2x^4-3x^3-24x^2+13x+12=0

Question

Solve set complex numbers

Answer 1

Solve: 2x^4 - 3x^3 - 24x^2 + 13x + 12 = 0

By inspection, x = 1 is a solution of 2x^4 - 3x^3 - 24x^2 + 13x + 12 = 0

Therefore, (x – 1) is a factor of 2x^4 - 3x^3 - 24x^2 + 13x + 12, giving

(x – 1)(2x^3 - x^2 – 25x – 12)

By inspection, x = 4 is a solution of 2x^3 - x^2 – 25x – 12 = 0

Therefore, (x – 4) is a factor of 2x^3 - x^2 – 25x – 12, giving

(x – 1)(x – 4)(2x^2 + 7x + 3)

The final quadratic factorises as: 2x^2 + 7x + 3 = (2x + 1)(x + 3)

The complete factorisation is: 2x^4 - 3x^3 - 24x^2 + 13x + 12 = (x – 1)(x – 4)(2x + 1)(x + 3) = 0

Solution is: x = 1, x = 4, x = -1/2, x = -3

 

Answer 2

By inspection, x=1 and -3 are solutions (zeroes). So (x-1) and (x+3) are factors. Using synthetic division, we get a quadratic: 

2x^2-7x-4=(2x+1)(x-4), so complete factorisation is (x-1)(x+3)(x-4)(2x+1). Solution is x=1, -3, 4, -1/2. There are no complex roots.