Mister Exam
Factor y^2+y+5 squared
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
$$\left(y^{2} + y\right) + 5$$
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
$$a = 1$$
$$b = 1$$
$$c = 5$$
Then
$$m = \frac{1}{2}$$
$$n = \frac{19}{4}$$
So,
$$\left(y + \frac{1}{2}\right)^{2} + \frac{19}{4}$$
$$\left(y^{2} + y\right) + 5$$
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
$$a = 1$$
$$b = 1$$
$$c = 5$$
Then
$$m = \frac{1}{2}$$
$$n = \frac{19}{4}$$
So,
$$\left(y + \frac{1}{2}\right)^{2} + \frac{19}{4}$$