# 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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