Question

How do you use the remainder theorem to see if the `n+5` is a factor of `n^5-25n^3-7n^2-37n-18`?

Answer 1

`(n+5)` is not a factor

Explanation:

Let `f(n)=n^5-25n^3-7n^2-37n-18`
To see if `(n+5)` is a factor, we calculate `f(-5)`
So `f(-5)=(-5)^5-25*(-5)^3-7(-5)^2-37*(-5)-18`
`=-3125+3125-175+185-18=-8`
`f(-5)!=0`
There is a remainder of `-8`

Therefore `(n+5)` is not a factor

Answer 2

Analyse the value of the polynomial for `n=-5` to find `(n+5)` is not a factor

Explanation:

Given:

`f(n) = n^5-25n^3-7n^2-37n-18`

The remainder theorem tells us that `(n+5)` is a factor of `f(n)` if and only if `f(-5) = 0`

Observe that all of the terms of `f(n)` except the constant term are divisible by `n`:

`f(n) = n^5-25n^3-7n^2-37n-18`

`color(white)(f(n)) = n(n^4-25n^2-7n-37)-18`

So if `n` is not a factor of the constant term then `f(n) != 0`...

`f(color(blue)(-5)) = (color(blue)(-5))((color(blue)(-5))^4-25(color(blue)(-5))^2-7(color(blue)(-5))-37)-18`

`color(white)(f(color(white)(-5))) = (color(blue)(-5))((color(blue)(-5))^4-25(color(blue)(-5))^2-7(color(blue)(-5))-37)-5*4+2`

`color(white)(f(color(white)(-5))) = 5k+2" "` for some integer `k`

`color(white)(f(color(white)(-5))) != 0`

Since `f(-5) != 0` we can deduce that `(n+5)` is not a factor.