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