When math students are taught factoring, usually the first thing they learn are special factoring identities such as:

1: a²-b²= (a-b)(a+b)
2: a³-b³ = (a-b)(a²+ab+b²)
3: a³+b³ = (a+b)(a²-ab+b²)

Find a factorization of the expression x⁴+x²+1 using only those identities.

If you check the formulas, you see a pattern in the trinomial a²+ab+b² in formula 2. So, feeling lucky, you can let a=x² and b=1 and plug these into the whole formula to get:

<o:p> </o:p>

x⁶-1=(x²-1)(x⁴+x²+1)

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Dividing both sides by (x²-1) and factoring the left side as a difference of 2 squares, we get

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(x³+1) (x³-1)/ (x²-1)= (x⁴+x²+1)

<o:p> </o:p>

Now, by factoring the two cubics on the left hand side as well as the difference of 2 squares in the denominator, there will be common factors of (x+1) and (x-1) that will cancel, leaving you with

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(x⁴+x²+1)=(x²-x+1)(x²+x+1)<o:p></o:p>

Posted by Mike C on 2007-03-22 14:32:29