# How do you determine the value of k if the remainder is 3 given (kx^3+3x+1) div(x+2)?

Jul 27, 2017

$k = - 1$

#### Explanation:

The Remainder Theorem states that:

if the polynomial $P \left(x\right) \text{ }$is divided by$\text{ "(x-a)" }$the remainder is $P \left(a\right)$

In this case we have

$P \left(x\right) = k {x}^{3} + 3 x + 1$

and the divisor $\left(x + 2\right) \implies a = - 2$

$\therefore P \left(- 2\right) = 3$

$\therefore P \left(- 2\right) = k {\left(- 2\right)}^{3} + 3 \times - 2 + 1 = 3$

$\implies - 8 k - 6 + 1 = 3$

$- 8 k - 5 = 3$

$- 8 k = 8$

$\implies k = - 1$