What are the solutions to the equation 3x^4+3x^3-17x^2+x-6=0 ?

The solutions to the equation 3x^4+3x^3-17x^2+x-6=0 are x=2,x=-3,x= 1/(sqrt(3))i,x=-1/(sqrt(3))i

Find the zeros of 3x^4+3x^3-17x^2+x-6=0

The zeros of 3x^4+3x^3-17x^2+x-6=0 are x=2,x=-3,x= 1/(sqrt(3))i,x=-1/(sqrt(3))i

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3 x^{4} + 3 x^{3} - 17 x^{2} + x - 6 = 0

x = 2, x = - 3, x = \frac{1}{3} i, x = - \frac{1}{3} i

Solution steps

3 x^{4} + 3 x^{3} - 17 x^{2} + x - 6 = 0

Factor

3 x^{4} + 3 x^{3} - 17 x^{2} + x - 6 : (x - 2) (x + 3) (3 x^{2} + 1)

(x - 2) (x + 3) (3 x^{2} + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 2 = 0 or x + 3 = 0 or 3 x^{2} + 1 = 0

Solve

x - 2 = 0 : x = 2

Solve

x + 3 = 0 : x = - 3

Solve

3 x^{2} + 1 = 0 : x = \frac{1}{3} i, x = - \frac{1}{3} i

The solutions are

x = 2, x = - 3, x = \frac{1}{3} i, x = - \frac{1}{3} i

t^{3} = 512

x^{3} - 5 x^{2} + 5 x + 7 = 7

x^{8} + 8 x + 16 = 0

9 + 3 a^{3} = 12

x^{4} = 2 x^{3} + 1

What are the solutions to the equation 3x^4+3x^3-17x^2+x-6=0 ?
The solutions to the equation 3x^4+3x^3-17x^2+x-6=0 are x=2,x=-3,x= 1/(sqrt(3))i,x=-1/(sqrt(3))i
Find the zeros of 3x^4+3x^3-17x^2+x-6=0
The zeros of 3x^4+3x^3-17x^2+x-6=0 are x=2,x=-3,x= 1/(sqrt(3))i,x=-1/(sqrt(3))i