I understand now, how to do the simple ones(hardly). The ^before a number indicates its to the __ power. ( x^2= x to the second power)

This one:

x^2(x^3+2) = x^5 + 2x^2

So I understand that you multiply x^2 by x^3 and 2.But what do you do when you have three or more things to multiply in the first part.Hereis the problem:

( 3x-1)(4x^2 - 2x- 1)

Do I first look for like terms? Or do I stick with trying to multiply like this - ( I changed eached number to a letter)

(ab- c)(de^2- fg- h)= ad- ae^2- af- ag- ah - bd- be^2- bf-bg-bh

(and so on and so forth)

Basiclly, what I am asking is, do I have to multiply A,B&C each by D,E^2,F,G&H ???? So, no matter how many variables there are in the first part, I just multiply each by problems in the second half??

Also, What do I do if there are variables outside of the parenthesis? Multiply these the same? And how do I know whether to put a multiplication/add/subtract sig between each part. I know you would put subtractions signs in some areas on this problem, but once you have finished multiplying A by everything on the other side, do you put a mult. symbol between that and the B`s that you multiply next?

sorry. please, if someone could explain, i would be so greatful

## Answers (2)

So let's start with a very simple application of this, without quite as many coefficients as you put in.

What you want to know is how can you expand

(a + b)(c + d + e)

Now, what you have to do is multiply each term in the first set of brackets (that is, a and b), by each term in the second set of brackets (in this case, c, d, and e)

So what we end up with is

ac + ad + ae + bc + bd + be

Inr esponse to your question of "do I have to multiply A,B&C each by D,E^2,F,G&H ????", I'd say not quite. You have to multiply both AB and C by by DE^2, FG, and H.

In terms of negative signs, include them as you do the multiplication. So in your example with the letters, whenever you multiply by C, actually multiply by (-C), and when you multiply by FG, multiply by (-FG).

Applying this to the question of

(3x - 1)(4x^2 - 2x - 1)

we get that it equals

(3x)(4x^2) + (3x)(-2x) + (3x)(-1) + (-1)(4x^2) + (-1)(-2x) + (-1)(-1)

= (12x^3) + (-6x^2) + (-3x) + (-4x^2) + (2x) + (1)

= 12x^3 - 6x^2 - 4x^2 - 3x + 2x + 1

= 12x^3 -10x^2 - x + 1

( 3x-1)(4x^2 - 2x- 1)

You do the problem like this

3x(4x^2 -2x -1) -1(4x^2 -2x -1)

12x^3 -6x^2-3x -4x^2 +2x +1

now add like terms

12x^3 -10x^2 -x +1

this is very simple know? There is no confusion at all.