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

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

Find the zeros of (2-x)^3-4+3x=0

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

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

x = 1, x = 4

Solution steps

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

Expand

(2 - x)^{3} - 4 + 3 x : - x^{3} + 6 x^{2} - 9 x + 4

- x^{3} + 6 x^{2} - 9 x + 4 = 0

Factor

- x^{3} + 6 x^{2} - 9 x + 4 : - (x - 1)^{2} (x - 4)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x - 4 = 0

Solve

x - 1 = 0 : x = 1

Solve

x - 4 = 0 : x = 4

The solutions are

x = 1, x = 4

x^{3} - 0.5 x^{2} - 8 x + 7.5 = 0

6 h^{5} = 12 h^{3}

2 x^{3} - 9 x^{2} + 12 x = 0

x^{3} + 15 x^{2} + 71 x + 105 = 0

x^{3} + x + 8 = 0

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