<PRE><B>Question 718551</B>


Looking at the expression {{{2x^2+13x+15}}}, we can see that the first coefficient is {{{2}}}, the second coefficient is {{{13}}}, and the last term is {{{15}}}.



Now multiply the first coefficient {{{2}}} by the last term {{{15}}} to get {{{(2)(15)=30}}}.



Now the question is: what two whole numbers multiply to {{{30}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{13}}}?



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 {{{30}}} (the previous product).



Factors of {{{30}}}:

1,2,3,5,6,10,15,30

-1,-2,-3,-5,-6,-10,-15,-30



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{30}}}.

1*30 = 30
2*15 = 30
3*10 = 30
5*6 = 30
(-1)*(-30) = 30
(-2)*(-15) = 30
(-3)*(-10) = 30
(-5)*(-6) = 30


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{13}}}:



&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;30&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+30=31&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;15&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;2+15=17&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;10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;3+10=13&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;6&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;5+6=11&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;-30&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+(-30)=-31&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;-15&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+(-15)=-17&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;-10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-3+(-10)=-13&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;-6&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-5+(-6)=-11&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{3}}} and {{{10}}} add to {{{13}}} (the middle coefficient).



So the two numbers {{{3}}} and {{{10}}} both multiply to {{{30}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{13}}}



Now replace the middle term {{{13x}}} with {{{3x+10x}}}. Remember, {{{3}}} and {{{10}}} add to {{{13}}}. So this shows us that {{{3x+10x=13x}}}.



{{{2x^2+highlight(3x+10x)+15}}} Replace the second term {{{13x}}} with {{{3x+10x}}}.



{{{(2x^2+3x)+(10x+15)}}} Group the terms into two pairs.



{{{x(2x+3)+(10x+15)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(2x+3)+5(2x+3)}}} 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+3)}}} Combine like terms. Or factor out the common term {{{2x+3}}}



===============================================================



Answer:



So {{{2x^2+13x+15}}} factors to {{{(x+5)(2x+3)}}}.



In other words, {{{2x^2+13x+15=(x+5)(2x+3)}}}.



Note: you can check the answer by expanding {{{(x+5)(2x+3)}}} to get {{{2x^2+13x+15}}} or by graphing the original expression and the answer (the two graphs should be identical).
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