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#### paulmdrdo

##### Active member

- May 13, 2013

- 386

how to force factor this into the difference of two squares.

$\displaystyle x^4 + 1$

$\displaystyle x^4 + 1$

- Thread starter paulmdrdo
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- Thread starter
- #1

- May 13, 2013

- 386

how to force factor this into the difference of two squares.

$\displaystyle x^4 + 1$

$\displaystyle x^4 + 1$

- Jan 17, 2013

- 1,667

\(\displaystyle x^4+1=x^4+2x^2+1-2x^2\)

or

\(\displaystyle x^4+1=x^4-i^2\)

The rest is for you

- Thread starter
- #3

- May 13, 2013

- 386

I would get this

$\displaystyle \displaystyle \begin{align*} x^4 + 1 &= x^4 + 2x^2 + 1 - 2x^2 \\& = \left( x^2 + 1 \right) ^2 - \left( \sqrt{2} \, x \right) ^2 \\& = \left( x^2 - \sqrt{2}\, x + 1 \right) \left( x^2 + \sqrt{2}\,x + 1 \right) \end{align*}$

but i want to know what's your reasoning by choosing the term 2x^2?

- Jan 17, 2013

- 1,667

The idea is complete the square , since if we have for example :I would get this

$\displaystyle \displaystyle \begin{align*} x^4 + 1 &= x^4 + 2x^2 + 1 - 2x^2 \\& = \left( x^2 + 1 \right) ^2 - \left( \sqrt{2} \, x \right) ^2 \\& = \left( x^2 - \sqrt{2}\, x + 1 \right) \left( x^2 + \sqrt{2}\,x + 1 \right) \end{align*}$

but i want to know what's your reasoning by choosing the term 2x^2?

\(\displaystyle x^2+1\) our first glance suggests converting it to $x^2+2x+1$ which is a complete square hence factorizing will be possible .