Question

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

Answer 1

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

`k = 1/2, -2` or `-3`

Explanation:

If you attempt to divide `f(x) = 2x^3+9x^2+7x+3` by `x-k` you end up with a remainder of `f(k)`...

So if the remainder is `9`, we are basically trying to solve `f(k) = 9`

`2k^3+9k^2+7k+3 = 9`

Subtract `9` from both sides to get:

`2k^3+9k^2+7k-6 = 0`

By the rational root theorem, any rational roots of this cubic will be of the form `p/q` in lowest terms, where `p, q in ZZ`, `q != 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:

`+-1/2`, `+-1`, `+-3/2`, `+-2`, `+-3`, `+-6`

Let's try the first one:

`f(1/2) = 1/4+9/4+7/2-6 = (1+9+14-24)/4 = 0`

so `k = 1/2` is a root and `(2k-1)` is a factor.

Divide by `(2k-1)` to find:

`2k^3+9k^2+7k-6 = (2k-1)(k^2+5k+6) = (2k-1)(k+2)(k+3)`

So the possible solutions are:

`k = 1/2`, `k = -2` and `k = -3`