Mister Exam
Factor polynomial z^2+4*z+13
The solution
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(z^{2} + 4 z\right) + 13$$
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
$$a = 1$$
$$b = 4$$
$$c = 13$$
Then
$$m = 2$$
$$n = 9$$
So,
$$\left(z + 2\right)^{2} + 9$$
$$\left(z^{2} + 4 z\right) + 13$$
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
$$a = 1$$
$$b = 4$$
$$c = 13$$
Then
$$m = 2$$
$$n = 9$$
So,
$$\left(z + 2\right)^{2} + 9$$
Factorization
[src]
(x + 2 + 3*I)*(x + 2 - 3*I)
$$\left(x + \left(2 - 3 i\right)\right) \left(x + \left(2 + 3 i\right)\right)$$
(x + 2 + 3*i)*(x + 2 - 3*i)