Question related to Polynomials: factoring x^2 + (1/x^2) - 11

sa.azeemm said:
How to factorize x^2 + (1/x^2) - 11 ?
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This is not an actual quadratic (or even a polynomial), because the middle term has the variable in the denominator. But the expression is of the quadratic "type". Here's the trick:

. . . . .\(\displaystyle x^2\, -\, 11\, +\, \dfrac{1}{x^2}\)

. . . . .\(\displaystyle (x)^2\, -\, 11\, +\, \left(\dfrac{1}{x}\right)^2\)

. . . . .\(\displaystyle (x)^2\, -\, 11(x)\left(\dfrac{1}{x}\right)\, +\, \left(\dfrac{1}{x}\right)^2\)

In other words, this is kind of like when you've had quadratics with \(\displaystyle \,x\,\) and \(\displaystyle \, y\,\) in them.

Unfortunately, this doesn't factor. Had the "1" in the fractional term been a "12", you'd have been golden. Oh, well. ;)
 
Denis said:
I'm headed for the corner, repeating "quartic...."....2^4 minutes.
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Wait one. If you had said it was a quadratic in x then I agree you head to the corner. But since it is a quadratic in x2 you might just make it 2^2 minutes [NOT! minuets] for being (very) slightly unclear.
 
sa.azeemm said:
How to factorize x^2 + (1/x^2) - 11 ?
Click to expand...
I would do as Denis suggested (I never learn) but would not set the expression equal to 0 (what were you thinking! Hmm, that assumes that you think).
x^2 + (1/x^2) - 11 = (1/x^2)(x^4 -11x^2 + 1). As lookagain pointed out this 2nd factor is a quartic, however we can think of it as a quadratic if we let u=x^2.
Then the 2nd factor becomes u^2 -11u +1. Use the quadratic formula to find the roots. Say they are c and d. So u^2-11u-1 =(u-c)(u-d) which is really (x^2-c)(x^2-d). If c and/or d is positive then you can factor x^2-c and/or x^2-d.
So x^2 + (1/x^2) - 11 = (1/x^2)(x^2-c)(x^2-d) and again you may still be able to factor more.
 
Hello, sa.azeemm!
Factor: \(\displaystyle \: x^2 + \dfrac{1}{x^2} - 11\)
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We have: \(\displaystyle \: x^2 - 2 + \dfrac{1}{x^2} - 9\). . . . . \(\displaystyle =\; \left(x - \dfrac{1}{x}\right)^2 - 3^2 \) . . . . difference of squares. . . . . \(\displaystyle =\;\left(x - \dfrac{1}{x} - 3\right)\,\left(x - \dfrac{1}{x} +3\right)\)
 
soroban said:
Hello, sa.azeemm!We have: \(\displaystyle \: x^2 - 2 + \dfrac{1}{x^2} - 9\). . . . . \(\displaystyle =\; \left(x - \dfrac{1}{x}\right)^2 - 3^2 \) . . . . difference of squares. . . . . \(\displaystyle =\;\left(x - \dfrac{1}{x} - 3\right)\,\left(x - \dfrac{1}{x} +3\right)\)
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That is genuinely impressive!!!!
 
soroban said:
Hello, sa.azeemm!We have: \(\displaystyle \: x^2 - 2 + \dfrac{1}{x^2} - 9\). . . . . \(\displaystyle =\; \left(x - \dfrac{1}{x}\right)^2 - 3^2 \) . . . . difference of squares. . . . . \(\displaystyle =\;\left(x - \dfrac{1}{x} - 3\right)\,\left(x - \dfrac{1}{x} +3\right)\)
Click to expand...
This is sweet! I am truly impressed. Who says that math is not beautiful?
 
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