Can someone please explain how to factor these algebra equations

Does anyone even know of a site? I have a quiz soon and I don't understand these.
14m^8+8m^5-10m^3

27y^3-9y^2+15y

xy-10-5y+2x

x^4-16

First of all, try to factor out anything that is in common with all terms. Like in the first one, all the coefficients, 14, 8, and 10 are even numbers, so what will divide an even number? answer 2. So we can factor out a 2 first of all.

So here we use the distributive property and rewrite it.
2(7m^8 + 4m^5 - 5m^3)

Next we notice the m's. We look for the highest power of m in common to all three, m^3. So we factor that out:
2m^3(m^5 + 4m^2 - 5)

We examine the three terms and see that there are no more factors in common. Also we see that it is not a quadratic situation, where you have the power of the first term being twice the power of the second term, like x^2 and x, or x^4 and x^2, or x^6 and x^3, and last term a number. This is as far as it can go, I believe.

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In second one, we do same thing, factor out highest common number to 27, -9, and 15, which is 3. We also can factor out y. So this leaves:
3y (3y^2 - 3y + 5)
So then we have to factor that quadratic. Using trial and error, we start with (3y )(y ). We will try to figure out what numbers should be put in those spaces.
The numbers of the two binomials must multiply to 5, so we guess that 5 and 1 are the two numbers. We notice, though, that the middle term (-3y) is negative, and there is no way to get a negative from using two positives for the numbers, so they must instead be -1 and -5. So we try both ways, using FOIL method as usual:
(3y - 5)(y - 1) = 3y^2 - 3y - 5y + 5 = 3y^2 - 8y + 5. So this is not the correct way

Trying other way:
(3y - 1)(y - 5) = 3y^2 - 15y - y + 5 = 3y^2 - 16y + 5. also not correct. So it can't be factored into binomials.

Checking quadratic formula, what is under radical is negative, so this is as far as it can be factored.