Question

Determine the value of k if P(x)=kx^3+4x^2+3x-4 and Q(x)=x^3-4x+k leaves the same remainder when divided by (x-3).

Answer 1

The remainder when dividing by \(x-3\) is exactly the value of the polynomial when evaluated at \(x=3\)... we want \(P(3)=Q(3)\) so:$$kx^3+4x^2+3x-4=x^3-4x+k\\27k+36+9-4=27-12+k\\26k+41=15\\26k=-26\\k=-1$$

Answer 2

The reason the above works follows from the idea behind polynomial division:$$\frac{P(x)}{x-3}=U(x)+\frac{R_1(x)}{x-3}\\\frac{Q(x)}{x-3}=V(x)+\frac{R_2(x)}{x-3}$$hence we observe $$P(x)=(x-3)U(x)+R_1(x)\\Q(x)=(x-3)V(x)+R_2(x)$$where \(R_1,R_2\) are our polynomial remainders. Observe, then, that at \(x=3\) our polynomials match their remainders:$$P(3)=R_1(3)\\Q(3)=R_2(3)$$ hence when their remainders are equal we have the functions themselves equal i.e. \(P(3)=Q(3)\) as we did above. this is known as the polynomial remainder theorem :-)

Answer 3

@oldrin.bataku thanks bro

Answer 4

very good explanation