Q25 of 32 Page 2

Let R1 and R2 are the remainders when polynomials x3 + 2x2 – 5ax – 7 and x3 + ax2 – 12x + 6 are divided by (x + 1) and (x – 2) respectively. If 2R1 + R2 = 6, find value of a.

Concept Used:


Remainder theorem: If a polynomial f(x) is divided by (x – a) then f(a) will give the remainder.


Explanation:


Let f(x) = x3 + 2x2 – 5ax – 7


And g(x) = x3 + ax2 – 12x + 6


Now, f(–1) = R1 and g(2) = R2


Therefore,


f(–1) = (–1)3 + 2(–1)2 – 5a(–1) – 7


f(–1) = –1 + 2 + 5a – 7


f(–1) = 5a – 6


And,


g(2) = (2)3 + a(2)2 – 12.2 + 6


g(2) = 8 + 4a – 24 + 6


g(2) = 4a – 10


It is given that,


2R1 + R2 = 6


2(5a – 6) + 4a – 10 = 6


10a – 12 + 4a – 10 = 6


14a – 22 = 6


14a = 28


a = 2


Hence, the value of a is 2.


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