# The remainder when 2x^3 + 9x^2 + 7x + 3 is divided by x - k is 9, how do you find k?

Oct 18, 2015

The remainder of dividing $f \left(x\right) = 2 {x}^{3} + 9 {x}^{2} + 7 x + 3$ by $\left(x - k\right)$ is $f \left(k\right)$, so solve $f \left(k\right) = 9$ using the rational root theorem and factoring to find:

$k = \frac{1}{2} , - 2$ or $- 3$

#### Explanation:

If you attempt to divide $f \left(x\right) = 2 {x}^{3} + 9 {x}^{2} + 7 x + 3$ by $x - k$ you end up with a remainder of $f \left(k\right)$...

So if the remainder is $9$, we are basically trying to solve $f \left(k\right) = 9$

$2 {k}^{3} + 9 {k}^{2} + 7 k + 3 = 9$

Subtract $9$ from both sides to get:

$2 {k}^{3} + 9 {k}^{2} + 7 k - 6 = 0$

By the rational root theorem, any rational roots of this cubic will be of the form $\frac{p}{q}$ in lowest terms, where $p , q \in \mathbb{Z}$, $q \ne 0$, $p$ a divisor of the constant term $- 6$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the possible rational roots are:

$\pm \frac{1}{2}$, $\pm 1$, $\pm \frac{3}{2}$, $\pm 2$, $\pm 3$, $\pm 6$

Let's try the first one:

$f \left(\frac{1}{2}\right) = \frac{1}{4} + \frac{9}{4} + \frac{7}{2} - 6 = \frac{1 + 9 + 14 - 24}{4} = 0$

so $k = \frac{1}{2}$ is a root and $\left(2 k - 1\right)$ is a factor.

Divide by $\left(2 k - 1\right)$ to find:

$2 {k}^{3} + 9 {k}^{2} + 7 k - 6 = \left(2 k - 1\right) \left({k}^{2} + 5 k + 6\right) = \left(2 k - 1\right) \left(k + 2\right) \left(k + 3\right)$

So the possible solutions are:

$k = \frac{1}{2}$, $k = - 2$ and $k = - 3$