(z^4 + 2z^3 - 4z^2 - 2z + 3)
=(z^3 - z^2 - z + 1)(z + 3)
=(z^2 - 2z + 1)(z + 1)(z + 3)
=(z - 1)^2(z + 1)(z + 3)
You have to set this equal to 0 and graph this on the calculator to find two whole number solutions, -3 and -1. Then I divided the polynomial by (z - -3) and (z - -1) or (z + 3) and (z + 1).
thank you very much! i never thought of doing it that way!
Hmm... There is a general solution of the quartic (but not of the quintic or higher), but it's very complicated.
For your problem,
z^4 + 2z^3 - 4z^2 - 2z + 3 = z^4 - 4z^2 + 3 + 2z^3 - 2z
= (z^2 - 3)(z^2 - 1) + 2z(z^2 - 1) = (z^2 - 1)(z^2 - 3 + 2z)
= (z^2 - 1)(z^2 + 2z - 3) = (z +1)(z - 1)(z + 3)(z - 1)
= (z - 1)^2(z + 1)(z + 3)
Is there a trick to factoring the quadratic:
ax(to the power of four) + bx³ + cx² + dx + e
My actual problem is: z^4 + 2z^3 - 4z^2 - 2z + 3.
All help appreaciated! :)