{Math}Second Degree Equations

My math teacher is horrible at explaining stuff, and has decided the easy way to make us learn second degree equations would be to make us do them with homework. And so, I call for help. I've managed to do two of the five problems he's given us. However, the following problems are a mystery to me. Can anyone bring light to these problems? The equations have to be factorized.

ab+3a-bc-3c

a(3)b-ab-ab(2)

x(2)-29x-81

Thank you in advance,

Wolf.

So do you need help with factorization? I assume by second degree, you mean a variable squared. Well... I'll do the first as an explanation.

ab + 3a - bc - 3c

1. Find a common factor between some numbers.
2. Group them together so they have a common factor. (Luckily, it's been done here.)
3. Factor out this common factor.

a(b+3) +c(-b-3)

4. Make sure the expression in parentheses are the same. You may have to start over if they are nowhere near the same.

a(b+3) -c(b+3) A negative has been taken out.

4. Treat the expression in parentheses as a single variable and factor that out.

(a-c)(b+3)

This is the way of factorization. Some can be factored entirely, like your second expression. I assume the parentheses show powers. By the way, this is third degree.

a^3b - ab - ab^2

1. Here the common factor, shared by all three, is ab.

ab(a^2 - 1 - b)

Factored!

Your third, though, is the infamous quadratic! It is in the form ax^2 + bx + c!

x^2 - 29x - 81.

1. Find the factors of -81.

1, 81
3, 23
9, 9

2. Add/ subtract to them to see if they equal -29.

1+81 = 90
1- 81 = -80
3+23 = 26
3-23 = -20
9+9= 18
9-9 = 0

None equal - 23!!

As of now... you should declare this... not factorable! Unless you know the quadratic formula or 'Complete the Square', leave it be.

Here is a general rule about quadratics.

ax^2 +bx +c.

The factors of c/a must add up to equal b. If this can't be done, then, (at your stage I assume) it is unfactorable.

Hope this helps.

Indeed it does; thanks a lot, man.