<PRE><B>Question 288024</B>


{{{10x^2+25x-15}}} Start with the given expression.



{{{5(2x^2+5x-3)}}} Factor out the GCF {{{5}}}.



Now let&#39;s try to factor the inner expression {{{2x^2+5x-3}}}



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Looking at the expression {{{2x^2+5x-3}}}, we can see that the first coefficient is {{{2}}}, the second coefficient is {{{5}}}, and the last term is {{{-3}}}.



Now multiply the first coefficient {{{2}}} by the last term {{{-3}}} to get {{{(2)(-3)=-6}}}.



Now the question is: what two whole numbers multiply to {{{-6}}} (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 {{{-6}}} (the previous product).



Factors of {{{-6}}}:

1,2,3,6

-1,-2,-3,-6



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-6}}}.

1*(-6) = -6
2*(-3) = -6
(-1)*(6) = -6
(-2)*(3) = -6


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;-6&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+(-6)=-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;2&lt;/font&gt;&lt;/td&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;2+(-3)=-1&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-1&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;6&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-1+6=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;-2&lt;/font&gt;&lt;/td&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;-2+3=1&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{-1}}} and {{{6}}} add to {{{5}}} (the middle coefficient).



So the two numbers {{{-1}}} and {{{6}}} both multiply to {{{-6}}} &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 {{{-x+6x}}}. Remember, {{{-1}}} and {{{6}}} add to {{{5}}}. So this shows us that {{{-x+6x=5x}}}.



{{{2x^2+highlight(-x+6x)-3}}} Replace the second term {{{5x}}} with {{{-x+6x}}}.



{{{(2x^2-x)+(6x-3)}}} Group the terms into two pairs.



{{{x(2x-1)+(6x-3)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(2x-1)+3(2x-1)}}} Factor out {{{3}}} 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+3)(2x-1)}}} Combine like terms. Or factor out the common term {{{2x-1}}}



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So {{{5(2x^2+5x-3)}}} then factors further to {{{5(x+3)(2x-1)}}}



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Answer:



So {{{10x^2+25x-15}}} completely factors to {{{5(x+3)(2x-1)}}}.



In other words, {{{10x^2+25x-15=5(x+3)(2x-1)}}}.



Note: you can check the answer by expanding {{{5(x+3)(2x-1)}}} to get {{{10x^2+25x-15}}} or by graphing the original expression and the answer (the two graphs should be identical).
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