What are the solutions to the equation x^3-4x^2-9x+12=0 ?

The solutions to the equation x^3-4x^2-9x+12=0 are x=1,x=(3+sqrt(57))/2 ,x=(3-sqrt(57))/2

Find the zeros of x^3-4x^2-9x+12=0

The zeros of x^3-4x^2-9x+12=0 are x=1,x=(3+sqrt(57))/2 ,x=(3-sqrt(57))/2

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x^{3} - 4 x^{2} - 9 x + 12 = 0

x = 1, x = \frac{3 + 57}{2}, x = \frac{3 - 57}{2}

Solution steps

x^{3} - 4 x^{2} - 9 x + 12 = 0

Factor

x^{3} - 4 x^{2} - 9 x + 12 : (x - 1) (x^{2} - 3 x - 12)

(x - 1) (x^{2} - 3 x - 12) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x^{2} - 3 x - 12 = 0

Solve

x - 1 = 0 : x = 1

Solve

x^{2} - 3 x - 12 = 0 : x = \frac{3 + 57}{2}, x = \frac{3 - 57}{2}

The solutions are

x = 1, x = \frac{3 + 57}{2}, x = \frac{3 - 57}{2}

x^{3} + 5 x^{2} - 16 x - 80 = 0

6 x^{8} + 6 = 13 x^{4}

x^{3} - 7 x + 4 = 0

y, x = y^{3} - 6

x^{3} - 6 x^{2} + 13 x - 12 = 0

What are the solutions to the equation x^3-4x^2-9x+12=0 ?
The solutions to the equation x^3-4x^2-9x+12=0 are x=1,x=(3+sqrt(57))/2 ,x=(3-sqrt(57))/2
Find the zeros of x^3-4x^2-9x+12=0
The zeros of x^3-4x^2-9x+12=0 are x=1,x=(3+sqrt(57))/2 ,x=(3-sqrt(57))/2