Question 195729


Looking at the expression {{{a^2+10a+21}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{10}}}, and the last term is {{{21}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{21}}} to get {{{(1)(21)=21}}}.



Now the question is: what two whole numbers multiply to {{{21}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{10}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{21}}} (the previous product).



Factors of {{{21}}}:

1,3,7,21

-1,-3,-7,-21



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{21}}}.

1*21
3*7
(-1)*(-21)
(-3)*(-7)


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{10}}}:



<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>21</font></td><td  align="center"><font color=black>1+21=22</font></td></tr><tr><td  align="center"><font color=red>3</font></td><td  align="center"><font color=red>7</font></td><td  align="center"><font color=red>3+7=10</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-21</font></td><td  align="center"><font color=black>-1+(-21)=-22</font></td></tr><tr><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>-7</font></td><td  align="center"><font color=black>-3+(-7)=-10</font></td></tr></table>



From the table, we can see that the two numbers {{{3}}} and {{{7}}} add to {{{10}}} (the middle coefficient).



So the two numbers {{{3}}} and {{{7}}} both multiply to {{{21}}} <font size=4><b>and</b></font> add to {{{10}}}



Now replace the middle term {{{10a}}} with {{{3a+7a}}}. Remember, {{{3}}} and {{{7}}} add to {{{10}}}. So this shows us that {{{3a+7a=10a}}}.



{{{a^2+highlight(3a+7a)+21}}} Replace the second term {{{10a}}} with {{{3a+7a}}}.



{{{(a^2+3a)+(7a+21)}}} Group the terms into two pairs.



{{{a(a+3)+(7a+21)}}} Factor out the GCF {{{a}}} from the first group.



{{{a(a+3)+7(a+3)}}} Factor out {{{7}}} 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.



{{{(a+7)(a+3)}}} Combine like terms. Or factor out the common term {{{a+3}}}


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Answer:



So {{{a^2+10a+21}}} factors to {{{(a+7)(a+3)}}}.