What are the solutions to the equation x^2(x+2)=16(x+2) ?

The solutions to the equation x^2(x+2)=16(x+2) are x=-2,x=-4,x=4

Find the zeros of x^2(x+2)=16(x+2)

The zeros of x^2(x+2)=16(x+2) are x=-2,x=-4,x=4

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x^{2} (x + 2) = 16 (x + 2)

x = - 2, x = - 4, x = 4

Solution steps

x^{2} (x + 2) = 16 (x + 2)

Move

16 (x + 2)

to the left side

x^{2} (x + 2) - 16 (x + 2) = 0

Factor

x^{2} (x + 2) - 16 (x + 2) : (x + 2) (x + 4) (x - 4)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 2 = 0 or x + 4 = 0 or x - 4 = 0

Solve

x + 2 = 0 : x = - 2

Solve

x + 4 = 0 : x = - 4

Solve

x - 4 = 0 : x = 4

The solutions are

x = - 2, x = - 4, x = 4

z^{3} = \frac{1}{8}

(x + 3)^{3} = 343

x^{3} + 5 x^{2} + 6 x + 0 = 0

3 x^{3} - 16 x^{2} + 23 x - 6 = 0

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

What are the solutions to the equation x^2(x+2)=16(x+2) ?
The solutions to the equation x^2(x+2)=16(x+2) are x=-2,x=-4,x=4
Find the zeros of x^2(x+2)=16(x+2)
The zeros of x^2(x+2)=16(x+2) are x=-2,x=-4,x=4