What are the solutions to the equation x^3-4x^2-x+1=(x-2)^3 ?

The solutions to the equation x^3-4x^2-x+1=(x-2)^3 are x=(13+sqrt(97))/4 ,x=(13-sqrt(97))/4

Find the zeros of x^3-4x^2-x+1=(x-2)^3

The zeros of x^3-4x^2-x+1=(x-2)^3 are x=(13+sqrt(97))/4 ,x=(13-sqrt(97))/4

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

x = \frac{13 + 97}{4}, x = \frac{13 - 97}{4}

Decimal Notation

x = 5.71221 \dots, x = 0.78778 \dots

Solution steps

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

Expand

(x - 2)^{3} : x^{3} - 6 x^{2} + 12 x - 8

x^{3} - 4 x^{2} - x + 1 = x^{3} - 6 x^{2} + 12 x - 8

Subtract

x^{3} - 6 x^{2} + 12 x - 8

from both sides

x^{3} - 4 x^{2} - x + 1 - (x^{3} - 6 x^{2} + 12 x - 8) = x^{3} - 6 x^{2} + 12 x - 8 - (x^{3} - 6 x^{2} + 12 x - 8)

Simplify

2 x^{2} - 13 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 2 \cdot 9}{2 \cdot 2}

(- 13)^{2} - 4 \cdot 2 \cdot 9 = 97

x_{1, 2} = \frac{- ( - 13 ) \pm 97}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 97}{2 \cdot 2}, x_{2} = \frac{- ( - 13 ) - 97}{2 \cdot 2}

x = \frac{- ( - 13 ) + 97}{2 \cdot 2} : \frac{13 + 97}{4}

x = \frac{- ( - 13 ) - 97}{2 \cdot 2} : \frac{13 - 97}{4}

The solutions to the quadratic equation are:

x = \frac{13 + 97}{4}, x = \frac{13 - 97}{4}

x, y^{2} - 1 = (x - 1) (x - 3)

y^{2} + 9 y - 10 = 0

3 z (1.2 - z) = 2.4

2 (x + 5) \cdot x = 300

x^{2} + 0.9 x - 0.36 = 0

What are the solutions to the equation x^3-4x^2-x+1=(x-2)^3 ?
The solutions to the equation x^3-4x^2-x+1=(x-2)^3 are x=(13+sqrt(97))/4 ,x=(13-sqrt(97))/4
Find the zeros of x^3-4x^2-x+1=(x-2)^3
The zeros of x^3-4x^2-x+1=(x-2)^3 are x=(13+sqrt(97))/4 ,x=(13-sqrt(97))/4