<PRE><B>Question 248077</B>


Looking at the expression {{{2x^2+7x+6}}}, we can see that the first coefficient is {{{2}}}, the second coefficient is {{{7}}}, and the last term is {{{6}}}.



Now multiply the first coefficient {{{2}}} by the last term {{{6}}} to get {{{(2)(6)=12}}}.



Now the question is: what two whole numbers multiply to {{{12}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{7}}}?



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 {{{12}}} (the previous product).



Factors of {{{12}}}:

1,2,3,4,6,12

-1,-2,-3,-4,-6,-12



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{12}}}.

1*12 = 12
2*6 = 12
3*4 = 12
(-1)*(-12) = 12
(-2)*(-6) = 12
(-3)*(-4) = 12


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{7}}}:



&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;12&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+12=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;2&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;2+6=8&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;4&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;3+4=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;-1&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;-1+(-12)=-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;-2&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;-2+(-6)=-8&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;-4&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-3+(-4)=-7&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{3}}} and {{{4}}} add to {{{7}}} (the middle coefficient).



So the two numbers {{{3}}} and {{{4}}} both multiply to {{{12}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{7}}}



Now replace the middle term {{{7x}}} with {{{3x+4x}}}. Remember, {{{3}}} and {{{4}}} add to {{{7}}}. So this shows us that {{{3x+4x=7x}}}.



{{{2x^2+highlight(3x+4x)+6}}} Replace the second term {{{7x}}} with {{{3x+4x}}}.



{{{(2x^2+3x)+(4x+6)}}} Group the terms into two pairs.



{{{x(2x+3)+(4x+6)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(2x+3)+2(2x+3)}}} Factor out {{{2}}} 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+2)(2x+3)}}} Combine like terms. Or factor out the common term {{{2x+3}}}



===============================================================



Answer:



So {{{2x^2+7x+6}}} factors to {{{(x+2)(2x+3)}}}.



In other words, {{{2x^2+7x+6=(x+2)(2x+3)}}}.



Note: you can check the answer by expanding {{{(x+2)(2x+3)}}} to get {{{2x^2+7x+6}}} or by graphing the original expression and the answer (the two graphs should be identical).
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