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

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

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

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

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

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

Solution steps

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

e^{2.5}

81 a^{7} b^{12} c^{9}

(2 a - 3 b) 0.4 b - 3 a (a - 2 b)

4 x^{2} - 2 x + 1 = 2 x

15 x^{2} - x < 2

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