Mister Exam
Factor polynomial z^2+2*z+5
The solution
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)
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(z^{2} + 2 z\right) + 5$$
To do this, let's use the formula
$$a z^{2} + b z + c = a \left(m + z\right)^{2} + n$$
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,
$$\left(z + 1\right)^{2} + 4$$
$$\left(z^{2} + 2 z\right) + 5$$
To do this, let's use the formula
$$a z^{2} + b z + c = a \left(m + z\right)^{2} + n$$
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,
$$\left(z + 1\right)^{2} + 4$$