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

The solutions to the equation x(x^2+4x+4)=3(x+2) are x=-2,x=1,x=-3

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

The zeros of x(x^2+4x+4)=3(x+2) are x=-2,x=1,x=-3

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

x = - 2, x = 1, x = - 3

Solution steps

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

Move

3 (x + 2)

to the left side

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

Factor

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

(x + 2) (x - 1) (x + 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 2 = 0 or x - 1 = 0 or x + 3 = 0

Solve

x + 2 = 0 : x = - 2

Solve

x - 1 = 0 : x = 1

Solve

x + 3 = 0 : x = - 3

The solutions are

x = - 2, x = 1, x = - 3

x^{3} - 9 x - 3240 = 0

2 x^{3} - 9 x^{2} + 12 x - 5 = 0

9 x^{3} - 12 x^{2} - 8 x - 1 = 0

4 x^{3} - x^{2} + 36 - 9 x = 0

20 x^{3} - 48 x^{2} + 24 x = 0

What are the solutions to the equation x(x^2+4x+4)=3(x+2) ?
The solutions to the equation x(x^2+4x+4)=3(x+2) are x=-2,x=1,x=-3
Find the zeros of x(x^2+4x+4)=3(x+2)
The zeros of x(x^2+4x+4)=3(x+2) are x=-2,x=1,x=-3