Mister Exam
Factor p^4+9*p^2+5 squared
The solution
Factorization
[src]
/ ____________\ / ____________\ / ____________\ / ____________\ | / ____ | | / ____ | | / ____ | | / ____ | | / 9 \/ 61 | | / 9 \/ 61 | | / 9 \/ 61 | | / 9 \/ 61 | |p + I* / - - ------ |*|p - I* / - - ------ |*|p + I* / - + ------ |*|p - I* / - + ------ | \ \/ 2 2 / \ \/ 2 2 / \ \/ 2 2 / \ \/ 2 2 /
$$\left(p - i \sqrt{\frac{9}{2} - \frac{\sqrt{61}}{2}}\right) \left(p + i \sqrt{\frac{9}{2} - \frac{\sqrt{61}}{2}}\right) \left(p + i \sqrt{\frac{\sqrt{61}}{2} + \frac{9}{2}}\right) \left(p - i \sqrt{\frac{\sqrt{61}}{2} + \frac{9}{2}}\right)$$
(((p + i*sqrt(9/2 - sqrt(61)/2))*(p - i*sqrt(9/2 - sqrt(61)/2)))*(p + i*sqrt(9/2 + sqrt(61)/2)))*(p - i*sqrt(9/2 + sqrt(61)/2))
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(p^{4} + 9 p^{2}\right) + 5$$
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
$$a = 1$$
$$b = 9$$
$$c = 5$$
Then
$$m = \frac{9}{2}$$
$$n = - \frac{61}{4}$$
So,
$$\left(p^{2} + \frac{9}{2}\right)^{2} - \frac{61}{4}$$
$$\left(p^{4} + 9 p^{2}\right) + 5$$
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
$$a = 1$$
$$b = 9$$
$$c = 5$$
Then
$$m = \frac{9}{2}$$
$$n = - \frac{61}{4}$$
So,
$$\left(p^{2} + \frac{9}{2}\right)^{2} - \frac{61}{4}$$
Combining rational expressions
[src]
2 / 2\ 5 + p *\9 + p /
$$p^{2} \left(p^{2} + 9\right) + 5$$
5 + p^2*(9 + p^2)