Problems on Remainder theorem

If  2x³ – (a – b)x² –(4b – 1)x + 4a is divisible by x² - 3x -28, then find the value of “a” and “b”.

Answer:

Factorizing x² - 3x -28 we get,

= x² - 7x + 4x -28

= x(x - 7) + 4(x - 7)

= (x- 7)(x + 4)

Let F(x) = 2x³ – (a – b)x² –(4b – 1)x + 4a

Now according to remainder theorem,  F(7) = 0 and F(-4) = 0 for complete divisibility.

Substituting x = 7 we get

686 –49 (a – b) – 7(4b – 1) + 4a = 0

Or,  693 – 45a + 21b = 0

Or, 45a – 21b = 693    -------------(1)

Similarly substituting x = 4, we get

-128 – 16(a – b) + 4(4b – 1) + 4a = 0

Or, -132 – 12a + 32b = 0

Or, 12a – 32b = -132  ---------(2)

Solving Equation 1 and 2 we get,

“a” = 21

“b” = 12