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

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

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

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

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

x = 1, x = - 1 + i, x = - 1 - i

Solution steps

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

Move

(x - 1) (3 x + 2)

to the left side

x (x - 1)^{2} + (x - 1) (3 x + 2) = 0

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 1 = 0 : x = 1

Solve

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

The solutions are

x = 1, x = - 1 + i, x = - 1 - i

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

x^{3} - 10 x = 0

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

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

4 x^{3} + 4 x^{2} - 7 x + 2 = 0

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