third-degree polynomial integer coefficients

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May 27th 2007, 04:11 PM
hsmath1918

third-degree polynomial integer coefficients

I have this equation x^3 + ax^2 + bx + c
I know x = 3/(7^(1/3) - 2) is a solution.
How can I find a, b, c when they are all integers?
May 27th 2007, 06:21 PM
ThePerfectHacker

Quote:

Originally Posted by hsmath1918

I have this equation x^3 + ax^2 + bx + c
I know x = 3/(7^(1/3) - 2) is a solution.
How can I find a, b, c when they are all integers?

You have that $x = \frac{3}{7^{1/3} - 2}$ is a solution.

So,
$\frac{3}{x} = 7^{1/3} - 2$

$\frac{3}{x} + 2 = 7^{1/3}$

Cube both sides,

$\frac{27}{x^3} + \frac{54}{x^2} + \frac{36}{x}+ 8 = 7$

Multiply by $x^3$ thus:

$27 + 54x+36x^2+8x^3 = 7x^3$

Thus, $x^3 + 36x^2 + 54x+27=0$
May 27th 2007, 06:41 PM
Soroban

Hello, hsmath1918!

Quote:

I have this equation: . $x^3 + ax^2 + bx + c \:=\:0$

I know $x = \frac{3}{\sqrt[3]{7} - 2}$ is a solution.

How can I find $a,\,b,\,c$ when they are all integers?

There may be many cubic equations with that solution . . . or not.
Here is one of them . . .

We have: . $x \:=\:\frac{3}{\sqrt[3]{7} -2}\quad\Rightarrow\quad\left(\sqrt[3]{7} - 2\right)x\:=\:3$

. . $\sqrt[3]{7}x - 2x\:=\:3\quad\Rightarrow\quad \sqrt[3]{7}x \:=\:2x + 3$

Cube both sides: . $\left(\sqrt[3]{7}x\right)^3\:=\:(2x + 3)^3\quad\Rightarrow\quad7x^3\:=\:8x^3 + 36x^2 + 54x + 27$

Therefore, an equation is: . $x^3 + 36x^2 + 54x + 27\;=\;0$

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