What are the solutions to the equation x^3-5x+4=0 ?

The solutions to the equation x^3-5x+4=0 are x=1,x=(-1+sqrt(17))/2 ,x=(-1-sqrt(17))/2

Find the zeros of x^3-5x+4=0

The zeros of x^3-5x+4=0 are x=1,x=(-1+sqrt(17))/2 ,x=(-1-sqrt(17))/2

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x^{3} - 5 x + 4 = 0

x = 1, x = \frac{- 1 + 17}{2}, x = \frac{- 1 - 17}{2}

Decimal Notation

x = 1, x = 1.56155 \dots, x = - 2.56155 \dots

Solution steps

x^{3} - 5 x + 4 = 0

Factor

x^{3} - 5 x + 4 : (x - 1) (x^{2} + x - 4)

(x - 1) (x^{2} + x - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x^{2} + x - 4 = 0

Solve

x - 1 = 0 : x = 1

Solve

x^{2} + x - 4 = 0 : x = \frac{- 1 + 17}{2}, x = \frac{- 1 - 17}{2}

The solutions are

x = 1, x = \frac{- 1 + 17}{2}, x = \frac{- 1 - 17}{2}

x^{3} - 4 x^{2} - 11 x + 30 = 0

x^{3} - 4 x - 1 = 0

3 x^{4} + 2 = 5 x^{2}

x + 6 = x^{3}

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

What are the solutions to the equation x^3-5x+4=0 ?
The solutions to the equation x^3-5x+4=0 are x=1,x=(-1+sqrt(17))/2 ,x=(-1-sqrt(17))/2
Find the zeros of x^3-5x+4=0
The zeros of x^3-5x+4=0 are x=1,x=(-1+sqrt(17))/2 ,x=(-1-sqrt(17))/2