factorising

Is there a trick to factorising 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! :)

http://en.wikipedia.org/wiki/Quartic_function#Solving_a_quartic_equation

quadratic, not quartic :).

Your problem is a quartic. :-)

try to use the rational roots property:  Get p from factors of 3 (that is, +/- 1 and +/-3)  and get q from factors of 1 (+/-1)...then apply synthetic division using p/q as possible rational roots.. whatever gives a remainder 0 will z-(p/q) as factor...carry out the process until you reduced the process(synthetic division) to its depressed equation that are coeffiecnts of a quadratic equation....If no such p/q gives 0 for a remainder then the roots of the given expression must be imaginary...you can still use synthetic division by using as p/q -- +/-i or +/-3i...

Good day and God bless...

(x -1)^2 (x +1) ( x -3)

Write down the coefficients, use synthetic division to try the possible rational zeroes of the function which are 1, 3, -1, -3.

1:      -1)   1  2  -4  -2  3

         -3)   1  1  -5   3 |0       (x + 1)(1x^3 +1x^2 -5x + 3)

                 1 -2  1 |0           (x + 1)(x + 3) (x ^2-2x +1)