<PRE><B>Question 678314</B>


Looking at the expression {{{x^2+9x+20}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{9}}}, and the last term is {{{20}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{20}}} to get {{{(1)(20)=20}}}.



Now the question is: what two whole numbers multiply to {{{20}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{9}}}?



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 {{{20}}} (the previous product).



Factors of {{{20}}}:

1,2,4,5,10,20

-1,-2,-4,-5,-10,-20



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{20}}}.

1*20 = 20
2*10 = 20
4*5 = 20
(-1)*(-20) = 20
(-2)*(-10) = 20
(-4)*(-5) = 20


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{9}}}:



&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;20&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+20=21&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;10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;2+10=12&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;4&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;4+5=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;-1&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;-1+(-20)=-21&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;-10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+(-10)=-12&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;-5&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-4+(-5)=-9&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{4}}} and {{{5}}} add to {{{9}}} (the middle coefficient).



So the two numbers {{{4}}} and {{{5}}} both multiply to {{{20}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{9}}}



Now replace the middle term {{{9x}}} with {{{4x+5x}}}. Remember, {{{4}}} and {{{5}}} add to {{{9}}}. So this shows us that {{{4x+5x=9x}}}.



{{{x^2+highlight(4x+5x)+20}}} Replace the second term {{{9x}}} with {{{4x+5x}}}.



{{{(x^2+4x)+(5x+20)}}} Group the terms into two pairs.



{{{x(x+4)+(5x+20)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(x+4)+5(x+4)}}} 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)(x+4)}}} Combine like terms. Or factor out the common term {{{x+4}}}



===============================================================



Answer:



So {{{x^2+9x+20}}} factors to {{{(x+5)(x+4)}}}.



In other words, {{{x^2+9x+20=(x+5)(x+4)}}}.



Note: you can check the answer by expanding {{{(x+5)(x+4)}}} to get {{{x^2+9x+20}}} or by graphing the original expression and the answer (the two graphs should be identical).
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