Mister Exam
Factor polynomial x+12+x^2
The solution
The perfect square
Let's highlight the perfect square of the square three-member
$$x^{2} + \left(x + 12\right)$$
To do this, let's use the formula
$$a x^{2} + b x + c = a \left(m + x\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = 1$$
$$b = 1$$
$$c = 12$$
Then
$$m = \frac{1}{2}$$
$$n = \frac{47}{4}$$
So,
$$\left(x + \frac{1}{2}\right)^{2} + \frac{47}{4}$$
$$x^{2} + \left(x + 12\right)$$
To do this, let's use the formula
$$a x^{2} + b x + c = a \left(m + x\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = 1$$
$$b = 1$$
$$c = 12$$
Then
$$m = \frac{1}{2}$$
$$n = \frac{47}{4}$$
So,
$$\left(x + \frac{1}{2}\right)^{2} + \frac{47}{4}$$
Factorization
[src]
/ ____\ / ____\ | 1 I*\/ 47 | | 1 I*\/ 47 | |x + - + --------|*|x + - - --------| \ 2 2 / \ 2 2 /
$$\left(x + \left(\frac{1}{2} - \frac{\sqrt{47} i}{2}\right)\right) \left(x + \left(\frac{1}{2} + \frac{\sqrt{47} i}{2}\right)\right)$$
(x + 1/2 + i*sqrt(47)/2)*(x + 1/2 - i*sqrt(47)/2)