# Using the rational zero theorem etc.

• May 15th 2010, 06:04 PM
ToXic01
Using 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.
$
x^3-7x^2+x-7=0$

Thanks
• May 15th 2010, 07:34 PM
Prove It
Quote:

Originally Posted by ToXic01
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.
$
x^3-7x^2+x-7=0$

Thanks

No need for any of that, you can factorise by grouping...

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

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

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

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

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

So $x = 7, i, -i$.
• May 15th 2010, 10:40 PM
integral
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:

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

But we could also find that using The fundamental theorem of algebra.
• May 16th 2010, 07:45 PM
ToXic01
the prof said we cant take short cuts (Itwasntme) so i can't take short cuts lol