Factorising (1 Viewer)

[Aznmichael92]

Hey, how can I factorise fully x^4 + 1

 

[tommykins]

hmmm....weird, i can only seem to do it using 4unit methods since it has complex roots.

 

[Aznmichael92]

gaaaa and its in a 3unit work sheet i was working on...

i guess i will just leave it as

(x2-1)^2 - 2x^2

 

[tommykins]

well another way is

x^4 + 1 = x^4 - 1 + 2 = (x²+1)(x²-1)+2 = (x²+1)(x-1)(x+1) +2

 

[Aznmichael92]

r u sure the questin is not x^4 -1?

positive as the question before it is x^4 - 1

 

[kaz1]

x⁴ +1
x⁴ +2x² + 1 - 2x²
(x² +1)² - 2x²
(x² +sqrt2x +1)(x²-sqrt2x +1)

 

[gurmies]

^ I don't think square roots in a factorising question make the expression any simpler?

 

[Trebla]

Factorising does NOT mean simplify and even then, simplify certainly doesn't necessarily mean there have to be integer or rational numbers involved.

If you were asked to factorise x² - 3, the answer would be (x - √3)(x + √3), it doesn't matter if there are irrational numbers involved, the expression is still factorised. So kaz1 is correct...

 

[gurmies]

oic...fair enough

 

[super.muppy]

(x-1)(x+1)(x^2+2)

 

[gurmies]

^wrong

that doesn't expand to x^4 + 1

 

[P.T.F.E]

Maybe (x^2+1)^2 - 2x^2

 

[Iruka]

I think kaz 1 had the correct answer - the expression has been factored into two polynomials of lower degree. Of course, the Q didn't specify which field we are supposed to factor over - the rationals, the reals or the complex numbers, so the use of radicals has not been ruled out (but then neither has the use of complex numbers, except that we would assume that a question in an MX1 textbook is not going to ask that).