Mister Exam
Factor y^4-4*y^2-12 squared
The solution
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(y^{4} - 4 y^{2}\right) - 12$$
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
$$a = 1$$
$$b = -4$$
$$c = -12$$
Then
$$m = -2$$
$$n = -16$$
So,
$$\left(y^{2} - 2\right)^{2} - 16$$
$$\left(y^{4} - 4 y^{2}\right) - 12$$
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
$$a = 1$$
$$b = -4$$
$$c = -12$$
Then
$$m = -2$$
$$n = -16$$
So,
$$\left(y^{2} - 2\right)^{2} - 16$$
Factorization
[src]
/ ___\ / ___\ / ___\ / ___\ \x + \/ 6 /*\x - \/ 6 /*\x + I*\/ 2 /*\x - I*\/ 2 /
$$\left(x - \sqrt{6}\right) \left(x + \sqrt{6}\right) \left(x + \sqrt{2} i\right) \left(x - \sqrt{2} i\right)$$
(((x + sqrt(6))*(x - sqrt(6)))*(x + i*sqrt(2)))*(x - i*sqrt(2))
Combinatorics
[src]
/ 2\ / 2\ \-6 + y /*\2 + y /
$$\left(y^{2} - 6\right) \left(y^{2} + 2\right)$$
(-6 + y^2)*(2 + y^2)
Combining rational expressions
[src]
2 / 2\ -12 + y *\-4 + y /
$$y^{2} \left(y^{2} - 4\right) - 12$$
-12 + y^2*(-4 + y^2)