Question 381165


Looking at the expression {{{x^2+7x+12}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{7}}}, and the last term is {{{12}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{12}}} to get {{{(1)(12)=12}}}.



Now the question is: what two whole numbers multiply to {{{12}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{7}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> 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's add up each pair of factors to see if one pair adds to the middle coefficient {{{7}}}:



<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td  align="center"><font color=black>1</font></td><td  align="center"><font color=black>12</font></td><td  align="center"><font color=black>1+12=13</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>6</font></td><td  align="center"><font color=black>2+6=8</font></td></tr><tr><td  align="center"><font color=red>3</font></td><td  align="center"><font color=red>4</font></td><td  align="center"><font color=red>3+4=7</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-12</font></td><td  align="center"><font color=black>-1+(-12)=-13</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>-6</font></td><td  align="center"><font color=black>-2+(-6)=-8</font></td></tr><tr><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>-4</font></td><td  align="center"><font color=black>-3+(-4)=-7</font></td></tr></table>



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}}} <font size=4><b>and</b></font> 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}}}.



{{{x^2+highlight(3x+4x)+12}}} Replace the second term {{{7x}}} with {{{3x+4x}}}.



{{{(x^2+3x)+(4x+12)}}} Group the terms into two pairs.



{{{x(x+3)+(4x+12)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(x+3)+4(x+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)(x+3)}}} Combine like terms. Or factor out the common term {{{x+3}}}



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Answer:



So {{{x^2+7x+12}}} factors to {{{(x+4)(x+3)}}}.



In other words, {{{x^2+7x+12=(x+4)(x+3)}}}.



Note: you can check the answer by expanding {{{(x+4)(x+3)}}} to get {{{x^2+7x+12}}} or by graphing the original expression and the answer (the two graphs should be identical).



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Jim