Mister Exam
Factor polynomial x^2-x-4
The solution
Factorization
[src]
/ ____\ / ____\ | 1 \/ 17 | | 1 \/ 17 | |x + - - + ------|*|x + - - - ------| \ 2 2 / \ 2 2 /
$$\left(x + \left(- \frac{1}{2} + \frac{\sqrt{17}}{2}\right)\right) \left(x + \left(- \frac{\sqrt{17}}{2} - \frac{1}{2}\right)\right)$$
(x - 1/2 + sqrt(17)/2)*(x - 1/2 - sqrt(17)/2)
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(x^{2} - x\right) - 4$$
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
$$a = 1$$
$$b = -1$$
$$c = -4$$
Then
$$m = - \frac{1}{2}$$
$$n = - \frac{17}{4}$$
So,
$$\left(x - \frac{1}{2}\right)^{2} - \frac{17}{4}$$
$$\left(x^{2} - x\right) - 4$$
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
$$a = 1$$
$$b = -1$$
$$c = -4$$
Then
$$m = - \frac{1}{2}$$
$$n = - \frac{17}{4}$$
So,
$$\left(x - \frac{1}{2}\right)^{2} - \frac{17}{4}$$