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

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

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

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

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

x = - 4, x = - 1, x = 1

Solution steps

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

Switch sides

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

Factor

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

(x + 4) (x + 1) (x - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 4 = 0 or x + 1 = 0 or x - 1 = 0

Solve

x + 4 = 0 : x = - 4

Solve

x + 1 = 0 : x = - 1

Solve

x - 1 = 0 : x = 1

The solutions are

x = - 4, x = - 1, x = 1

x^{8} - 2 x^{6} + 2 x^{4} - 2 x^{2} + 1 = 0

x^{3} + 4 x^{2} - 21 x = 0

z^{4} + 4 \cdot z^{2} - 45 = 0

x^{3} - 9 x^{2} + 24 x - 19 = 0

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

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