What are the solutions to the equation x(x+1)(x^2-36)=0 ?

The solutions to the equation x(x+1)(x^2-36)=0 are x=0,x=-1,x=-6,x=6

Find the zeros of x(x+1)(x^2-36)=0

The zeros of x(x+1)(x^2-36)=0 are x=0,x=-1,x=-6,x=6

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x (x + 1) (x^{2} - 36) = 0

x = 0, x = - 1, x = - 6, x = 6

Solution steps

x (x + 1) (x^{2} - 36) = 0

Factor

x (x + 1) (x^{2} - 36) : x (x + 1) (x + 6) (x - 6)

x (x + 1) (x + 6) (x - 6) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x + 1 = 0 or x + 6 = 0 or x - 6 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x + 6 = 0 : x = - 6

Solve

x - 6 = 0 : x = 6

The solutions are

x = 0, x = - 1, x = - 6, x = 6

x^{3} - 3 x^{2} - 50 = 0

72 x^{4} - x^{2} - 1 = 0

(1 + x)^{5} = 1.24

6 x^{9} - 72 x^{7} - 7 x^{2} + 96 x - 63 = 0

- x^{3} - 13 x^{2} - 12 x + 10 = 0

What are the solutions to the equation x(x+1)(x^2-36)=0 ?
The solutions to the equation x(x+1)(x^2-36)=0 are x=0,x=-1,x=-6,x=6
Find the zeros of x(x+1)(x^2-36)=0
The zeros of x(x+1)(x^2-36)=0 are x=0,x=-1,x=-6,x=6