What are the solutions to the equation (1-x)(1-x)(-1-x)-i(-i)(-1-x)=0 ?

The solutions to the equation (1-x)(1-x)(-1-x)-i(-i)(-1-x)=0 are x=0,x=-1,x=2

Find the zeros of (1-x)(1-x)(-1-x)-i(-i)(-1-x)=0

The zeros of (1-x)(1-x)(-1-x)-i(-i)(-1-x)=0 are x=0,x=-1,x=2

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(1 - x) (1 - x) (- 1 - x) - i (- i) (- 1 - x) = 0

x = 0, x = - 1, x = 2

Solution steps

(1 - x) (1 - x) (- 1 - x) - i (- i) (- 1 - x) = 0

Expand

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

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

Find one solution for

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

using Newton-Raphson:

x \approx 0.00000 \dots

Apply long division:

\frac{- x ^{3} + x ^{2} + 2 x}{x} = - x^{2} + x + 2

- x^{2} + x + 2 \approx 0

Find one solution for

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

using Newton-Raphson:

x \approx - 1.00000 \dots

Apply long division:

\frac{- x ^{2} + x + 2}{x + 1} = - x + 2

- x + 2 \approx 0

x \approx 2

Approximate to algebraic form

x = 2

Approximate to algebraic form

x \approx 0.00000 \dots : x = 0

x \approx - 1.00000 \dots : x = - 1

x = 0, x = - 1, x = 2

The solutions are

x = 0, x = - 1, x = 2

36 x^{4} - 25 x^{2} + 4 = 0

165 = 5 x^{3}

2 = 6 x^{5}

0 = - x^{3} - 3 x^{2} + 54 x

x^{3} + 7 x - 6 = 0

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