Rom

When you say complex, do you mean non-real?

This is a 4th degree polynomial.  There are clearly 4 complex zeros.

However it looks like some of them might be real.

$x^4+5x^2+6 = (x^2+3)(x^2+2)$

Nope.  None of these zeros are real.  So there are 4 complex zeros.

Given that factors can you quickly list off what they are?

$n = \dbinom{9}{2}\dbinom{12}{3} = 7920$

It's all the same idea.  Most likely your real problem is you just (like most people) can't see powers of small primes at a glance.

$2\sqrt[5]{320} - \sqrt[6]{320} - 2\sqrt[5]{384} = \\ 2\sqrt[5]{2^6 \cdot 5}- \sqrt[6]{2^6\cdot 5} - 2\sqrt[5]{2^7\cdot 3}\\ 2\sqrt[5]{2^5\cdot 30}-2\sqrt[6]{5}-2\cdot 2\sqrt[5]{2^2\cdot 3} =\\ 4\sqrt[5]{30}-2\sqrt[6]{5}-4\sqrt[5]{12}$

$\sqrt[3]{16} = \sqrt[3]{2^4} = 2\sqrt[3]{2} \\ \\ \sqrt[3]{16}+ \sqrt[3]{5}+\sqrt[3]{2} = \\ \\ 2\sqrt[3]{2} +\sqrt[3]{5}+\sqrt[3]{2} = \\ \\ 3\sqrt[3]{2} + \sqrt[3]{5}$

$567 = 81 \cdot 7 = 3^4 \cdot 7$

$6\sqrt[4]{567x^4} = \\ 6\sqrt[4]{3^4x^4\cdot 7} =\\ 6\cdot 3x\sqrt[4]{7} = \\ 18x\sqrt[4]{7}$

$4\sqrt[3]{64n^8} = \\ 4\sqrt[3]{2^6 n^6 \cdot n^2} = \\ 4\cdot 2^2 n^2 \sqrt[3]{n^2} = \\ 16n^2 \sqrt[3]{n^2}$

$x^2 + 24 = 0\\ x^2 = -24\\ x = \pm i\sqrt{24} = \pm 2i\sqrt{6}\\ x = 2i\sqrt{6} \text{ or }x= -2i\sqrt{6}$

$\text{the key is that } \dfrac a b \div \dfrac c d = \dfrac a b \times \dfrac d c\\ -\dfrac{21}{10}\div \dfrac{56}{24} = -\dfrac{21}{10}\times \dfrac{24}{56} = -\dfrac{9}{10}$

well they've highlighted two points on this line so I guess they want you to use them.

$\text{the two points are }(3,21) \text{ and }(6,12) \\ \text{we calculate the slope as }m = \dfrac{y_2-y_1}{x_2-x_1} \\ m = \dfrac{12 - 21}{6-3} = \dfrac{-9}{3}=-3 \\ \text{using the first point, point slope form is thus}\\ (y-21)=-3(x-3) \\ \text{and with a bit of rearranging we get }\\ y = -3x + 30 \text{ which is slope intercept form}$

for (b) just plug in x=0 for the y intercept

we already found the slope as m=-3

so his delivery time decreased 3 minutes per day