Mister Exam
Factor -y^2-y-8 squared
The solution
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(- y^{2} - y\right) - 8$$
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
$$a = -1$$
$$b = -1$$
$$c = -8$$
Then
$$m = \frac{1}{2}$$
$$n = - \frac{31}{4}$$
So,
$$- \left(y + \frac{1}{2}\right)^{2} - \frac{31}{4}$$
$$\left(- y^{2} - y\right) - 8$$
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
$$a = -1$$
$$b = -1$$
$$c = -8$$
Then
$$m = \frac{1}{2}$$
$$n = - \frac{31}{4}$$
So,
$$- \left(y + \frac{1}{2}\right)^{2} - \frac{31}{4}$$
Factorization
[src]
/ ____\ / ____\ | 1 I*\/ 31 | | 1 I*\/ 31 | |x + - + --------|*|x + - - --------| \ 2 2 / \ 2 2 /
$$\left(x + \left(\frac{1}{2} - \frac{\sqrt{31} i}{2}\right)\right) \left(x + \left(\frac{1}{2} + \frac{\sqrt{31} i}{2}\right)\right)$$
(x + 1/2 + i*sqrt(31)/2)*(x + 1/2 - i*sqrt(31)/2)