Thread: solve polynomial

Hi.

Originally Posted by gwiggy

Solve the equation: z^3 - 3z^2 + 6z -4=0

Please Help!

I guess you don't know newton's method to find zeros. Therefor you have to find a zero by guessing

f(z) :=z^3 - 3z^2 + 6z -4

very soon you find that f(1) = 1-3+6-4 = 0

Then use polynomial long division to get

[z^3-3z^2+6z-4] / [z-1] = z^2 - 2z + 4

I guess you do know how to find the complex zeros of this polynom.

Any questions?

Regards,
Rapha

Originally Posted by gwiggy

Solve the equation: z^3 - 3z^2 + 6z -4=0

Apply the Rational Roots Test to get a list of possible rational (integer or fraction) solutions: -4, -2, -1, 1, 2, and 4.

Then apply long polynomial division or synthetic division to test each possible solution, and see which work. You may find it helpful to graph the related functon, f(x) = x^3 - 3x^2 + 6x - 4, in your calculator, so you can "see" which is the best possibility to check.

Once you've found one solution, you'll be left with a quadratic, to which you can apply the Quadratic Formula.

Have fun!