<PRE><B>Question 687136</B>


Looking at the expression {{{2w^2+13w+20}}}, we can see that the first coefficient is {{{2}}}, the second coefficient is {{{13}}}, and the last term is {{{20}}}.



Now multiply the first coefficient {{{2}}} by the last term {{{20}}} to get {{{(2)(20)=40}}}.



Now the question is: what two whole numbers multiply to {{{40}}} (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 {{{40}}} (the previous product).



Factors of {{{40}}}:

1,2,4,5,8,10,20,40

-1,-2,-4,-5,-8,-10,-20,-40



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{40}}}.

1*40 = 40
2*20 = 40
4*10 = 40
5*8 = 40
(-1)*(-40) = 40
(-2)*(-20) = 40
(-4)*(-10) = 40
(-5)*(-8) = 40


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;40&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+40=41&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;20&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;2+20=22&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;4&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;4+10=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;5&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;8&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;5+8=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;-1&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-40&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+(-40)=-41&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;-20&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+(-20)=-22&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-4&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;-4+(-10)=-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;-5&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-8&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-5+(-8)=-13&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{5}}} and {{{8}}} add to {{{13}}} (the middle coefficient).



So the two numbers {{{5}}} and {{{8}}} both multiply to {{{40}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{13}}}



Now replace the middle term {{{13w}}} with {{{5w+8w}}}. Remember, {{{5}}} and {{{8}}} add to {{{13}}}. So this shows us that {{{5w+8w=13w}}}.



{{{2w^2+highlight(5w+8w)+20}}} Replace the second term {{{13w}}} with {{{5w+8w}}}.



{{{(2w^2+5w)+(8w+20)}}} Group the terms into two pairs.



{{{w(2w+5)+(8w+20)}}} Factor out the GCF {{{w}}} from the first group.



{{{w(2w+5)+4(2w+5)}}} Factor out {{{4}}} 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.



{{{(w+4)(2w+5)}}} Combine like terms. Or factor out the common term {{{2w+5}}}



===============================================================



Answer:



So {{{2w^2+13w+20}}} factors to {{{(w+4)(2w+5)}}}.



In other words, {{{2w^2+13w+20=(w+4)(2w+5)}}}.



Note: you can check the answer by expanding {{{(w+4)(2w+5)}}} to get {{{2w^2+13w+20}}} or by graphing the original expression and the answer (the two graphs should be identical).
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