Mister Exam
Factor y^2-2*y+5 squared
The solution
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(y^{2} - 2 y\right) + 5$$
To do this, let's use the formula
$$a y^{2} + b y + c = a \left(m + y\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = 1$$
$$b = -2$$
$$c = 5$$
Then
$$m = -1$$
$$n = 4$$
So,
$$\left(y - 1\right)^{2} + 4$$
$$\left(y^{2} - 2 y\right) + 5$$
To do this, let's use the formula
$$a y^{2} + b y + c = a \left(m + y\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = 1$$
$$b = -2$$
$$c = 5$$
Then
$$m = -1$$
$$n = 4$$
So,
$$\left(y - 1\right)^{2} + 4$$
Factorization
[src]
(x + -1 + 2*I)*(x + -1 - 2*I)
$$\left(x + \left(-1 - 2 i\right)\right) \left(x + \left(-1 + 2 i\right)\right)$$
(x - 1 + 2*i)*(x - 1 - 2*i)