Polynomial P ( x ) contains only terms of odd degree. When P ( x ) is divided by ( x -3), the remainder is 6. If P ( x ) is divided by ( x 2-9) then remainder is g ( x ). Find the value of g (2).

Question

Polynomial P ( x ) contains only terms of odd degree. When P ( x ) is divided by ( x -3), the remainder is 6. If P ( x ) is divided by ( x 2-9) then remainder is g ( x ). Find the value of g (2). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

As P(x) is an odd function hence P(-x)=-P(x) ⇒ P(-3)=-P(3)=-6 let P ( x )= Q ( x 2-9)+ ax + b (where Q is quotient and ( ax + b )= g ( x )= remainder) now P (3)=3 a + b =6....(1) P(-3)=-3 a+b=-6.....(2) hence b=0 and a=2 hence g(x)=2 x ⇒ g(2)=4

Q. Polynomial P(x) contains only terms of odd degree. When P(x) is divided by (x−3), the remainder is 6. If P(x) is divided by (x2−9) then remainder is g(x). Find the value of g(2).

As P(x) is an odd function hence P(−x)=−P(x)⇒P(−3)=−P(3)=−6 let P(x)=Q(x2−9)+ax+b (where Q is quotient and (ax+b)=g(x)= remainder) now P(3)=3a+b=6....(1) P(−3)=−3a+b=−6.....(2) hence b=0 and a=2 hence g(x)=2x⇒g(2)=4

Q. Polynomial $P (x)$ contains only terms of odd degree. When $P (x)$ is divided by $(x - 3)$ , the remainder is 6. If $P (x)$ is divided by $(x^{2} - 9)$ then remainder is $g (x)$ . Find the value of $g (2)$ .