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

The solutions to the equation x^3-x^2+x-2=4 are x=2,x=-1/2+i(sqrt(11))/2 ,x=-1/2-i(sqrt(11))/2

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

The zeros of x^3-x^2+x-2=4 are x=2,x=-1/2+i(sqrt(11))/2 ,x=-1/2-i(sqrt(11))/2

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

x = 2, x = - \frac{1}{2} + i \frac{11}{2}, x = - \frac{1}{2} - i \frac{11}{2}

Solution steps

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

Move

4

to the left side

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 2 = 0 : x = 2

Solve

x^{2} + x + 3 = 0 : x = - \frac{1}{2} + i \frac{11}{2}, x = - \frac{1}{2} - i \frac{11}{2}

The solutions are

x = 2, x = - \frac{1}{2} + i \frac{11}{2}, x = - \frac{1}{2} - i \frac{11}{2}

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

(x + 5) x^{2} = - (x + 5)

20 = (5 + \frac{0.12 x}{10})^{20}

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

x^{3} + 2 x^{2} + 3 x + 6 = 0

What are the solutions to the equation x^3-x^2+x-2=4 ?
The solutions to the equation x^3-x^2+x-2=4 are x=2,x=-1/2+i(sqrt(11))/2 ,x=-1/2-i(sqrt(11))/2
Find the zeros of x^3-x^2+x-2=4
The zeros of x^3-x^2+x-2=4 are x=2,x=-1/2+i(sqrt(11))/2 ,x=-1/2-i(sqrt(11))/2