Question 539493


{{{15x^2-84x-36}}} Start with the given expression.



{{{3(5x^2-28x-12)}}} Factor out the GCF {{{3}}}.



Now let's try to factor the inner expression {{{5x^2-28x-12}}}



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Looking at the expression {{{5x^2-28x-12}}}, we can see that the first coefficient is {{{5}}}, the second coefficient is {{{-28}}}, and the last term is {{{-12}}}.



Now multiply the first coefficient {{{5}}} by the last term {{{-12}}} to get {{{(5)(-12)=-60}}}.



Now the question is: what two whole numbers multiply to {{{-60}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{-28}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{-60}}} (the previous product).



Factors of {{{-60}}}:

1,2,3,4,5,6,10,12,15,20,30,60

-1,-2,-3,-4,-5,-6,-10,-12,-15,-20,-30,-60



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



These factors pair up and multiply to {{{-60}}}.

1*(-60) = -60
2*(-30) = -60
3*(-20) = -60
4*(-15) = -60
5*(-12) = -60
6*(-10) = -60
(-1)*(60) = -60
(-2)*(30) = -60
(-3)*(20) = -60
(-4)*(15) = -60
(-5)*(12) = -60
(-6)*(10) = -60


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{-28}}}:



<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>-60</font></td><td  align="center"><font color=black>1+(-60)=-59</font></td></tr><tr><td  align="center"><font color=red>2</font></td><td  align="center"><font color=red>-30</font></td><td  align="center"><font color=red>2+(-30)=-28</font></td></tr><tr><td  align="center"><font color=black>3</font></td><td  align="center"><font color=black>-20</font></td><td  align="center"><font color=black>3+(-20)=-17</font></td></tr><tr><td  align="center"><font color=black>4</font></td><td  align="center"><font color=black>-15</font></td><td  align="center"><font color=black>4+(-15)=-11</font></td></tr><tr><td  align="center"><font color=black>5</font></td><td  align="center"><font color=black>-12</font></td><td  align="center"><font color=black>5+(-12)=-7</font></td></tr><tr><td  align="center"><font color=black>6</font></td><td  align="center"><font color=black>-10</font></td><td  align="center"><font color=black>6+(-10)=-4</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>60</font></td><td  align="center"><font color=black>-1+60=59</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>30</font></td><td  align="center"><font color=black>-2+30=28</font></td></tr><tr><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>20</font></td><td  align="center"><font color=black>-3+20=17</font></td></tr><tr><td  align="center"><font color=black>-4</font></td><td  align="center"><font color=black>15</font></td><td  align="center"><font color=black>-4+15=11</font></td></tr><tr><td  align="center"><font color=black>-5</font></td><td  align="center"><font color=black>12</font></td><td  align="center"><font color=black>-5+12=7</font></td></tr><tr><td  align="center"><font color=black>-6</font></td><td  align="center"><font color=black>10</font></td><td  align="center"><font color=black>-6+10=4</font></td></tr></table>



From the table, we can see that the two numbers {{{2}}} and {{{-30}}} add to {{{-28}}} (the middle coefficient).



So the two numbers {{{2}}} and {{{-30}}} both multiply to {{{-60}}} <font size=4><b>and</b></font> add to {{{-28}}}



Now replace the middle term {{{-28x}}} with {{{2x-30x}}}. Remember, {{{2}}} and {{{-30}}} add to {{{-28}}}. So this shows us that {{{2x-30x=-28x}}}.



{{{5x^2+highlight(2x-30x)-12}}} Replace the second term {{{-28x}}} with {{{2x-30x}}}.



{{{(5x^2+2x)+(-30x-12)}}} Group the terms into two pairs.



{{{x(5x+2)+(-30x-12)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(5x+2)-6(5x+2)}}} 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.



{{{(x-6)(5x+2)}}} Combine like terms. Or factor out the common term {{{5x+2}}}



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So {{{3(5x^2-28x-12)}}} then factors further to {{{3(x-6)(5x+2)}}}



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Answer:



So {{{15x^2-84x-36}}} completely factors to {{{3(x-6)(5x+2)}}}.



In other words, {{{15x^2-84x-36=3(x-6)(5x+2)}}}.



Note: you can check the answer by expanding {{{3(x-6)(5x+2)}}} to get {{{15x^2-84x-36}}} or by graphing the original expression and the answer (the two graphs should be identical).



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