Question 914765


Looking at the expression {{{3n^2+7n+4}}}, we can see that the first coefficient is {{{3}}}, the second coefficient is {{{7}}}, and the last term is {{{4}}}.



Now multiply the first coefficient {{{3}}} by the last term {{{4}}} to get {{{(3)(4)=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 {{{7n}}} with {{{3n+4n}}}. Remember, {{{3}}} and {{{4}}} add to {{{7}}}. So this shows us that {{{3n+4n=7n}}}.



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



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



{{{3n(n+1)+(4n+4)}}} Factor out the GCF {{{3n}}} from the first group.



{{{3n(n+1)+4(n+1)}}} 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.



{{{(3n+4)(n+1)}}} Combine like terms. Or factor out the common term {{{n+1}}}



===============================================================



Answer:



So {{{3n^2+7n+4}}} factors to {{{(3n+4)(n+1)}}}.



In other words, {{{3n^2+7n+4=(3n+4)(n+1)}}}.



Note: you can check the answer by expanding {{{(3n+4)(n+1)}}} to get {{{3n^2+7n+4}}} or by graphing the original expression and the answer (the two graphs should be identical).



Let me know if you need more help or if you need me to explain a step in more detail.
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Thanks,


Jim