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Try different values of x to observe the behaviour of p(x).

x=0: p(0)=0, because 40 mod (x+4)=40 mod 4=0.

x=1: p(1)=k+24, because 40 mod 5=0. If k+24=-28, then k=-52.

x=-1: p(-1)=5-k, because 40 mod 3=1. If 5-k=-28, then k=33.

x=2: p(2)=8k+74, because 40 mod 6=4. If 8k+74=-28, then k=-102/8=-51/4.

These give different values for k, so there is no single solution for k. However, if the question is supposed to read: p(x)=(kx^3+15x^2+9x-40) mod (x+4)=-28, the situation is quite different.

Using synthetic division, we can work out the remainder: -64k+164=-28. The modulo function only requires the remainder, which we know is -28; therefore 64k=192 and k=3.

Details of division:

-4 | k......15..........9...........-40

......k.....-4k 16k-60 -64k+204

......k 15-4k 16k-51 -64k+164 (=-28)


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