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

Mar 18, 2017

The equation becomes:
$f \left(x\right) = \left(x - 9\right) \left(x - 7\right) \left(x - 2\right)$

#### Explanation:

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

First, you're given that $f \left(9\right) = 0$
and
$f \left(x\right) = {x}^{3} - 18 {x}^{2} + 95 x - 126$

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

$f \left(9\right) = {9}^{3} - 18 \cdot {9}^{2} + 95 \cdot 9 - 126$
i.e.
$f \left(9\right) =$"big numbers", darn.

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

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:

$g \left(x\right) = x - 9$

This give $g \left(9\right) = 0$, 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:

$f \left(x\right) = \left(x - 9\right) \left(a {x}^{2} + b x + c\right)$
where $a$, $b$, and $c$ are real numbers.

Now, let's open the parentheses and let's gather all similar terms together:
$f \left(x\right) = a {x}^{3} + b {x}^{2} + c x - 9 a {x}^{2} - 9 b x - 9 c$
i.e.
$f \left(x\right) = a {x}^{3} + \left(b - 9 a\right) {x}^{2} + \left(c - 9 b\right) x - 9 c$

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

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
$\left(b - 9 \cdot 1\right) = - 18$
i.e.
$b = - 9$

and finally,
$c = - \frac{126}{-} 9$ i.e. $c = 14$

So, our new equation looks like the following:
$f \left(x\right) = \left(x - 9\right) \left({x}^{2} - 9 x + 14\right)$

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.
${x}^{2} - 9 x + 14 = 0$
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:
${x}^{2} - 9 x + 14 = \left(x - 7\right) \left(x - 2\right) = 0$

So, finally, the first equation becomes:
$f \left(x\right) = \left(x - 9\right) \left(x - 7\right) \left(x - 2\right)$

Q.E.D.