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

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

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

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

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

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

Solution steps

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 1 = 0 : x = - 1

Solve

x + 3 = 0 : x = - 3

Solve

x - 1 = 0 : x = 1

The solutions are

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

4 x^{3} - 24 x^{2} - x + 6 = 0

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

16 x^{4} - 41 x^{2} + 25 = 0

x^{6} - 124 x^{3} - 125 = 0

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

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