# How do you find the remainder when x^5-2x^4-3x^3+4x^2+3x+6 is divided by x-1?

Jun 7, 2017

Remainder is $9$

#### Explanation:

$x - 1 = 0 \therefore x = 1$

${x}^{5} - 2 {x}^{4} - 3 {x}^{3} + 4 {x}^{2} + 3 x + 6$

By putting $x = 1$ in the equation we can find remainder (R).

$R = {1}^{5} - 2 \cdot {1}^{4} - 3 \cdot {1}^{3} + 4 \cdot {1}^{2} + 3 \cdot 1 + 6 = 1 - 2 - 3 + 4 + 3 + 6 = 14 - 5 = 9$ [Ans]