Mister Exam
Factor -y^4+2*y^2-1 squared
The solution
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(- y^{4} + 2 y^{2}\right) - 1$$
To do this, let's use the formula
$$a y^{4} + b y^{2} + c = a \left(m + y^{2}\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = -1$$
$$b = 2$$
$$c = -1$$
Then
$$m = -1$$
$$n = 0$$
So,
$$- \left(y^{2} - 1\right)^{2}$$
$$\left(- y^{4} + 2 y^{2}\right) - 1$$
To do this, let's use the formula
$$a y^{4} + b y^{2} + c = a \left(m + y^{2}\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = -1$$
$$b = 2$$
$$c = -1$$
Then
$$m = -1$$
$$n = 0$$
So,
$$- \left(y^{2} - 1\right)^{2}$$
Combinatorics
[src]
2 2 -(1 + y) *(-1 + y)
$$- \left(y - 1\right)^{2} \left(y + 1\right)^{2}$$
-(1 + y)^2*(-1 + y)^2
Combining rational expressions
[src]
2 / 2\ -1 + y *\2 - y /
$$y^{2} \left(2 - y^{2}\right) - 1$$
-1 + y^2*(2 - y^2)