What are the solutions to the equation (x^2+3)(x+1)+x=-1 ?

The solutions to the equation (x^2+3)(x+1)+x=-1 are x=-1,x=2i,x=-2i

Find the zeros of (x^2+3)(x+1)+x=-1

The zeros of (x^2+3)(x+1)+x=-1 are x=-1,x=2i,x=-2i

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

x = - 1, x = 2 i, x = - 2 i

Solution steps

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

Expand

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

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

Move

1

to the left side

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

Factor

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

(x + 1) (x^{2} + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or x^{2} + 4 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x^{2} + 4 = 0 : x = 2 i, x = - 2 i

The solutions are

x = - 1, x = 2 i, x = - 2 i

3.8 y^{5} = 43.8

2 x^{4} - 8 = 2584

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

16 x^{4} + 24 x^{2} + 9 x = 0

- x^{3} + 11 x^{2} - 38 x + 44 = 0

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