Question

the polynomial p(x)= x^3+2x^2-5ax-7 and q(x)= x^3+ax^2-12x+6 leave same remainder when divided by (x+1) and (x-2) respectively. Find thevalue of a.

Answer 1

(X+1)=0( using remainder theorem)
X+1=0
X=-1

P(x) = x ^3+2×^2+5ax-7.
P(-1)=-1 ^3+2×-1^2+5a×-1 -7( substitute x=-1)
= -1 +2 -5a-7

-1+2-5a-7= 0
1+5a-7=0
-6+5a=0
5a =6
a= 6\5




(X-2)=0( using remainder theorem)
X-2=0
X= 2

q(x) = x^3 +ax^2-12x+6(substitute x=2)
= 2^3 +a2^2 - 12 ×2+6
= 8+4a -24+6
= 14-24+4a
= -10+4a

-10+4a = 0
4a = 10
a= 10\4
By simplifying 10\4 we get 5\2
a=5\2