Question

How many solutions over the complex number system does this polynomial have?

2x^4 – 3x^3 – 24x^2 + 13x + 12 = 0

Answer 1

$f(x) =0\\ 2x^4 - 3x^3- 24x^2 + 13x + 12=0\\ 2x^4 - 2x^3-x^3+x^2- 25x^2 + 25x -12x+ 12=0\\ 2x^3(x-1) - x^2(x-1)-25x(x-1)-12(x-1)=0\\ (x-1)(2x^3- x^2-25x-12)=0\\ (x-1)(2x^3+6x^2-7 x^2-21x-4x-12)=0\\ (x-1)(2x^2(x+3)-7 x(x+3)-4(x+3)=0\\ (x-1)(x+3)(2x^2-7x-4)=0\\ (x-1)(x+3)(2x^2-8x+x-4)=0\\ (x-1)(x+3)(2x(x-4)+1(x-4))=0\\ (x-1)(x+3)(x-4)(2x+1)=0\\ \text{Thus, only real roots i.e., -3, -0.5, 1 and 4. Since, real number set is a subset of complex number. Therefore, given polynomial has 4 root over complex number system.}$