How do you solve equations over a set of complex numbers?

Question

2x^4 - 3x^3 - 24x^2 + 13x +12? 

Answer 1

The equation 2x⁴- 3x³-24x² + 13x+12 = 0

Rational Root Theorem, if a rational number in simplest form p/q is a root of the polynomial equation a_nxⁿ + a_n – 1 x^n–1 + ... + a₁x + a₀ = 0, then p is a factor of a₀ and q is a factor if a_n.

If p/q is a rational zero, then p  is a factor of 2 and q  is a factor of 12.

The possible values of p  are   ± 1 and ± 2.

The possible values for q are ± 1, ±2, ±3, ±4, ±6 and ±12.

Possible p/q values are ± 1, ± 1/2, ± 1/3, ± 1/4, ± 1/6, ± 1/12, ± 2, ± 2/3, ± 2/6, ± 2/12

Let us consider a test point p/q = ± 1

Make a table for the synthetic division and test possible real zeros.

p/q	2	-3	-24	13	12
1	2	-1	-25	-12	0
-1	2	-5	-19	32	-20

Since f (1) = 0,  x = 1 is a zero. The reduced polynomial is  2x³ - x ² - 25x - 12= 0.

Consider the polynomial x³ - x ² - 25x - 12= 0.

If p/q is a rational zero, then p  is a factor of 1 and q  is a factor of 12.

The possible values of p  are   ± 1.

The possible values for q are ± 1, ±2, ±3, ±4, ±6 and ±12.

Possible p/q values are ± 1, ± 1/2, ± 1/3, ± 1/4, ± 1/6 and ± 1/12

Let us consider a test point p/q = ±1 and - 1/2

Make a table for the synthetic division and test possible real zeros.

p/q	2	-1	-25	-12
1	2	1	-24	-36
- 1	2	3	-22	-3
- 1/2	2	-2	-24	0

Since f (-1/2) = 0,  x = -1/2 is a zero. The reduced polynomial is  2x ² - 2x - 24= 0.

2x ² - 2x - 24= 0.

2(x² - x - 12)= 0

x² - 4x + 3x - 12 = 0

x(x - 4) + 3(x - 4) = 0

(x + 3)(x - 4) = 0

x = -3 and 4

Therefore the values of x are 1, -1/2, -3 and 4.