Question

Use the rational zero theorem, Descartes rules of signs, and the theorem of bounds as aids in finding all real and imaginary roots to the equation.

$\displaystyle

x^3-7x^2+x-7=0$

Answer: 7,-i,i

Help Please

Thanks

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- May 15th 2010, 06:04 PMToXic01Using the rational zero theorem etc.
Question

Use the rational zero theorem, Descartes rules of signs, and the theorem of bounds as aids in finding all real and imaginary roots to the equation.

$\displaystyle

x^3-7x^2+x-7=0$

Answer: 7,-i,i

Help Please

Thanks - May 15th 2010, 07:34 PMProve It
No need for any of that, you can factorise by grouping...

$\displaystyle x^3 - 7x^2 + x - 7 = 0$

$\displaystyle x^3 + x - 7x^2 - 7 = 0$

$\displaystyle x(x^2 + 1) - 7(x^2 + 1) = 0$

$\displaystyle (x - 7)(x^2 + 1) = 0$.

Therefore $\displaystyle x - 7 = 0$ or $\displaystyle x^2 + 1 = 0$

So $\displaystyle x = 7, i, -i$. - May 15th 2010, 10:40 PMintegral
Descartes rules of signs:

States, that the number of transitions from positive to negative (or vise versa) in a polynomial will be the number of roots a polynomial contains.

Here is the signs of the function:

$\displaystyle +,-,+,-$ I count 3 there.

But we could also find that using The fundamental theorem of algebra. - May 16th 2010, 07:45 PMToXic01
the prof said we cant take short cuts (Itwasntme) so i can't take short cuts lol