Mister Exam
Factor q^2+q+4 squared
The solution
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(q^{2} + q\right) + 4$$
To do this, let's use the formula
$$a q^{2} + b q + c = a \left(m + q\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = 1$$
$$b = 1$$
$$c = 4$$
Then
$$m = \frac{1}{2}$$
$$n = \frac{15}{4}$$
So,
$$\left(q + \frac{1}{2}\right)^{2} + \frac{15}{4}$$
$$\left(q^{2} + q\right) + 4$$
To do this, let's use the formula
$$a q^{2} + b q + c = a \left(m + q\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = 1$$
$$b = 1$$
$$c = 4$$
Then
$$m = \frac{1}{2}$$
$$n = \frac{15}{4}$$
So,
$$\left(q + \frac{1}{2}\right)^{2} + \frac{15}{4}$$
Factorization
[src]
/ ____\ / ____\ | 1 I*\/ 15 | | 1 I*\/ 15 | |q + - + --------|*|q + - - --------| \ 2 2 / \ 2 2 /
$$\left(q + \left(\frac{1}{2} - \frac{\sqrt{15} i}{2}\right)\right) \left(q + \left(\frac{1}{2} + \frac{\sqrt{15} i}{2}\right)\right)$$
(q + 1/2 + i*sqrt(15)/2)*(q + 1/2 - i*sqrt(15)/2)