Question 183211
First, multiply the first coefficient 1 and the last term 50 to get 50. Now list the factors of 50





Factors of 50:

1,2,5,10,25,50

-1,-2,-5,-10,-25,-50



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



These factors pair up and multiply to {{{50}}}.

1*50
2*25
5*10
(-1)*(-50)
(-2)*(-25)
(-5)*(-10)


Now add up each paired factor (ie 1+50=51, 2+25=27, etc..):



<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>50</font></td><td  align="center"><font color=black>1+50=51</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>25</font></td><td  align="center"><font color=black>2+25=27</font></td></tr><tr><td  align="center"><font color=black>5</font></td><td  align="center"><font color=black>10</font></td><td  align="center"><font color=black>5+10=15</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-50</font></td><td  align="center"><font color=black>-1+(-50)=-51</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>-25</font></td><td  align="center"><font color=black>-2+(-25)=-27</font></td></tr><tr><td  align="center"><font color=black>-5</font></td><td  align="center"><font color=black>-10</font></td><td  align="center"><font color=black>-5+(-10)=-15</font></td></tr></table>



All of the numbers in the last column are possible values for the value of "m". Remember, you can only factor {{{ax^2+bx+c}}} only if the factors of {{{ac}}} add to "b"



So the possible values for "m" are: 51, 27, 15, -51, -27, -15



Note: all of these values of "m" are straight from that third column.



This means that the following quadratics are factorable:


{{{y^2+51y+50}}}, 


{{{y^2+27y+50}}},


{{{y^2+15y+50}}},


{{{y^2-51y+50}}},


{{{y^2-27y+50}}}, and 


{{{y^2-15y+50}}}