Mister Exam
Factor polynomial y^2+8*y+16
The solution
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(y^{2} + 8 y\right) + 16$$
To do this, let's use the formula
$$a y^{2} + b y + c = a \left(m + y\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = 1$$
$$b = 8$$
$$c = 16$$
Then
$$m = 4$$
$$n = 0$$
So,
$$\left(y + 4\right)^{2}$$
$$\left(y^{2} + 8 y\right) + 16$$
To do this, let's use the formula
$$a y^{2} + b y + c = a \left(m + y\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = 1$$
$$b = 8$$
$$c = 16$$
Then
$$m = 4$$
$$n = 0$$
So,
$$\left(y + 4\right)^{2}$$