Question

How do you factor given that `f(4)=0` and `f(x)=x^3-14x^2+47x-18`?

Answer 1

`f(x) = (x-9)(x-5/2-sqrt(17)/2)(x-5/2+sqrt(17)/2)`

Explanation:

The difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

We use this later with `a=(x-5/2)` and `b=sqrt(17)/2`.

By the rational root theorem, any rational zeros of `f(x)` must be expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-18` and `q` a divisor of the coefficient `1` of the leading term.

So the only possible rational zeros are:

`+-1, +-2, +-3, +-6, +-9, +-18`

In particular `f(4) != 0` since `4` is not a factor of `18`.

Trying each of these in turn, we (eventually) find:

`f(9) = 729-1134+423-18 = 0`

So `(x-9)` is a factor of `f(x)`:

`x^3-14x^2+47x-18`

`= (x-9)(x^2-5x+2)`

`= (x-9)((x-5/2)^2-25/4+2)`

`= (x-9)((x-5/2)^2-(sqrt(17)/2)^2)`

`= (x-9)(x-5/2-sqrt(17)/2)(x-5/2+sqrt(17)/2)`