What are the solutions to the equation (2x+1)(x+1)=5(x+1) ?

The solutions to the equation (2x+1)(x+1)=5(x+1) are x=-1,x=2

Find the zeros of (2x+1)(x+1)=5(x+1)

The zeros of (2x+1)(x+1)=5(x+1) are x=-1,x=2

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(2 x + 1) (x + 1) = 5 (x + 1)

x = - 1, x = 2

Solution steps

(2 x + 1) (x + 1) = 5 (x + 1)

Move

5 (x + 1)

to the left side

(2 x + 1) (x + 1) - 5 (x + 1) = 0

Factor

(2 x + 1) (x + 1) - 5 (x + 1) : 2 (x + 1) (x - 2)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or x - 2 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 2 = 0 : x = 2

The solutions to the quadratic equation are:

x = - 1, x = 2

x^{3} + x^{2} - 2

- 8 - 3 x = - x - 6

(\frac{9}{8})^{- 2}

4 (2 x + 3) = x + 12

4 x^{2} + 49 = 0

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