Question

Please HElp!
Solve the equation in the real number system

x^4-2x^3+6x^2-18x-27=0

The real solutions of the equation are x=____
(type each answer only once; do not duplicate answers in the case of repeated roots.) Type in exact answer, using radicals as needed. Use integers or fractions for any numbers in the expression)
OR
There are no real solutions.

Answer 1

Do you know the Rational Root Theorem?

Answer 2

I dont know much about it since I learned it two days ago

Answer 3

It says that IF there are any rational roots, they are of the form: \(\dfrac{dividers~of~ \pm 27}{dividers~of~\pm1}\).

Here, the -27 is the constant term at the end, and the 1 is the coefficient of the first term (1x^4, so it is 1).

Dividers of 27 are 1, 3, 9 and 27, and 1 can only be divided 1.

So any rational roots are of the form \(\dfrac{\pm 1,\pm3,\pm9,\pm27}{\pm1}\).

So there are not so many possibilities: \(\pm1,\pm3,\pm9,\pm27\).

What good is this information? After all, you have to give all real solutions.
The possible solutions mentioined above are real, but so is \(\sqrt{2}\)!
That is true, but in these kind of questions, you can usually trust there are some "nice" solutions.

Answer 4

do you just plug in those number for x then??

Answer 5

How to find out if one of the candidates ±1,±3,±9,±27 is actually a solution?
Just put them in the equation. If you get 0, you've found one.

To help you: I found x=3 to be a solution.
This means you can factor x^4-2x^3+6x^2-18x-27=0; it will be:

(x-3)(x^3 .......)=0.

This is because when you put in 3 here, you get (3-3)(...........)=0(.........)=0.
How to find out what's on the dots?

That is where your Synthetic Division comes in!

So set up a Synth. Div. :

3 | 1 -2 6 -18 -27
| 3 3
-------------------- +
1 1 9

Etc.

You can do the rest yourself.
The remainder will be 0!

Answer 6

will there be 4 answers?

Answer 7

In theory, there could be, but we do not know that yet.

We have got so far:

\(x^4-2x^3+6x^2-18x-27=0 \Leftrightarrow (x-3)(x^3+x^2+9x+9)=0\)

We could now use the Rational Root Theorem on \(x^3+x^2+9x+9\) again, but there is a quicker way: we can try to factor.

\(x^3+x^2+9x+9=x^2(x+1)+9(x+1)=(x^2+9)(x+1)\).

Do you see what I did here?

Answer 8

no really. but I got -1, 3 for the synthetic division way

Answer 9

OK, so we have 3 and -1 as real solutions.
We can now factor the thing like this:
\(x^4−2x^3+6x^2−18x−27=0⇔(x−3)(x+1)(x^2+9)=0\).

That last factor, \(x^2+9\) cannot be factored anymore, because its zeroes would be \(\pm \sqrt{-9}\) and those are not real.

Conclusion: -1 and 3 are the only real solutions (and there are 2 more complex solutions).

Answer 10

Thank you SUPER_DUPER MUCH!!!!

Answer 11

YW!
It was fun, don't you agree?

Answer 12

yes it was when I understood it!