Mister Exam
Factor polynomial c^2+9+c
The solution
Factorization
[src]
/ ____\ / ____\ | 1 I*\/ 35 | | 1 I*\/ 35 | |c + - + --------|*|c + - - --------| \ 2 2 / \ 2 2 /
$$\left(c + \left(\frac{1}{2} - \frac{\sqrt{35} i}{2}\right)\right) \left(c + \left(\frac{1}{2} + \frac{\sqrt{35} i}{2}\right)\right)$$
(c + 1/2 + i*sqrt(35)/2)*(c + 1/2 - i*sqrt(35)/2)
The perfect square
Let's highlight the perfect square of the square three-member
$$c + \left(c^{2} + 9\right)$$
To do this, let's use the formula
$$a c^{2} + b c + c = a \left(c + m\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = 1$$
$$b = 1$$
$$c = 9$$
Then
$$m = \frac{1}{2}$$
$$n = \frac{35}{4}$$
So,
$$\left(c + \frac{1}{2}\right)^{2} + \frac{35}{4}$$
$$c + \left(c^{2} + 9\right)$$
To do this, let's use the formula
$$a c^{2} + b c + c = a \left(c + m\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = 1$$
$$b = 1$$
$$c = 9$$
Then
$$m = \frac{1}{2}$$
$$n = \frac{35}{4}$$
So,
$$\left(c + \frac{1}{2}\right)^{2} + \frac{35}{4}$$