What is the answer to (x^2+1)(y^2+y)=0 ?

The answer to (x^2+1)(y^2+y)=0 is y=0,y=-1

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(x^2+1)(y^2+y)=0

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

y = 0, y = - 1

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x^{2} + 1 = 0 : x = i, x = - i

Solve

y^{2} + y = 0 : y = 0, y = - 1

The solutions are

Remove complex solutions:

x = i, x = - i

y = 0, y = - 1

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

0 = 3 y - 2 x y - 3 y, 0 = 3 x - x^{2} - 6 x y

(x + 2) (x - 5) \geq = 0

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

2 x^{7} - 256 x^{4.6} = 0

What is the answer to (x^2+1)(y^2+y)=0 ?
The answer to (x^2+1)(y^2+y)=0 is y=0,y=-1