A.maths about binomial

(1-kx)^4(1-4x)^3
=[1-4kx+6k^2x^2-4k^3x^3+k^4x^4][1-12x+48x^2-64x^3]
=1-12x-4kx+48x^2+48kx^2+6k^2x^2
=1-(12+4k)x+(6k^2+48k+48)x^2
the coefficient of x is 0
12+4k=0
k=-3
the cofficient of x^2
=6(-3)^2+48(-3)+48
=54-96
=-42