Nine Standard >> Application of Remainder theorem | Part -2

Question:

\(P(x)=2x^{3}+kx^{2}-3x+5\) and \(q(x)=x^{3}+2x^{2}-x+k\) when divided by (x-2) leave remainder \(r_1\) and \(r_2\) respectively. Find k if \(r_1-r_2=0\)

Solution:

x-2 is the divison. Remainder P(2)=\(r_1\)

\(P(2)=2.2^3+k.2^2-3.2+5\)
       =16+4k-6+5
       =4k+15

\(r_1\)=4k+15-----(1)
\(q(x)=x^3+2x^2-x+k\) is divided by x-2
\(q(2)=2^3+2.2^2-2+k\)
       =8+8-2+k=14+k
  \(r_2\)=14+k------(2)
Putting in \(r_1-r_2=0\)
               =>4k+15-14-k=0
               =>3k+1=0
               =>k=-\(\frac{1}{3}\)