What are the solutions to the equation (2x+1)(4x^2-2x+1)-1=1.6x^2(5x-2) ?

The solutions to the equation (2x+1)(4x^2-2x+1)-1=1.6x^2(5x-2) are x=0

Find the zeros of (2x+1)(4x^2-2x+1)-1=1.6x^2(5x-2)

The zeros of (2x+1)(4x^2-2x+1)-1=1.6x^2(5x-2) are x=0

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(2 x + 1) (4 x^{2} - 2 x + 1) - 1 = 1.6 x^{2} (5 x - 2)

x = 0

Solution steps

(2 x + 1) (4 x^{2} - 2 x + 1) - 1 = 1.6 x^{2} (5 x - 2)

Expand

(2 x + 1) (4 x^{2} - 2 x + 1) - 1 : 8 x^{3}

Expand

1.6 x^{2} (5 x - 2) : 8 x^{3} - 3.2 x^{2}

8 x^{3} = 8 x^{3} - 3.2 x^{2}

Switch sides

8 x^{3} - 3.2 x^{2} = 8 x^{3}

Move

8 x^{3}

to the left side

- 3.2 x^{2} = 0

Divide both sides by

- 3.2

x = 0

9 x^{2} - 12 x + 5

x^{2} + 6 x + 1 = - 2 x + 1

2 p^{2} - 7 p = 5

3 (x - 4)^{2} = 27

x^{2} - 20 x + 90 = - x

What are the solutions to the equation (2x+1)(4x^2-2x+1)-1=1.6x^2(5x-2) ?
The solutions to the equation (2x+1)(4x^2-2x+1)-1=1.6x^2(5x-2) are x=0
Find the zeros of (2x+1)(4x^2-2x+1)-1=1.6x^2(5x-2)
The zeros of (2x+1)(4x^2-2x+1)-1=1.6x^2(5x-2) are x=0