What are the solutions to the equation -3x^3+4x^2-12x+16=0 ?

The solutions to the equation -3x^3+4x^2-12x+16=0 are x= 4/3 ,x=2i,x=-2i

Find the zeros of -3x^3+4x^2-12x+16=0

The zeros of -3x^3+4x^2-12x+16=0 are x= 4/3 ,x=2i,x=-2i

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- 3 x^{3} + 4 x^{2} - 12 x + 16 = 0

x = \frac{4}{3}, x = 2 i, x = - 2 i

Solution steps

- 3 x^{3} + 4 x^{2} - 12 x + 16 = 0

Factor

- 3 x^{3} + 4 x^{2} - 12 x + 16 : - (3 x - 4) (x^{2} + 4)

- (3 x - 4) (x^{2} + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 x - 4 = 0 or x^{2} + 4 = 0

Solve

3 x - 4 = 0 : x = \frac{4}{3}

Solve

x^{2} + 4 = 0 : x = 2 i, x = - 2 i

The solutions are

x = \frac{4}{3}, x = 2 i, x = - 2 i

a^{3} + 3 a^{2} + 2 a = 0

y^{4} - 3 y^{2} - 4 = 0

x^{4} - 6 x^{3} + 7 x^{2} + 6 x - 3 = 0

2 y^{2} - 2 y = 12 y^{2} - 12 y^{3}

2 x^{4} + 12 x^{2} + 16 = 0

What are the solutions to the equation -3x^3+4x^2-12x+16=0 ?
The solutions to the equation -3x^3+4x^2-12x+16=0 are x= 4/3 ,x=2i,x=-2i
Find the zeros of -3x^3+4x^2-12x+16=0
The zeros of -3x^3+4x^2-12x+16=0 are x= 4/3 ,x=2i,x=-2i