Mister Exam
Factor 2*p^2+3*p-2 squared
The solution
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(2 p^{2} + 3 p\right) - 2$$
To do this, let's use the formula
$$a p^{2} + b p + c = a \left(m + p\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = 2$$
$$b = 3$$
$$c = -2$$
Then
$$m = \frac{3}{4}$$
$$n = - \frac{25}{8}$$
So,
$$2 \left(p + \frac{3}{4}\right)^{2} - \frac{25}{8}$$
$$\left(2 p^{2} + 3 p\right) - 2$$
To do this, let's use the formula
$$a p^{2} + b p + c = a \left(m + p\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = 2$$
$$b = 3$$
$$c = -2$$
Then
$$m = \frac{3}{4}$$
$$n = - \frac{25}{8}$$
So,
$$2 \left(p + \frac{3}{4}\right)^{2} - \frac{25}{8}$$
Factorization
[src]
(p + 2)*(p - 1/2)
$$\left(p - \frac{1}{2}\right) \left(p + 2\right)$$
(p + 2)*(p - 1/2)
Combining rational expressions
[src]
-2 + p*(3 + 2*p)
$$p \left(2 p + 3\right) - 2$$
-2 + p*(3 + 2*p)
Combinatorics
[src]
(-1 + 2*p)*(2 + p)
$$\left(p + 2\right) \left(2 p - 1\right)$$
(-1 + 2*p)*(2 + p)