Question

If `f(x)=x^2+3x+k` is divided by `x+k`, remainder is `0`. What is the value of `k`?

Answer 1

`k= 0 and 2`

Explanation:

`(x^2 + 3 x + k )/ (x + k)`

we use long division to solve it.
`x^2 + 3 x + k -> x * (x + k)`
`-(x^2 +kx)`
........................
`3 x - kx + k`
`(3 - k)x + k ->(3 - k) * (x + k)`
`-((3 - k)x + (3 - k)k)`
.......................
`k - (3 - k)k = k -3k + k^2 = -2k + k^2`

to be a divisible,
`-2k + k^2 = 0`
`k(-2 + k) = 0`

`k= 0 and 2`

Answer 2

`k=0` or `k=2`

Explanation:

We can use Remainder Theorem here. According to this theorem, if a polynomial `f(x)` is divided by `x-a`, the remainder is `f(a)`.

In other words, if `f(x)` is divisible by `x-a`, `f(a)=0`.

Now `f(x)=x^2+3x+k` to be divisible by `x+k` or `x-(-k)`,

we should have `f(-k)=(-k)^2+3xx(-k)+k=0`

or `k^2-3k+k=0`

i.e. `k(k-2)=0`

For this we must have either `k=0` or `k-2=0`

i.e. either `k=0` or `k=2`