<PRE><B>Question 257298</B>
The GCF of {{{x^2 + 5x + 6}}} is 1, but it is trivial to factor it out. So we don&#39;t have to worry about the GCF.




Looking at the expression {{{x^2+5x+6}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{5}}}, and the last term is {{{6}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{6}}} to get {{{(1)(6)=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=7&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;3&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;2+3=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;-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)=-7&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)=-5&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{2}}} and {{{3}}} add to {{{5}}} (the middle coefficient).



So the two numbers {{{2}}} and {{{3}}} 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 {{{2x+3x}}}. Remember, {{{2}}} and {{{3}}} add to {{{5}}}. So this shows us that {{{2x+3x=5x}}}.



{{{x^2+highlight(2x+3x)+6}}} Replace the second term {{{5x}}} with {{{2x+3x}}}.



{{{(x^2+2x)+(3x+6)}}} Group the terms into two pairs.



{{{x(x+2)+(3x+6)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(x+2)+3(x+2)}}} 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)(x+2)}}} Combine like terms. Or factor out the common term {{{x+2}}}



===============================================================



Answer:



So {{{x^2+5x+6}}} factors to {{{(x+3)(x+2)}}}.



In other words, {{{x^2+5x+6=(x+3)(x+2)}}}.



Note: you can check the answer by expanding {{{(x+3)(x+2)}}} to get {{{x^2+5x+6}}} or by graphing the original expression and the answer (the two graphs should be identical).
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