Find k.

Since $kx^3-(k+3)x^2+13 \equiv 157 mod (x-4)$ , this means that $kx^3-\left(k+3\right)x^2+13-157$ is divisible by $(x-4)$ . Let $f\left(x\right)=kx^3-\left(k+3\right)x^2+13-157$ . By remainder factor theorem, if $f\left(x\right)$ is divisible by $(x-k)$ , then $f\left(k\right)=0$ . This means $f\left(4\right)=0$ .

$kx^3-\left(k+3\right)x^2+13-157=0$ $k\left(4\right)^3-\left(k+3\right)\left(4\right)^2-144=0$ $48k=144+48$ $k=4$

Sorry for the messy solution. I'm not very good at Latex.