<PRE><B>Question 191767</B>


Looking at the expression {{{2x^2+3x-5}}}, we can see that the first coefficient is {{{2}}}, the second coefficient is {{{3}}}, and the last term is {{{-5}}}.



Now multiply the first coefficient {{{2}}} by the last term {{{-5}}} to get {{{(2)(-5)=-10}}}.



Now the question is: what two whole numbers multiply to {{{-10}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{3}}}?



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 {{{-10}}} (the previous product).



Factors of {{{-10}}}:

1,2,5,10

-1,-2,-5,-10



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-10}}}.

1*(-10)
2*(-5)
(-1)*(10)
(-2)*(5)


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{3}}}:



&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;-10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+(-10)=-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;2&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;2+(-5)=-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;10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+10=9&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;5&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-2+5=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 {{{5}}} add to {{{3}}} (the middle coefficient).



So the two numbers {{{-2}}} and {{{5}}} both multiply to {{{-10}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{3}}}



Now replace the middle term {{{3x}}} with {{{-2x+5x}}}. Remember, {{{-2}}} and {{{5}}} add to {{{3}}}. So this shows us that {{{-2x+5x=3x}}}.



{{{2x^2+highlight(-2x+5x)-5}}} Replace the second term {{{3x}}} with {{{-2x+5x}}}.



{{{(2x^2-2x)+(5x-5)}}} Group the terms into two pairs.



{{{2x(x-1)+(5x-5)}}} Factor out the GCF {{{2x}}} from the first group.



{{{2x(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.



{{{(2x+5)(x-1)}}} Combine like terms. Or factor out the common term {{{x-1}}}


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Answer:



So {{{2x^2+3x-5}}} factors to {{{(2x+5)(x-1)}}}.



Note: you can check the answer by FOILing {{{(2x+5)(x-1)}}} to get {{{2x^2+3x-5}}} or by graphing the original expression and the answer (the two graphs should be identical).
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