<PRE><B>Question 473453</B>


Looking at the expression {{{2x^2+11x+12}}}, we can see that the first coefficient is {{{2}}}, the second coefficient is {{{11}}}, and the last term is {{{12}}}.



Now multiply the first coefficient {{{2}}} by the last term {{{12}}} to get {{{(2)(12)=24}}}.



Now the question is: what two whole numbers multiply to {{{24}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{11}}}?



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 {{{24}}} (the previous product).



Factors of {{{24}}}:

1,2,3,4,6,8,12,24

-1,-2,-3,-4,-6,-8,-12,-24



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{24}}}.

1*24 = 24
2*12 = 24
3*8 = 24
4*6 = 24
(-1)*(-24) = 24
(-2)*(-12) = 24
(-3)*(-8) = 24
(-4)*(-6) = 24


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{11}}}:



&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;24&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+24=25&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;12&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;2+12=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;3&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;3+8=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;4&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;4+6=10&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;-24&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+(-24)=-25&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;-12&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+(-12)=-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;-3&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;-3+(-8)=-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;-4&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;-4+(-6)=-10&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{3}}} and {{{8}}} add to {{{11}}} (the middle coefficient).



So the two numbers {{{3}}} and {{{8}}} both multiply to {{{24}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{11}}}



Now replace the middle term {{{11x}}} with {{{3x+8x}}}. Remember, {{{3}}} and {{{8}}} add to {{{11}}}. So this shows us that {{{3x+8x=11x}}}.



{{{2x^2+highlight(3x+8x)+12}}} Replace the second term {{{11x}}} with {{{3x+8x}}}.



{{{(2x^2+3x)+(8x+12)}}} Group the terms into two pairs.



{{{x(2x+3)+(8x+12)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(2x+3)+4(2x+3)}}} 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.



{{{(x+4)(2x+3)}}} Combine like terms. Or factor out the common term {{{2x+3}}}



===============================================================



Answer:



So {{{2x^2+11x+12}}} factors to {{{(x+4)(2x+3)}}}.



In other words, {{{2x^2+11x+12=(x+4)(2x+3)}}}.



Note: you can check the answer by expanding {{{(x+4)(2x+3)}}} to get {{{2x^2+11x+12}}} or by graphing the original expression and the answer (the two graphs should be identical).
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