polynomials (1 Viewer)

9876543210 · Aug 14, 2009

i dont quite get this example in the cambridge textbook.
it's on page 134, example 19.
can someone please explain it?
thx!
i dont quite get how you find the argument = pie/4

gurmies · Aug 14, 2009

9876543210 · Aug 14, 2009

oh yeah...
i forgot the negative sign
thx!!
now i can keep going...

gurmies · Aug 14, 2009

No worries, best of luck.

9876543210 · Aug 15, 2009

more questions.
1. If P(x)=x^4 -8x^3 +30x^2 -56x +49 has a non-real double zero, solve the equation P(x)=0 over the complex field and factorise P(x) fully over the real field.

2. The equation x^3 +3x +2=0 has roots a,b and c. Find the equation with roots a+1/a, b+1/b, c+1/c.

thx

gurmies · Aug 15, 2009

Question 2

$x^{3}+3x+2=0 \\\\ x=\alpha+\frac{1}{\alpha}\Rightarrow x\alpha=\alpha^{2}+1\Rightarrow \alpha^{2}-x\alpha+1=0 \\\\ \alpha=\frac{x\pm\sqrt{x^{2}-4}}{2}\\\\(\frac{x\pm \sqrt{x^{2}-4}}{2})^{3}+3(\frac{x\pm\sqrt{x^{2}-4}}{2})+2=0 \\\\ (x\pm \sqrt{x^{2}-4})^{3}+12(x\pm\sqrt{x^{2}-4})+16=0 \\\\ x^{3}+3x^{2}\sqrt{x^{2}-4}+3x(x^{2}-4)+(x^{2}-4)\sqrt{x^{2}-4}+12x+12\sqrt{x^{2}-4}+16=0 \\\\ x^{3}+3x^{3}-12x+12x+16+\sqrt{x^{2}-4}(3x^{2}+x^{2}-4+12)=0 \\\\ 4x^{3}+16+\sqrt{x^{2}-4}(4x^{2}+8)=0 \\\\ x^{3}+4=-\sqrt{x^{2}-4}(x^{2}+2) \\\\ (x^{3}+4)^{2}=(x^{2}-4)(x^{2}+2)^{2} \\\\ x^{6}+8x^{3}+16=x^{6}+4x^{4}+4x^{2}-4x^{4}-16x^{2}-16 \\\\ 8x^{3}+12x^{2}+32=0\Rightarrow 2x^{3}+3x^{2}+8=0$

Eventually I just get rid of that plus/minus and I can't really account for it, but I assume using the negative also yields the same equation...

9876543210 · Aug 15, 2009

yay thx!
i dont like it when the numbers are weird with roots and stuff...
but thx again!

gurmies · Aug 15, 2009

9876543210 said:
yay thx!
i dont like it when the numbers are weird with roots and stuff...
but thx again!

To avoid that, you could have written the new equation as x^3-(sum of roots)
x^2+(sum of product of roots taken 2 at a time)x - (product of roots) and you'd be able to work that out, but I don't usually do it that way =]

9876543210 · Aug 15, 2009

ok

i'll survive through the roots XD

untouchablecuz · Aug 15, 2009

9876543210 said:
more questions.
1. If P(x)=x^4 -8x^3 +30x^2 -56x +49 has a non-real double zero, solve the equation P(x)=0 over the complex field and factorise P(x) fully over the real field.

2. The equation x^3 +3x +2=0 has roots a,b and c. Find the equation with roots a+1/a, b+1/b, c+1/c.

thx

number 2

edit: in the picture, it should be "since z is a complex root z_ is also a root" in the third line

polynomials (1 Viewer)

9876543210

everything's fuzzy..

gurmies

Drover

9876543210

everything's fuzzy..

gurmies

Drover

9876543210

everything's fuzzy..

gurmies

Drover

9876543210

everything's fuzzy..

gurmies

Drover

9876543210

everything's fuzzy..

untouchablecuz

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