# Factorisation

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• Dec 3rd 2009, 09:06 PM
Paymemoney
Factorisation
Can someone tell me if i have factorized correctly:

Equation is this:$\displaystyle x^4-6x^2+1$
First i completed the square:
$\displaystyle x^4-6x^2+1$
$\displaystyle x^4-6x^2+1-9x^2$
$\displaystyle (x^2-3)^2-(3x)^2$
$\displaystyle (x^2-3-3x)(x^2-3+3x)$
Now i use the difference of two squares where $\displaystyle a^2-b^2=(a-b)(a+b)$
$\displaystyle x^2-3x-3$
$\displaystyle (x-3)^2-(\sqrt3)^2$
$\displaystyle (x-3-\sqrt3)(x-3+\sqrt3)$
$\displaystyle x^2+3x-3$
$\displaystyle (x+3)^2-(\sqrt3)^2$
$\displaystyle (x+3-\sqrt3)(x+3+\sqrt3)$
Therefore answer is:
$\displaystyle (x-3-\sqrt3)(x-3+\sqrt3)(x+3-\sqrt3)(x+3+\sqrt3)$

P.S
• Dec 3rd 2009, 09:52 PM
Bacterius
Use the \sqrt{number} command to make a square root :D
• Dec 3rd 2009, 10:56 PM
Paymemoney
Quote:

Originally Posted by Bacterius
Use the \sqrt{number} command to make a square root :D

oh ok
• Dec 3rd 2009, 11:24 PM
dedust
i got this

$\displaystyle x^4-6x^2+1=(x^2-3)^2-{\sqrt{8}}^2 = (x^2-3-\sqrt{8})(x^2-3+\sqrt{8})$
• Dec 3rd 2009, 11:43 PM
Prove It
Quote:

Originally Posted by Paymemoney
Can someone tell me if i have factorized correctly:

Equation is this:$\displaystyle x^4-6x^2+1$
First i completed the square:
$\displaystyle x^4-6x^2+1$
$\displaystyle x^4-6x^2+1-9x^2$
$\displaystyle (x^2-3)^2-(3x)^2$
$\displaystyle (x^2-3-3x)(x^2-3+3x)$
Now i use the difference of two squares where $\displaystyle a^2-b^2=(a-b)(a+b)$
$\displaystyle x^2-3x-3$
$\displaystyle (x-3)^2-(\sqrt3)^2$
$\displaystyle (x-3-\sqrt3)(x-3+\sqrt3)$
$\displaystyle x^2+3x-3$
$\displaystyle (x+3)^2-(\sqrt3)^2$
$\displaystyle (x+3-\sqrt3)(x+3+\sqrt3)$
Therefore answer is:
$\displaystyle (x-3-\sqrt3)(x-3+\sqrt3)(x+3-\sqrt3)(x+3+\sqrt3)$

P.S

I'd use a dummy variable $\displaystyle X = x^2$ to turn it into a quadratic.

So $\displaystyle x^4 - 6x^2 + 1 = X^2 - 6X + 1$

$\displaystyle = X^2 - 6X + (-3)^2 - (-3)^2 + 1$

$\displaystyle = (X - 3)^2 - 8$

$\displaystyle = (X - 3)^2 - (\sqrt{8})^2$

$\displaystyle = (X - 3)^2 - (2\sqrt{2})^2$

$\displaystyle = (X - 3 + 2\sqrt{2})(X - 3 - 2\sqrt{2})$

Now remembering that $\displaystyle X = x^2$

$\displaystyle = (x^2 - 3 + 2\sqrt{2})(x^2 - 3 - 2\sqrt{2})$.

You could now use the difference of two squares again.

$\displaystyle = [x^2 - (3 - 2\sqrt{2})][x^2 - (3 + 2\sqrt{2})]$

$\displaystyle = \left[x^2 - \left(\sqrt{3 - 2\sqrt{2}}\right)^2\right]\left[x^2 - \left(\sqrt{3 + 2\sqrt{2}}\right)^2\right]$

$\displaystyle = \left(x + \sqrt{3 - 2\sqrt{2}}\right)\left(x - \sqrt{3 - 2\sqrt{2}}\right)\left(x + \sqrt{3 + 2\sqrt{2}}\right)\left(x - \sqrt{3 + 2\sqrt{2}}\right)$.

Ugly, yes, but fully factorised, yes.
• Dec 4th 2009, 12:03 AM
Paymemoney
^
i understand the first part, however i don't understand why $\displaystyle 3-2\sqrt2$ is square rooted?
• Dec 4th 2009, 12:07 AM
Prove It
Quote:

Originally Posted by Paymemoney
^
i understand the first part, however i don't understand why is $\displaystyle = \left[x^2 - \left(\sqrt{3 - 2\sqrt{2}}\right)^2\right]\left[x^2 - \left(\sqrt{3 + 2\sqrt{2}}\right)^2\right]$ is square rooted?

Because $\displaystyle 3 - 2\sqrt{2} = \left(\sqrt{3 - 2\sqrt{2}}\right)^2$ and $\displaystyle 3 + 2\sqrt{2} = \left(\sqrt{3 + 2\sqrt{2}}\right)^2$.

We had to do this to create another difference of two squares.
• Dec 4th 2009, 12:16 AM
Paymemoney
Quote:

Originally Posted by Prove It
Because $\displaystyle 3 - 2\sqrt{2} = \left(\sqrt{3 - 2\sqrt{2}}\right)^2$ and $\displaystyle 3 + 2\sqrt{2} = \left(\sqrt{3 + 2\sqrt{2}}\right)^2$.

We had to do this to create another difference of two squares.

ok thanks