Question

How do you factor given that `f(9)=0` and `f(x)=x^3-18x^2+95x-126`?

Answer 1

The equation becomes:
`f(x) = (x-9)(x-7)(x-2)`

Explanation:

Hmm, I don't know... Let's try it out, shall we?

First, you're given that `f(9) = 0`
and
`f(x) = x^3 - 18x^2 +95x -126`

Well, let's check if that's true!

`f(9) = 9^3 - 18*9^2 +95*9 -126`
i.e.
`f(9) =`"big numbers", darn.

Since I'm too lazy to do the calculations,
let's say they are correct, and assume that `f(9) = 0`

What's another equation that we know of, that also has 9 as its solution?
Can you think of a simple one?
This one for example:

`g(x) = x-9`

This give `g(9) = 0`, exactly the same as our first equation.
This also means that "if" 9 is really a solution to our equation, we can factor it out, and the equation should look like:

`f(x) = (x-9)(ax^2 + bx + c)`
where `a`, `b`, and `c` are real numbers.

Now, let's open the parentheses and let's gather all similar terms together:
`f(x) = ax^3 + bx^2 + cx - 9ax^2 -9bx -9c`
i.e.
`f(x) = ax^3 + (b-9a)x^2 + (c-9b)x -9c`

Great!
Now, we can relate this to our first equation.
This means that
`a=1`,
`(b-9a) = -18`,
`(c-9b)=95`, and
`-9c = -126`

Let's solve for `a`, `b`, and `c`. Well `a=1` so that's taken care of.
Moving to the second line, we then have
`(b-9*1) = -18`
i.e.
`b=-9`

and finally,
`c=-126/-9` i.e. `c=14`

So, our new equation looks like the following:
`f(x) = (x-9) (x^2 -9x +14)`

But let's not stop here! Maybe we can factor the right side one more time! For that, let's see if we can solve it if we set it to 0.
`x^2 -9x +14 = 0`
We solve it using the quadratic formula (see other posts!),
and we get:
`x_1 = 7` and `x_2 = 2`
which means we can rewrite the equation as such:
`x^2 -9x + 14 = (x-7)(x-2) = 0`

So, finally, the first equation becomes:
`f(x) = (x-9)(x-7)(x-2)`

Q.E.D.