Question 681947


Looking at the expression {{{4b^2-7b-15}}}, we can see that the first coefficient is {{{4}}}, the second coefficient is {{{-7}}}, and the last term is {{{-15}}}.



Now multiply the first coefficient {{{4}}} by the last term {{{-15}}} to get {{{(4)(-15)=-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 {{{-7}}}?



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 {{{-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>-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=red>5</font></td><td  align="center"><font color=red>-12</font></td><td  align="center"><font color=red>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 {{{5}}} and {{{-12}}} add to {{{-7}}} (the middle coefficient).



So the two numbers {{{5}}} and {{{-12}}} both multiply to {{{-60}}} <font size=4><b>and</b></font> add to {{{-7}}}



Now replace the middle term {{{-7b}}} with {{{5b-12b}}}. Remember, {{{5}}} and {{{-12}}} add to {{{-7}}}. So this shows us that {{{5b-12b=-7b}}}.



{{{4b^2+highlight(5b-12b)-15}}} Replace the second term {{{-7b}}} with {{{5b-12b}}}.



{{{(4b^2+5b)+(-12b-15)}}} Group the terms into two pairs.



{{{b(4b+5)+(-12b-15)}}} Factor out the GCF {{{b}}} from the first group.



{{{b(4b+5)-3(4b+5)}}} Factor out {{{3}}} 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.



{{{(b-3)(4b+5)}}} Combine like terms. Or factor out the common term {{{4b+5}}}



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Answer:



So {{{4b^2-7b-15}}} factors to {{{(b-3)(4b+5)}}}.



In other words, {{{4b^2-7b-15=(b-3)(4b+5)}}}.



Note: you can check the answer by expanding {{{(b-3)(4b+5)}}} to get {{{4b^2-7b-15}}} or by graphing the original expression and the answer (the two graphs should be identical).