If x

^{4}– kx^{3}+ 10x^{2}+ 1x – 3 is divisible by (x – 1), and if when it is divided by (x + 2) the remainder is 27, find the constants k and 1-
**A.**

k = -7, 1 = -15 -
**B.**

k = -15, 1 = -7 -
**C.**

k = (frac{15}{3}) , 1 = -7 -
**D.**

k = (frac{7}{3}) , 1 = -17

##### Correct Answer: Option A

##### Explanation

If k = -7 is put as -15, the equation x^{4} – kx^{3} + 10x^{2} + 1x – 3 becomes x^{4} – (7x^{3}) + 10x^{2} + (15)^{-3} = x^{4} + 7x^{3} + 10x^{2} – 15x – 3

This equation is divisible by (x – 1) and (x + 2) with the remainder as 27

k = -7, 1 = -15