What are the solutions to the equation x^3-5x^2+4x+6=0 ?

The solutions to the equation x^3-5x^2+4x+6=0 are x=3,x=1+sqrt(3),x=1-sqrt(3)

Find the zeros of x^3-5x^2+4x+6=0

The zeros of x^3-5x^2+4x+6=0 are x=3,x=1+sqrt(3),x=1-sqrt(3)

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x^{3} - 5 x^{2} + 4 x + 6 = 0

x = 3, x = 1 + 3, x = 1 - 3

Decimal Notation

x = 3, x = 2.73205 \dots, x = - 0.73205 \dots

Solution steps

x^{3} - 5 x^{2} + 4 x + 6 = 0

Factor

x^{3} - 5 x^{2} + 4 x + 6 : (x - 3) (x^{2} - 2 x - 2)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 3 = 0 : x = 3

Solve

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

The solutions are

x = 3, x = 1 + 3, x = 1 - 3

3 b = 9 (b^{3} + 3)

x, x^{4} + 4 y^{2} = z^{2}

y^{(10)} = 0

27 x^{3} + 27 x^{2} + 9 x = - 1

6 x^{3} + 43 x^{2} + 43 x + 6 = 0

What are the solutions to the equation x^3-5x^2+4x+6=0 ?
The solutions to the equation x^3-5x^2+4x+6=0 are x=3,x=1+sqrt(3),x=1-sqrt(3)
Find the zeros of x^3-5x^2+4x+6=0
The zeros of x^3-5x^2+4x+6=0 are x=3,x=1+sqrt(3),x=1-sqrt(3)