math: factoring? Options

[jennyjenny]

How do you do some of these type of problems? It's my homework and I didn't really understand in class.

a^3-4a^2-a+4

or

cx+cy+bx+by

it's algebra 1 or algebra 2, btw.

[Gigi]

Okay, for the cubic equation, have you guys learned the Remainder or Factor Theorem yet? I'll assume you have, for now.

So basically, you're going to look for a factor that works in a^3-4a^2-a+4.

Since the last number is 4, that must mean that the three factors' (a-k) last numbers (k) must multiply to 4.

Which means a^3-4a^2-a+4 must have factors that are (a-k), k being a factor of 4. Which means we only need to try out (a-1), (a+1), (a-2), (a+2), and (a-4),(a+4).

The Factor Theorem states that f(k) must equal to 0 (f(k)=0) if the number is a factor of the polynomial.

So let's try out (a-1); that means that a=1. Plug that into the equation.
f(a) = a^3 - 4a^2 - a + 4
f(1) = 1^3 - 4(1^2) - 1 +4 <---- This must equal to 0 for (a-1) to be a factor.

Sooo...
1^3 - 4(1^2) - 1 +4 = 0
1 - 4 - 1 + 4 = 0
-3 +3 = 0

And since that equals to zero, it must mean that (a-1) factors into a^3 - 4a^2 - a + 4.

Now divide a^3 - 4a^2 - a + 4 by (x-1), and I'm going to assume that you know how to do that, also.

You end up with:
a^2 - 3a - 4.

And since that's a quadratic equation, you can factor that easily just by looking at it, using the quadratic formula, or completing the square.

Factor the quadratic. You'll get:
a^2 - 3a - 4 = (a-4)(a+1)

Take the factor you got from guessing and checking, which was (a-1), and put (a-4)(a+1), factors you got from factoring the quadratic, and put them together.

a^3 - 4a^2 - a + 4 = (a-1)(a-4)(a+1)

Check if it works!



Oh god, that's long. I hope that wasn't confusing...if you haven't learned any of that stuff I'll help you with that too.