# Math Help - Factoring A Third Degree Polynomial

1. ## Factoring A Third Degree Polynomial

So it's been quite a while since I've worked with third degree polynomials, and I'm pretty rusty ...any help would be greatly appreciated. This is the equation:

"Find all solutions to the equation correct to two decimal places."
x^3 - 9x^2 - 4 = 0

I tried synthetic division but it didn't work out (I think).

2. Originally Posted by maryanna91
So it's been quite a while since I've worked with third degree polynomials, and I'm pretty rusty ...any help would be greatly appreciated. This is the equation:

"Find all solutions to the equation correct to two decimal places."
x^3 - 9x^2 - 4 = 0

I tried synthetic division but it didn't work out (I think).
You will have to use Newton's method.

3. Originally Posted by maryanna91
So it's been quite a while since I've worked with third degree polynomials, and I'm pretty rusty ...any help would be greatly appreciated. This is the equation:

"Find all solutions to the equation correct to two decimal places."
x^3 - 9x^2 - 4 = 0

I tried synthetic division but it didn't work out (I think).
The fact that you are NOT asked to find exact solutions should suggest something to you .... You are probably expected to use technology. What context did the question appear in?

4. Originally Posted by maryanna91
So it's been quite a while since I've worked with third degree polynomials, and I'm pretty rusty ...any help would be greatly appreciated. This is the equation:

"Find all solutions to the equation correct to two decimal places."
x^3 - 9x^2 - 4 = 0

I tried synthetic division but it didn't work out (I think).
Use the rational roots theorem to find any rational roots (it has none).

Using Descartes' rule of signs you will find that this has one positive root and no negative roots.

It is also evident that the root is less than $10$ (Cauchy's bound tells us this).

The cubic has different signs at $x=0$ and $x=10$, so the bisection method can be used to localise the root.

CB

5. Originally Posted by maryanna91
So it's been quite a while since I've worked with third degree polynomials, and I'm pretty rusty ...any help would be greatly appreciated. This is the equation:

"Find all solutions to the equation correct to two decimal places."
x^3 - 9x^2 - 4 = 0

I tried synthetic division but it didn't work out (I think).

Alternatively

$\displaystyle\ x^3-9x^2-4=0\Rightarrow\ x^3-9x^2=4\Rightarrow\ x^2(x-9)=4$

Both factors must be positive, hence $x>9$

which also implies (from both factors) that $x<10$

6. I've always wondered about that phrase "using techology". Pencils and paper don't grow out of the ground, do they? They have to be manufactured, so using pencil and paper is "using technology"!

(I wouldn't recommend using that argument with a teacher.)

$x^2(x-9)=4$

$9

where "y" is the decimal part.

This leads to a cubic in "y"

$(9+y)^2y=4\Rightarrow\ 81y+18y^2+y^3=4$

for which the 81y term is dominant.

$f(0)=0<4$

$f(1)=99>4$

$f(0.1)=8.1+..>4$

$f(0.001)=0.81+..<4$

$f(0.05)=4.05+..$ is very close to 4.

Hence the only real solution is very near x=9.05

8. Originally Posted by HallsofIvy
I've always wondered about that phrase "using techology". Pencils and paper don't grow out of the ground, do they? They have to be manufactured, so using pencil and paper is "using technology"!

(I wouldn't recommend using that argument with a teacher.)
I have to admit that in the context of mathematics I use the word 'technology' as a catch all word for 'graphics calculator or Computer Algebra System such as a CAS calculator or appropriate computer software'.

9. Originally Posted by HallsofIvy
I've always wondered about that phrase "using techology". Pencils and paper don't grow out of the ground, do they? They have to be manufactured, so using pencil and paper is "using technology"!

(I wouldn't recommend using that argument with a teacher.)
One of the best books dealing with the history of a technology is "The Pencil: A History of Design and Circumstance" by Henry Petroski

CB