Mister Exam
Factor y^4-8*y^2-9 squared
The solution
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(y^{4} - 8 y^{2}\right) - 9$$
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
$$a = 1$$
$$b = -8$$
$$c = -9$$
Then
$$m = -4$$
$$n = -25$$
So,
$$\left(y^{2} - 4\right)^{2} - 25$$
$$\left(y^{4} - 8 y^{2}\right) - 9$$
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
$$a = 1$$
$$b = -8$$
$$c = -9$$
Then
$$m = -4$$
$$n = -25$$
So,
$$\left(y^{2} - 4\right)^{2} - 25$$
Factorization
[src]
(x + 3)*(x - 3)*(x + I)*(x - I)
$$\left(x - 3\right) \left(x + 3\right) \left(x + i\right) \left(x - i\right)$$
(((x + 3)*(x - 3))*(x + i))*(x - i)
Combinatorics
[src]
/ 2\ \1 + y /*(-3 + y)*(3 + y)
$$\left(y - 3\right) \left(y + 3\right) \left(y^{2} + 1\right)$$
(1 + y^2)*(-3 + y)*(3 + y)
Combining rational expressions
[src]
2 / 2\ -9 + y *\-8 + y /
$$y^{2} \left(y^{2} - 8\right) - 9$$
-9 + y^2*(-8 + y^2)