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

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

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

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

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

x = 0, x = - \frac{3}{2}, x = \frac{3}{2}

Decimal Notation

x = 0, x = - 0.86602 \dots, x = 0.86602 \dots

Solution steps

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

Expand

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

4 x^{3} - 3 x + 1 = 1

Move

1

to the left side

4 x^{3} - 3 x = 0

Factor

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

x (2 x + 3) (2 x - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 2 x + 3 = 0 or 2 x - 3 = 0

Solve

2 x + 3 = 0 : x = - \frac{3}{2}

Solve

2 x - 3 = 0 : x = \frac{3}{2}

The solutions are

x = 0, x = - \frac{3}{2}, x = \frac{3}{2}

x^{3} + 16 x = 0

- 2 x^{4} + 7 = 2 x - 1

2 x^{4} - 72 x^{2} = 0

D^{3} - 6 D^{2} + 11 D - 6 = 0

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

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