What are the solutions to the equation x^3-5x^2-8x+12=0 ?

The solutions to the equation x^3-5x^2-8x+12=0 are x=1,x=-2,x=6

Find the zeros of x^3-5x^2-8x+12=0

The zeros of x^3-5x^2-8x+12=0 are x=1,x=-2,x=6

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x^{3} - 5 x^{2} - 8 x + 12 = 0

x = 1, x = - 2, x = 6

Solution steps

x^{3} - 5 x^{2} - 8 x + 12 = 0

Factor

x^{3} - 5 x^{2} - 8 x + 12 : (x - 1) (x + 2) (x - 6)

(x - 1) (x + 2) (x - 6) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x + 2 = 0 or x - 6 = 0

Solve

x - 1 = 0 : x = 1

Solve

x + 2 = 0 : x = - 2

Solve

x - 6 = 0 : x = 6

The solutions are

x = 1, x = - 2, x = 6

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

2 x^{3} - x - 15 = 0

5 x^{4} = 6 x^{2}

2 x^{3} + x^{2} + 18 x + 9 = 0

64 a^{6} - 112 a^{4} + 56 a^{2} - 7 = 0

What are the solutions to the equation x^3-5x^2-8x+12=0 ?
The solutions to the equation x^3-5x^2-8x+12=0 are x=1,x=-2,x=6
Find the zeros of x^3-5x^2-8x+12=0
The zeros of x^3-5x^2-8x+12=0 are x=1,x=-2,x=6