Mister Exam
Factor 2*p^4+p^2-2 squared
The solution
Factorization
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$$\left(p - i \sqrt{\frac{1}{4} + \frac{\sqrt{17}}{4}}\right) \left(p + i \sqrt{\frac{1}{4} + \frac{\sqrt{17}}{4}}\right) \left(p + \sqrt{- \frac{1}{4} + \frac{\sqrt{17}}{4}}\right) \left(p - \sqrt{- \frac{1}{4} + \frac{\sqrt{17}}{4}}\right)$$
(((p + i*sqrt(1/4 + sqrt(17)/4))*(p - i*sqrt(1/4 + sqrt(17)/4)))*(p + sqrt(-1/4 + sqrt(17)/4)))*(p - sqrt(-1/4 + sqrt(17)/4))
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(2 p^{4} + p^{2}\right) - 2$$
To do this, let's use the formula
$$a p^{4} + b p^{2} + c = a \left(m + p^{2}\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = 2$$
$$b = 1$$
$$c = -2$$
Then
$$m = \frac{1}{4}$$
$$n = - \frac{17}{8}$$
So,
$$2 \left(p^{2} + \frac{1}{4}\right)^{2} - \frac{17}{8}$$
$$\left(2 p^{4} + p^{2}\right) - 2$$
To do this, let's use the formula
$$a p^{4} + b p^{2} + c = a \left(m + p^{2}\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = 2$$
$$b = 1$$
$$c = -2$$
Then
$$m = \frac{1}{4}$$
$$n = - \frac{17}{8}$$
So,
$$2 \left(p^{2} + \frac{1}{4}\right)^{2} - \frac{17}{8}$$
Combining rational expressions
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2 / 2\ -2 + p *\1 + 2*p /
$$p^{2} \left(2 p^{2} + 1\right) - 2$$
-2 + p^2*(1 + 2*p^2)