Mister Exam
Factor polynomial z^2-z+5
The solution
Factorization
[src]
/ ____\ / ____\ | 1 I*\/ 19 | | 1 I*\/ 19 | |x + - - + --------|*|x + - - - --------| \ 2 2 / \ 2 2 /
$$\left(x + \left(- \frac{1}{2} - \frac{\sqrt{19} i}{2}\right)\right) \left(x + \left(- \frac{1}{2} + \frac{\sqrt{19} i}{2}\right)\right)$$
(x - 1/2 + i*sqrt(19)/2)*(x - 1/2 - i*sqrt(19)/2)
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(z^{2} - z\right) + 5$$
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
$$a = 1$$
$$b = -1$$
$$c = 5$$
Then
$$m = - \frac{1}{2}$$
$$n = \frac{19}{4}$$
So,
$$\left(z - \frac{1}{2}\right)^{2} + \frac{19}{4}$$
$$\left(z^{2} - z\right) + 5$$
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
$$a = 1$$
$$b = -1$$
$$c = 5$$
Then
$$m = - \frac{1}{2}$$
$$n = \frac{19}{4}$$
So,
$$\left(z - \frac{1}{2}\right)^{2} + \frac{19}{4}$$