Mister Exam
Factor y^4-9*y^2+8 squared
The solution
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(y^{4} - 9 y^{2}\right) + 8$$
To do this, let's use the formula
$$a y^{4} + b y^{2} + c = a \left(m + y^{2}\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = 1$$
$$b = -9$$
$$c = 8$$
Then
$$m = - \frac{9}{2}$$
$$n = - \frac{49}{4}$$
So,
$$\left(y^{2} - \frac{9}{2}\right)^{2} - \frac{49}{4}$$
$$\left(y^{4} - 9 y^{2}\right) + 8$$
To do this, let's use the formula
$$a y^{4} + b y^{2} + c = a \left(m + y^{2}\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = 1$$
$$b = -9$$
$$c = 8$$
Then
$$m = - \frac{9}{2}$$
$$n = - \frac{49}{4}$$
So,
$$\left(y^{2} - \frac{9}{2}\right)^{2} - \frac{49}{4}$$
Factorization
[src]
/ ___\ / ___\ (x + 1)*(x - 1)*\x + 2*\/ 2 /*\x - 2*\/ 2 /
$$\left(x - 1\right) \left(x + 1\right) \left(x + 2 \sqrt{2}\right) \left(x - 2 \sqrt{2}\right)$$
(((x + 1)*(x - 1))*(x + 2*sqrt(2)))*(x - 2*sqrt(2))
Combinatorics
[src]
/ 2\ (1 + y)*(-1 + y)*\-8 + y /
$$\left(y - 1\right) \left(y + 1\right) \left(y^{2} - 8\right)$$
(1 + y)*(-1 + y)*(-8 + y^2)
Combining rational expressions
[src]
2 / 2\ 8 + y *\-9 + y /
$$y^{2} \left(y^{2} - 9\right) + 8$$
8 + y^2*(-9 + y^2)