Question 192309


Looking at the expression {{{n^2-11n-80}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{-11}}}, and the last term is {{{-80}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{-80}}} to get {{{(1)(-80)=-80}}}.



Now the question is: what two whole numbers multiply to {{{-80}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{-11}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{-80}}} (the previous product).



Factors of {{{-80}}}:

1,2,4,5,8,10,16,20,40,80

-1,-2,-4,-5,-8,-10,-16,-20,-40,-80



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



These factors pair up and multiply to {{{-80}}}.

1*(-80)
2*(-40)
4*(-20)
5*(-16)
8*(-10)
(-1)*(80)
(-2)*(40)
(-4)*(20)
(-5)*(16)
(-8)*(10)


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{-11}}}:



<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>-80</font></td><td  align="center"><font color=black>1+(-80)=-79</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>-40</font></td><td  align="center"><font color=black>2+(-40)=-38</font></td></tr><tr><td  align="center"><font color=black>4</font></td><td  align="center"><font color=black>-20</font></td><td  align="center"><font color=black>4+(-20)=-16</font></td></tr><tr><td  align="center"><font color=red>5</font></td><td  align="center"><font color=red>-16</font></td><td  align="center"><font color=red>5+(-16)=-11</font></td></tr><tr><td  align="center"><font color=black>8</font></td><td  align="center"><font color=black>-10</font></td><td  align="center"><font color=black>8+(-10)=-2</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>80</font></td><td  align="center"><font color=black>-1+80=79</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>40</font></td><td  align="center"><font color=black>-2+40=38</font></td></tr><tr><td  align="center"><font color=black>-4</font></td><td  align="center"><font color=black>20</font></td><td  align="center"><font color=black>-4+20=16</font></td></tr><tr><td  align="center"><font color=black>-5</font></td><td  align="center"><font color=black>16</font></td><td  align="center"><font color=black>-5+16=11</font></td></tr><tr><td  align="center"><font color=black>-8</font></td><td  align="center"><font color=black>10</font></td><td  align="center"><font color=black>-8+10=2</font></td></tr></table>



From the table, we can see that the two numbers {{{5}}} and {{{-16}}} add to {{{-11}}} (the middle coefficient).



So the two numbers {{{5}}} and {{{-16}}} both multiply to {{{-80}}} <font size=4><b>and</b></font> add to {{{-11}}}



Now replace the middle term {{{-11n}}} with {{{5n-16n}}}. Remember, {{{5}}} and {{{-16}}} add to {{{-11}}}. So this shows us that {{{5n-16n=-11n}}}.



{{{n^2+highlight(5n-16n)-80}}} Replace the second term {{{-11n}}} with {{{5n-16n}}}.



{{{(n^2+5n)+(-16n-80)}}} Group the terms into two pairs.



{{{n(n+5)+(-16n-80)}}} Factor out the GCF {{{n}}} from the first group.



{{{n(n+5)-16(n+5)}}} Factor out {{{16}}} 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.



{{{(n-16)(n+5)}}} Combine like terms. Or factor out the common term {{{n+5}}}


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Answer:



So {{{n^2-11n-80}}} factors to {{{(n-16)(n+5)}}}.



Note: you can check the answer by FOILing {{{(n-16)(n+5)}}} to get {{{n^2-11n-80}}} or by graphing the original expression and the answer (the two graphs should be identical).