If (mx-3)(nx+2)=6x^2+kx-6 and m+n=5, what are the possible values of k?
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(mx-3)(nx+2)=6x^2+kx-6  we work on the first side and we get  mnx^2+2mx-3nx-6

so it is 6x^2 = mnx^2  --->>  mn=6  then 2mx-3nx =(2m-3n)x  = kx then  2m-3n=k

we know that m+n=5  since we know that mn=6 and m+n=5 m and n are the roots of the equation

x^2-5x+6  solving this equation we have the roots, they are 3 and 2 so it will be m=3 and n=2

or m=2 and n=3 substituting these values to the equation 2m-3n = k we get k=-5 and k=0
by Level 5 User (13.1k points)

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