What are the solutions to the equation 1x^3-4x^2+1x+6=0 ?

The solutions to the equation 1x^3-4x^2+1x+6=0 are x=-1,x=2,x=3

Find the zeros of 1x^3-4x^2+1x+6=0

The zeros of 1x^3-4x^2+1x+6=0 are x=-1,x=2,x=3

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1 x^{3} - 4 x^{2} + 1 x + 6 = 0

x = - 1, x = 2, x = 3

Solution steps

1 \cdot x^{3} - 4 x^{2} + 1 \cdot x + 6 = 0

Refine

x^{3} - 4 x^{2} + x + 6 = 0

Factor

x^{3} - 4 x^{2} + x + 6 : (x + 1) (x - 2) (x - 3)

(x + 1) (x - 2) (x - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or x - 2 = 0 or x - 3 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 2 = 0 : x = 2

Solve

x - 3 = 0 : x = 3

The solutions are

x = - 1, x = 2, x = 3

x^{3} - 2 x^{2} - 15 x + 30 = 0

x^{5} - x^{3} + 3 x - 5 = 0

(x - 1)^{7} = x^{5}

12 x^{3} + 21 x^{2} + 6 x = 0

x^{3} - 7 x^{2} + 17 x - 15 = 0

What are the solutions to the equation 1x^3-4x^2+1x+6=0 ?
The solutions to the equation 1x^3-4x^2+1x+6=0 are x=-1,x=2,x=3
Find the zeros of 1x^3-4x^2+1x+6=0
The zeros of 1x^3-4x^2+1x+6=0 are x=-1,x=2,x=3