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

1 Answer
Mar 18, 2017

Answer:

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.