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
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
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 $\displaystyle x = 7, i, -i$.
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.