Question 336150
I'll do the first one to get you started. Please only post one problem at a time. Thanks.




Looking at the expression {{{a^2-10a+24}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{-10}}}, and the last term is {{{24}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{24}}} to get {{{(1)(24)=24}}}.



Now the question is: what two whole numbers multiply to {{{24}}} (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 {{{24}}} (the previous product).



Factors of {{{24}}}:

1,2,3,4,6,8,12,24

-1,-2,-3,-4,-6,-8,-12,-24



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



These factors pair up and multiply to {{{24}}}.

1*24 = 24
2*12 = 24
3*8 = 24
4*6 = 24
(-1)*(-24) = 24
(-2)*(-12) = 24
(-3)*(-8) = 24
(-4)*(-6) = 24


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>24</font></td><td  align="center"><font color=black>1+24=25</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>12</font></td><td  align="center"><font color=black>2+12=14</font></td></tr><tr><td  align="center"><font color=black>3</font></td><td  align="center"><font color=black>8</font></td><td  align="center"><font color=black>3+8=11</font></td></tr><tr><td  align="center"><font color=black>4</font></td><td  align="center"><font color=black>6</font></td><td  align="center"><font color=black>4+6=10</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-24</font></td><td  align="center"><font color=black>-1+(-24)=-25</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>-12</font></td><td  align="center"><font color=black>-2+(-12)=-14</font></td></tr><tr><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>-8</font></td><td  align="center"><font color=black>-3+(-8)=-11</font></td></tr><tr><td  align="center"><font color=red>-4</font></td><td  align="center"><font color=red>-6</font></td><td  align="center"><font color=red>-4+(-6)=-10</font></td></tr></table>



From the table, we can see that the two numbers {{{-4}}} and {{{-6}}} add to {{{-10}}} (the middle coefficient).



So the two numbers {{{-4}}} and {{{-6}}} both multiply to {{{24}}} <font size=4><b>and</b></font> add to {{{-10}}}



Now replace the middle term {{{-10a}}} with {{{-4a-6a}}}. Remember, {{{-4}}} and {{{-6}}} add to {{{-10}}}. So this shows us that {{{-4a-6a=-10a}}}.



{{{a^2+highlight(-4a-6a)+24}}} Replace the second term {{{-10a}}} with {{{-4a-6a}}}.



{{{(a^2-4a)+(-6a+24)}}} Group the terms into two pairs.



{{{a(a-4)+(-6a+24)}}} Factor out the GCF {{{a}}} from the first group.



{{{a(a-4)-6(a-4)}}} Factor out {{{6}}} 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-6)(a-4)}}} Combine like terms. Or factor out the common term {{{a-4}}}



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Answer:



So {{{a^2-10a+24}}} factors to {{{(a-6)(a-4)}}}.



In other words, {{{a^2-10a+24=(a-6)(a-4)}}}.



Note: you can check the answer by expanding {{{(a-6)(a-4)}}} to get {{{a^2-10a+24}}} or by graphing the original expression and the answer (the two graphs should be identical).



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Jim