If r and s be the remainders when the polynomials (x³ + 2x² - 5ax - 7) and (x³ + ax² - 12x + 6) are divided by (x + 1) and (x - 2) respectively and 2r + s = 6 find the value of a.

If we divide f(x) = (x³ + 2x² - 5ax - 7) by (x + 1) remainder can be find at value of –

(x + 1) = 0

Or x = -1

So, we will put x = -1 in f(x) = (x³ + 2x² - 5ax - 7)

f(-1) = ((-1)³ + 2(-1)² - 5a(-1)-7)

= -6 + 5a

So, remainder = r = -6 + 5a

Also if we divide f(x) = (x³ + ax² - 12x + 6) by (x - 2) remainder can be find at value of –

(x - 2) = 0

Or x = 2

So we will put x = 2 in f(x) = (x³ + ax² - 12x + 6)

f(2) = (2³ + a2² - 12(2) + 6)

= 4a - 10

So, remainder = s = 4a - 10

Also it is given that 2r + s = 6

So putting r and s from above expressions-

2(-6 + 5a) + (4a - 10) = 6

= 14a = 28

= a = 2