<PRE><B>Question 629843</B>


{{{4x^3-10x^2-50x}}} Start with the given expression.



{{{2x(2x^2-5x-25)}}} Factor out the GCF {{{2x}}}.



Now let&#39;s try to factor the inner expression {{{2x^2-5x-25}}}



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Looking at the expression {{{2x^2-5x-25}}}, we can see that the first coefficient is {{{2}}}, the second coefficient is {{{-5}}}, and the last term is {{{-25}}}.



Now multiply the first coefficient {{{2}}} by the last term {{{-25}}} to get {{{(2)(-25)=-50}}}.



Now the question is: what two whole numbers multiply to {{{-50}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{-5}}}?



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 {{{-50}}} (the previous product).



Factors of {{{-50}}}:

1,2,5,10,25,50

-1,-2,-5,-10,-25,-50



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-50}}}.

1*(-50) = -50
2*(-25) = -50
5*(-10) = -50
(-1)*(50) = -50
(-2)*(25) = -50
(-5)*(10) = -50


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{-5}}}:



&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;-50&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+(-50)=-49&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;-25&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;2+(-25)=-23&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&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;-10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;5+(-10)=-5&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;50&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+50=49&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;25&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+25=23&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;10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-5+10=5&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{5}}} and {{{-10}}} add to {{{-5}}} (the middle coefficient).



So the two numbers {{{5}}} and {{{-10}}} both multiply to {{{-50}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{-5}}}



Now replace the middle term {{{-5x}}} with {{{5x-10x}}}. Remember, {{{5}}} and {{{-10}}} add to {{{-5}}}. So this shows us that {{{5x-10x=-5x}}}.



{{{2x^2+highlight(5x-10x)-25}}} Replace the second term {{{-5x}}} with {{{5x-10x}}}.



{{{(2x^2+5x)+(-10x-25)}}} Group the terms into two pairs.



{{{x(2x+5)+(-10x-25)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(2x+5)-5(2x+5)}}} 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)(2x+5)}}} Combine like terms. Or factor out the common term {{{2x+5}}}



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So {{{2x(2x^2-5x-25)}}} then factors further to {{{2x(x-5)(2x+5)}}}



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Answer:



So {{{4x^3-10x^2-50x}}} completely factors to {{{2x(x-5)(2x+5)}}}.



In other words, {{{4x^3-10x^2-50x=2x(x-5)(2x+5)}}}.



Note: you can check the answer by expanding {{{2x(x-5)(2x+5)}}} to get {{{4x^3-10x^2-50x}}} or by graphing the original expression and the answer (the two graphs should be identical).


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