbers multiply to {{{-15}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{2}}}?



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 {{{-15}}} (the previous product).



Factors of {{{-15}}}:

1,3,5,15

-1,-3,-5,-15



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-15}}}.

1*(-15) = -15
3*(-5) = -15
(-1)*(15) = -15
(-3)*(5) = -15


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{2}}}:



&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;-15&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+(-15)=-14&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;-5&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;3+(-5)=-2&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;15&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+15=14&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-3&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;5&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-3+5=2&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{-3}}} and {{{5}}} add to {{{2}}} (the middle coefficient).



So the two numbers {{{-3}}} and {{{5}}} both multiply to {{{-15}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{2}}}



Now replace the middle term {{{2x}}} with {{{-3x+5x}}}. Remember, {{{-3}}} and {{{5}}} add to {{{2}}}. So this shows us that {{{-3x+5x=2x}}}.



{{{3x^2+highlight(-3x+5x)-5}}} Replace the second term {{{2x}}} with {{{-3x+5x}}}.



{{{(3x^2-3x)+(5x-5)}}} Group the terms into two pairs.



{{{3x(x-1)+(5x-5)}}} Factor out the GCF {{{3x}}} from the first group.



{{{3x(x-1)+5(x-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.



{{{(3x+5)(x-1)}}} Combine like terms. Or factor out the common term {{{x-1}}}



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So {{{4x(3x^2+2x-5)}}} then factors further to {{{4x(3x+5)(x-1)}}}



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Answer:



So {{{12x^3+8x^2-20x}}} completely factors to {{{4x(3x+5)(x-1)}}}.



In other words, {{{12x^3+8x^2-20x=4x(3x+5)(x-1)}}}.



Note: you can check the answer by expanding {{{4x(3x+5)(x-1)}}} to get {{{12x^3+8x^2-20x}}} or by graphing the original expression and the answer (the two graphs should be identical).
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