i need helps with some algebra

[Ceist]

i didn't take notes well enough, now i'm asking for a tutor from the blforums.
help pls.




I'm not asking for answers, I just want to see how I can do these. I used a few sources that my own teacher provided but got me no where. I also want you to tell me if any of these are wrong.

I get extremely stressed about homework.

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[Jairo]

loving hell I used to know how to factor expressions

god dammit mrs. catts why can't you teach

[BuildaBox]

Nice handwriting ;P

To factor an expression you take coefficients and numbers/multiple/whatever that are in every number in a problem or whatever and put them outside of a parenthesis. Sorry, don't know how to properly explain it lol.

For example...

3x + 4x would factor to x(3 + 4)
4 + 2 would factor to 2(2 + 1)
4x + 2x would factor to 2x(2 + 1)

[Jairo]

Nice handwriting ;P

To factor an expression you take coefficients and numbers/multiple/whatever that are in every number in a problem or whatever and put them outside of a parenthesis. Sorry, don't know how to properly explain it lol.

For example...

3x + 4x would factor to x(3 + 4)
4 + 2 would factor to 2(2 + 1)

he's doing trinomials though, not binomials

EDIT: the second paper is trinomials

[BuildaBox]

he's doing trinomials though, not binomials

EDIT: the second paper is trinomials

It's the same concept though isn't it?

[Jake White]

It's the same concept though isn't it?

Try factoring a quadratic with that principle. See how far you get.

[BuildaBox]

Try factoring a quadratic with that principle. See how far you get.

What's a quadratic

[Headcrab Zombie]

What's a quadratic

Ax^2+bx+c

My teacher taught me a method of factoring quadratic that was extremely easy, I don't remember it though

Oh half the stuff on the second page is quadratic, though

[BuildaBox]

Ax^2+bx+c

Alright... So then how would it not work with them?

Like number twenty two would be 3(x^2 - 5x + 6) right?

[Treynolds416]

Not bothering to check your existing answers, but this has been the easiest way for me to figure out quadratics over the past 4-5 years:

Ax^2 + Bx + C
If A is a prime number, write it down on your paper like this:
(Ax    )(x    )
I'll use #22 as an example
(3x    )(x    )
If the operator between the B and C terms is positive, you know that both of the factors have the same sign. In the case of #22, the operator between A and B is negative, and the one between B and C is positive, so you should have this written on your paper:
(3x -  )(x -  )
Now you have to separate the C term into a number of possible factors. For the number 18, the possible factors are 1 and 18, 2 and 9, 3 and 6. You want the two factors (with one of them multiplied by the A coefficient) to add up to 15, because that's the B term. This is the hard part for most people, because there are quite a few combinations possible, but it will start to become easier the more you do it as you make better guesses and then check them. For the sake of learning, I calculated all the possibilities below:

1 * 3 + 18 = 21
1 + 18 * 3 = 55
2 * 3 + 9 = 15
2 + 9 * 3 = 29
3 * 3 + 6 = 15
3 + 6 * 3 = 21

As you can see, there are two solutions (even though they're really the same solution). It doesn't matter which one you pick to use at this point.
The term you want multiplied against the A term must be in the opposite factor
(3x -  )(x - 2) or (3x -  )(x - 3)
And then fill in the other factor of 18
(3x - 9)(x - 2) or (3x -  6)(x - 3)
If you check, you'll notice that both of these evaluate back to 3(x - 3)(x - 2) or 3x^2 - 15x + 18

I'll post again in a second when I write one up for a different problem that has an even A term and a negative B-C operator

[Headcrab Zombie]

Only half the stuff on the second page is quadratic, the rest is cubic
I don't remember ever factoring cubic equations

[Bloody Mary]

My completely rusty opinion:

1 seems okay. 0.5 is the same thing as 1/2, there's no need to write both.
2 seems fine
3 seems fine
4 seems okay I guess. Honestly I hate mixed numbers so I kind of glossed over it.
5 is screwed up. When you multiplied -3 by (-1/3), you for some reason made the subtraction multiplication instead of addition. Your answer of (2(1)/4) is not correct.
6 seems fine
7 is fine I think
8 seems okay
I can't tell if you did 9 correctly because I can't read your handwriting well enough; the Ns and Ms are confusing me
10 suffers from the same problem as 9
11 is messed up. When you split the large fraction into two separate ones, you did not preserve the sign between them.
12 seems fine
13 I can hardly read
14 isn't factored completely.
15 is screwed up. -6m + 6m is not m.
16 isn't factored properly.
17 isn't factored properly.
18 is not factored properly

The rest aren't even done. If you have a specific question feel free to ask; I'm not going to go through a bunch of not even tried problems and tell you how to do them.

[Treynolds416]

#33:

Ax^2 + Bx + C I'm using x as the variable for this one too because forget "a" as a variable

QUICK NOTE: I've been using incorrect terminology when I say "B-C operator", I should've said "the sign of the C term"

Anyway, the first thing I'll do is factor out some of the big numbers to make it easier to work with
60x^3 + 54x^2 - 6x becomes 6x(10x^2 + 9x - 1)
Now you're just going to factor the expression in the parentheses, just like before. There are now two different setups you could use though, because A is even.
6x(10x    )(x    ) or 6x(5x    )(2x    )
Because the C term is negative, there are now four different combinations you could try
6x(10x +  )(x -  ) or 6x(10x -  )(x +  ) or 6x(5x +  )(2x -  ) or 6x(5x -  )(2x +  )
You may be a bit overwhelmed, but don't worry it will get easier. Because the C term is 1, the only possible factors are 1 and 1.

10 * -1 + 1 * 1 = -9
10 * 1 + 1 * -1 = 9
5 * -1 + 2 * 1 = -3
5 * 1 + 2 * -1 = 3

The correct combination is the second one, so your paper should look like this
6x(10x - 1)(x + 1)
Which factors back out to the original problem.

Sorry for explaining it so terribly, I did it quickly. Just ask me questions about anything you don't understand

Only half the stuff on the second page is quadratic, the rest is cubic
I don't remember ever factoring cubic equations

They're not real cubics, the constant is 0, so you can just factor out one of the x's. On a graph, it means that the curve intersects the x axis at x = 0.

[Zanaran2]

just use google.
google can actually solve math equations for you

[Jairo]

just use google.
google can actually solve math equations for you

you mean wolfram alpha?