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

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

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

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

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

x = 0, x = - 1

Solution steps

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

Move

(x + 1) (x + 2)

to the left side

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

Factor

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

- x (x + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x + 1 = 0

Solve

x + 1 = 0 : x = - 1

The solutions to the quadratic equation are:

x = 0, x = - 1

3 q^{2} + q - 14 = 0

x^{4} + 3 x^{3} - 5 x^{2} - 1 = 0

\frac{k}{6} = \frac{4}{3}

(- 4 x^{3} y^{5})^{4}

∣ 3 x - 5 ∣ + 2 = 4 x

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