1. (x+1)(x+2)
2. (x+2)(x-1)
3. (x-1)(x-2)
4. (x+3)(x-3)
5. (x-7)squared
6. xsquared + 4x + 3
7. xsquared + 7x + 12
8. xsquared - 5x + 4
9. xsquared + 21x + 20
10. xsquared + 2x - 15
11. xsquared - x - 12
12. xsquared + 5x - 36

2. Originally Posted by JordanJH1993

1. (x+1)(x+2)
2. (x+2)(x-1)
3. (x-1)(x-2)
4. (x+3)(x-3)
5. (x-7)squared
6. xsquared + 4x + 3
7. xsquared + 7x + 12
8. xsquared - 5x + 4
9. xsquared + 21x + 20
10. xsquared + 2x - 15
11. xsquared - x - 12
12. xsquared + 5x - 36
Well, first off, these are not equations. They are expressions. I guess you intended to set each expression to 0.

Second, that looks like an awful lot of work for us tutors. Can you supply any of the work on your own?

3. Yes I have done the most of them myself. I just want to know if I am correct.

I don't want to hand in an incorrect homework...

4. Originally Posted by JordanJH1993
Yes I have done the most of them myself. I just want to know if I am correct.

I don't want to hand in an incorrect homework...
Supply us with the answers you have, and we'll check 'em.

5. 1. Xsquared + 3x + 3
2. Xsquared + 1x -3
3. Xsquared - 3x -3
4. Xsquared + 0 -6
5. 49
6. (x+1)(x+3)
7. (x+3)(x+4)
8. (x-4)(x-1)

The rest I do not know...

6. Originally Posted by JordanJH1993

1. (x+1)(x+2)
2. (x+2)(x-1)
3. (x-1)(x-2)
4. (x+3)(x-3)
5. (x-7)squared
6. xsquared + 4x + 3
7. xsquared + 7x + 12
8. xsquared - 5x + 4
9. xsquared + 21x + 20
10. xsquared + 2x - 15
11. xsquared - x - 12
12. xsquared + 5x - 36
I guess I'm confused now about what you want to accomplish. In your last post, all you did was multiply the two binomials together. And, I might add, you had a little problem with that.

So, let me ask this. For the first 5, do you just want to multiply these binomials together? Or do you want to set them = 0 and solve for x?

{1}

$\displaystyle (x+1)(x+2)=x^2+3x+2$

or

$\displaystyle (x+1)(x+2)=0$

$\displaystyle x+1= \ \ or \ \ x+2=0$

$\displaystyle x=-1 \ \ or \ \ x=-2$

And for 6-12, do you just want to factor them or set them = 0 and solve for x?

7. I am not too sure how to answer what you asked really.

The easiest way I could put it is that I want the answers for 1-4 to look like the questions for 6-12 and vice versa.

So that would basically be the first example you said.

I hope that makes it clearer for you.

8. Originally Posted by JordanJH1993

1. (x+1)(x+2)
2. (x+2)(x-1)
3. (x-1)(x-2)
4. (x+3)(x-3)
5. (x-7)squared
6. xsquared + 4x + 3
7. xsquared + 7x + 12
8. xsquared - 5x + 4
9. xsquared + 21x + 20
10. xsquared + 2x - 15
11. xsquared - x - 12
12. xsquared + 5x - 36
Ok, I got it now. You just want to multiply these binomials together in 1-5 and factor 6-12.

I will demonstrate a technique for 1-5 using #1.

$\displaystyle (x+1)(x+2)$

We are going to expand this rascal using the distributive property. The general form is this:

$\displaystyle (a+b)(c+d)=a(c+d)+b(c+d)$

Specifically for #1, we have:

$\displaystyle x(x+2)+1(x+2)$

$\displaystyle x^2+2x+x+2$

$\displaystyle x^2+3x+2$

Now work the other 4 just like that.

For 6-12, we need to factor the trinomial (3 term expression) into the product of two binomals (2 term expressions).

We'll use #6

$\displaystyle x^2+4x+3$

Now this is where you have to use your head a little. You must think of a pair of numbers whose product is 3 and whose sum is 4. The 3 comes from the last term in the trinomial, and the 4 comes from the middle term.

How many ways can you multiply two whole numbers to get 3? Remember their sum must be 4.

3 X 1 = 3
3 + 1 = 4

So, the numbers must be 3 and 1.

$\displaystyle x^2+4x+3=(x+3)(x+1)$

Now you try some.

9. Thanks for the help. I have gotten the grasp of it now.

The examples made me understand it far clearer.

I will be sure to use this site when I am stuck now !

Thanks a million!