Mister Exam
Factor polynomial x^2-8*x+15
The solution
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(x^{2} - 8 x\right) + 15$$
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
$$a = 1$$
$$b = -8$$
$$c = 15$$
Then
$$m = -4$$
$$n = -1$$
So,
$$\left(x - 4\right)^{2} - 1$$
$$\left(x^{2} - 8 x\right) + 15$$
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
$$a = 1$$
$$b = -8$$
$$c = 15$$
Then
$$m = -4$$
$$n = -1$$
So,
$$\left(x - 4\right)^{2} - 1$$