What are the solutions to the equation x^3+8-(x+2)(x+4)=0 ?

The solutions to the equation x^3+8-(x+2)(x+4)=0 are x=0,x=-2,x=3

Find the zeros of x^3+8-(x+2)(x+4)=0

The zeros of x^3+8-(x+2)(x+4)=0 are x=0,x=-2,x=3

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

x = 0, x = - 2, x = 3

Solution steps

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

Expand

x^{3} + 8 - (x + 2) (x + 4) : x^{3} - x^{2} - 6 x

x^{3} - x^{2} - 6 x = 0

Factor

x^{3} - x^{2} - 6 x : x (x + 2) (x - 3)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 2 = 0 : x = - 2

Solve

x - 3 = 0 : x = 3

The solutions are

x = 0, x = - 2, x = 3

x, - y = (- x)^{3}

t^{3} - 2 t = 0

(1 + x^{2}) (1 - x^{3}) = 0

x^{4} - 25 x^{2} - 144 = 0

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

What are the solutions to the equation x^3+8-(x+2)(x+4)=0 ?
The solutions to the equation x^3+8-(x+2)(x+4)=0 are x=0,x=-2,x=3
Find the zeros of x^3+8-(x+2)(x+4)=0
The zeros of x^3+8-(x+2)(x+4)=0 are x=0,x=-2,x=3