Question 590755


Looking at the expression {{{20s^2+11s-3}}}, we can see that the first coefficient is {{{20}}}, the second coefficient is {{{11}}}, and the last term is {{{-3}}}.



Now multiply the first coefficient {{{20}}} by the last term {{{-3}}} to get {{{(20)(-3)=-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 {{{11}}}?



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 {{{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>-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><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=red>-4</font></td><td  align="center"><font color=red>15</font></td><td  align="center"><font color=red>-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 {{{-4}}} and {{{15}}} add to {{{11}}} (the middle coefficient).



So the two numbers {{{-4}}} and {{{15}}} both multiply to {{{-60}}} <font size=4><b>and</b></font> add to {{{11}}}



Now replace the middle term {{{11s}}} with {{{-4s+15s}}}. Remember, {{{-4}}} and {{{15}}} add to {{{11}}}. So this shows us that {{{-4s+15s=11s}}}.



{{{20s^2+highlight(-4s+15s)-3}}} Replace the second term {{{11s}}} with {{{-4s+15s}}}.



{{{(20s^2-4s)+(15s-3)}}} Group the terms into two pairs.



{{{4s(5s-1)+(15s-3)}}} Factor out the GCF {{{4s}}} from the first group.



{{{4s(5s-1)+3(5s-1)}}} 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.



{{{(4s+3)(5s-1)}}} Combine like terms. Or factor out the common term {{{5s-1}}}



===============================================================



Answer:



So {{{20s^2+11s-3}}} factors to {{{(4s+3)(5s-1)}}}.



In other words, {{{20s^2+11s-3=(4s+3)(5s-1)}}}.



Note: you can check the answer by expanding {{{(4s+3)(5s-1)}}} to get {{{20s^2+11s-3}}} or by graphing the original expression and the answer (the two graphs should be identical).


<font color="red">--------------------------------------------------------------------------------------------------------------</font>
If you need more help, email me at <a href="mailto:jim_thompson5910@hotmail.com?Subject=I%20Need%20Algebra%20Help">jim_thompson5910@hotmail.com</a>


Also, please consider visiting my website: <a href="http://www.freewebs.com/jimthompson5910/home.html">http://www.freewebs.com/jimthompson5910/home.html</a> and making a donation. Thank you


Jim
<font color="red">--------------------------------------------------------------------------------------------------------------</font>