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

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

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

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

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(x + 2) (2 x - 1) = (x + 2) (x - 1)

x = 0, x = - 2

Solution steps

(x + 2) (2 x - 1) = (x + 2) (x - 1)

Move

(x + 2) (x - 1)

to the left side

(x + 2) (2 x - 1) - (x + 2) (x - 1) = 0

Factor

(x + 2) (2 x - 1) - (x + 2) (x - 1) : x (x + 2)

x (x + 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x + 2 = 0

Solve

x + 2 = 0 : x = - 2

The solutions to the quadratic equation are:

x = 0, x = - 2

\frac{9}{5} 30 + 32

3150

(\frac{2}{3})^{3} - 2 (\frac{2}{3})^{2} - 4 (\frac{2}{3}) + 2

c + (4 - 3 c) - 2 = 0

3 a^{2} + 16 a - 35

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