Question

When we divide `4x^3-2x^2-18x+16` by `2x-4`, what is the remainder and the constant term in quotient?

(A) `0` and `-4`
(B) `2` and `-8`
(C) `2` and `-3`
(D) `4` and `-1`

Answer 1

Answer is (C)

Explanation:

I have tried solving without carrying out any long or synthetic division.

When we divide a rational polynomial `f(x)` (with coefficient of leading term being `1`) by `(x-a)` remaindeer is #f(a).

Here we have to divide `4x^3-2x^2-18x+16` by `2x-4`.

Halving the two we get `2x^3-x^2-9x+8` and `x-2` and if we divide former by latter, while quotient will be the same remainder will be halved

As remainder is `f(2)=2xx2^3-2^2-9xx2+8=16-4-18+8=2`

Hence remainder on dividing `4x^3-2x^2-18x+16` by `2x-4` will be `4`

As such the dividend that will be divisible by `2x-4` will be `4x^3-2x^2-18x+16-4=4x^3-2x^2-18x+12`

and hence constant term in quotient will be `12/(4)=-3`

Hence answer is (C)