Question

If 2x³ + ax² + bx – 6 has (x-1) as a factor and leaves a remainder 2 when divided by (x-2).

Find the values of ‘a’ and ‘b’.​

Answer 1

$\huge\pink{Answer}$

Let P(x) = 2x³ + ax² + bx - 6

If x-1 is a factor of P(x) then the sum of the coefficients of P(x) = 0

So 2 + a + b - 6 = 0 => a+b = 4 ……..(1)

When P(x) is divided by x-2 then the remainder is 2.

Then by remainder theorem P(2) = 2

2×2³ + a×2²+2b-6 = 2

=> 4a + 2b = -8 => 2a + b = -4 ……(2)

(2) - (1) gives a = -8 and using this in (1) we get b = 12

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