What are the solutions to the equation (2x+3)(x+2)^2=4(2x+3) ?

The solutions to the equation (2x+3)(x+2)^2=4(2x+3) are x=0,x=-4,x=-3/2

Find the zeros of (2x+3)(x+2)^2=4(2x+3)

The zeros of (2x+3)(x+2)^2=4(2x+3) are x=0,x=-4,x=-3/2

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(2 x + 3) (x + 2)^{2} = 4 (2 x + 3)

x = 0, x = - 4, x = - \frac{3}{2}

Decimal Notation

x = 0, x = - 4, x = - 1.5

Solution steps

(2 x + 3) (x + 2)^{2} = 4 (2 x + 3)

Move

4 (2 x + 3)

to the left side

(2 x + 3) (x + 2)^{2} - 4 (2 x + 3) = 0

Factor

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

x (x + 4) (2 x + 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x + 4 = 0 or 2 x + 3 = 0

Solve

x + 4 = 0 : x = - 4

Solve

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

The solutions are

x = 0, x = - 4, x = - \frac{3}{2}

- x^{3} + 3 x^{2} + 62 x - 184 = 0

x^{3} - 2 x^{2} + 8 - 4 x = 0

1 x^{3} = 2 (1) + 3

4 x^{3} = 2

18 x^{3} + 5 x = 8 x^{3}

What are the solutions to the equation (2x+3)(x+2)^2=4(2x+3) ?
The solutions to the equation (2x+3)(x+2)^2=4(2x+3) are x=0,x=-4,x=-3/2
Find the zeros of (2x+3)(x+2)^2=4(2x+3)
The zeros of (2x+3)(x+2)^2=4(2x+3) are x=0,x=-4,x=-3/2