# When 3x^2+6x-10 is divided by x+k, the remainder is 14. How do you determine the value of k?

Feb 19, 2017

The values of $k$ are $\left\{- 4 , 2\right\}$

#### Explanation:

We apply the remainder theorem

When a polynomial $f \left(x\right)$ is divided by $\left(x - c\right)$, we get

$f \left(x\right) = \left(x - c\right) q \left(x\right) + r \left(x\right)$

When $x = c$

$f \left(c\right) = 0 + r$

Here,

$f \left(x\right) = 3 {x}^{2} + 6 x - 10$

$f \left(k\right) = 3 {k}^{2} + 6 k - 10$

which is also equal to $14$

therefore,

$3 {k}^{2} + 6 k - 10 = 14$

$3 {k}^{2} + 6 k - 24 = 0$

We solve this quadratic equation for $k$

$3 \left({k}^{2} + 2 k - 8\right) = 0$

$3 \left(k + 4\right) \left(k - 2\right) = 0$

So,

$k = - 4$

or

$k = 2$