If (mx-3)(nx+2)=6x²+kx-6 and m+n=5, what are the possible values of k?

Question

If (mx-3)(nx+2)=6x^2+kx-6 and m+n=5, what are the possible values of k?

Answer 1

 

(mx-3)(nx+2) = mnx² + 2mx -3nx - 6 =  (mn)x² + (2m-3n)x -6 

 

Comparing the rightmost expression with  6x² + kx - 6   

 

Tells you  mn = 6   and   2m-3n = k

 

You also know m+n = 5   so  (using mn=6 and m+n=5)  m=3 or m=2  (if m=3, n=2;   if m=2, n=3)

 

Since k=2m-3n  you can plug in the two possibilities for m and n to find the possible values of k:

 

m=2,n=3:  k = 2(2)-3(3) = -5

m=3,n=2:  k = 2(3)-3(2) = 0

 

 

Answer 2

Using the FOIL method we find that m·n=6 (F).  This means that the possible values of m,n are:  1,6; 6,1; 2,3; 3,2; they could also all be negative too.  Since m+n=5, the only sets that work are 2,3 and 3,2.  Put these in and calculate kx from OI.

 

m=2, n=3, k=-5

m=3, n=2, k=0