Mister Exam
Factor polynomial z^2-16+64/8-z
The solution
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2 z - 16 + 8 - z
$$- z + \left(\left(z^{2} - 16\right) + 8\right)$$
z^2 - 16 + 8 - z
Factorization
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/ ____\ / ____\ | 1 \/ 33 | | 1 \/ 33 | |x + - - + ------|*|x + - - - ------| \ 2 2 / \ 2 2 /
$$\left(x + \left(- \frac{1}{2} + \frac{\sqrt{33}}{2}\right)\right) \left(x + \left(- \frac{\sqrt{33}}{2} - \frac{1}{2}\right)\right)$$
(x - 1/2 + sqrt(33)/2)*(x - 1/2 - sqrt(33)/2)
The perfect square
Let's highlight the perfect square of the square three-member
$$- z + \left(\left(z^{2} - 16\right) + 8\right)$$
To do this, let's use the formula
$$a z^{2} + b z + c = a \left(m + z\right)^{2} + n$$
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
$$m = - \frac{1}{2}$$
$$n = - \frac{33}{4}$$
So,
$$\left(z - \frac{1}{2}\right)^{2} - \frac{33}{4}$$
$$- z + \left(\left(z^{2} - 16\right) + 8\right)$$
To do this, let's use the formula
$$a z^{2} + b z + c = a \left(m + z\right)^{2} + n$$
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
$$m = - \frac{1}{2}$$
$$n = - \frac{33}{4}$$
So,
$$\left(z - \frac{1}{2}\right)^{2} - \frac{33}{4}$$