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