Question

the polynomial ky^3+3y^2-3 and 2y^3-5y+k when divided by (y-5) leave the same remainder in each case. Find the value of k.

Answer 1

Answer:

$k=\frac{153}{124}$

Step-by-step explanation:

According to the Remainder Theorem; when P(y) is divided by y-a, the remainder is p(a).

The first polynomial is :

$p(y)=ky^3+3y^2-3$

When p(y) is divided by y-5, the remainder is

$p(5)=k(5)^3+3(5)^2-3$

$p(5)=125k+75-3$

$p(5)=125k+72$

When the second polynomial:

$m(y)= 2y^3-5y+k$ is divided by y-5.

The remainder is;

$m(5)= 2(5)^3-5(5)+k$

$m(5)= 225+k$

The two remainders are equal;

$\implies 125k+72=225+k$

$\implies 125k-k=225-72$

$\implies 124k=153$

$k=\frac{153}{124}$