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.

[lKVNiiKINKYl]

If you are just factoring, wouldn't it be

a^3 - 4a^2 -a + 4
A x A x A -2A x 2A -a +4

Gosh, I should try to stay awake in math -.-

[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.

[jennyjenny]

i don't mean to sound liek a bitch but i have no clue what you guys are talking about ;x

we haven't learned that yet, but it's probably right.

i can remember some of what she taught us, and it was called grouping or something?

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

and then she did something like (a^3-4a^s)(-a+4)

or something.

now i forget what to do after that.

[Gigi]

^ Er, I dont' know how that works...

about cx+cy+bx+by, just factor out the like terms. You need to separate them.

So factor out the c from cx+cy

c(x+y)+bx+by

Then factor out the b from bx+by

c(x+y)+b(x+y)

And you can see that (x+y) is the common factor of c(x+y) and b(x+y), so you can factor that out.

(x+y)(c+b)

And that's your answer for that.


btw, do you take notes in class?

[silver-rain]

Ok, you first can separate the equation the way that Gigi said:
(a^3 - 4a^2) - (a - 4). Factor out the first part: a^2(a - 4) - (a - 4)
You notice that (a - 4) is a common factor of a^2(a - 4) - (a - 4), so you factor that out and get (a^2 - 1)(a - 4). And, you can then further factor the first part into (a - 1)(a + 1).

[*wind&fire*]

oops just like the one on the top.... ^how did you get the last step?
=a^3-4a^2-a+4

=a^2(a-4)-(a-4)

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

[Gigi]

^ (a^2 - 1), I forgot what that's called but it's just a quadratic. The "a" is gone because they've canceled each other out, so basically you're looking for two numbers that multiply to -1 and add to 0. And that's easy, because that's just 1 and -1. So you'll get (a - 1)(a + 1)

[silver-rain]

^^ Yeah, it's called the difference of two squares. It's a formula that's really useful. (a^2 - b^2) = (a - b)(a + b).

[jennyjenny]

Thanks, you guys.

Yeah it's called the difference of two squares, and we're also doing difference of two cubes and common factoring and blah. i don't really get it, which is why i'm going for extra help tomorrow. and yeah, i take notes but i missed it on those.

this can be closed-- it was due today.

[*mzkandi*]

Problem solved.

Topic Closed.