factoring by grouping: 8x - 4x - 6x + 6 and x(x-1) - 3(x-1)

[chasann29]

1. 8x - 4x - 6x + 6
(8x -4x) - (6x +6)
x(8-4) - 6(x+6)

I got this far, but I don't even know if this is the correct way to complete the problem. Please help with figuring this out factoring by grouping.

2. x(x-1) - 3(x-1)

This is the same type of factoring by grouping. I just need help solving both of these problems.

chasann29

 

[tkhunny]

Re: factoring by grouping

1. 8x - 4x - 6x + 6
(8x -4x) - (6x +6)
x(8-4) - 6(x+6)

I got this far, but I don't even know if this is the correct way to complete the problem. Please help with figuring this out factoring by grouping.

2. x(x-1) - 3(x-1)

This is the same type of factoring by grouping. I just need help solving both of these problems.

chasann29

The second one is already done for you. Just tidy it up. (x-3)(x-1)

This also gives a hint how to solve the first one. You must get EXACTLY the same thing in the parentheses. See how it is 'x-1' in BOTH on the second?

This, x(8-4) - 6(x+6), is no good because the parentheses contain different things. They must be EXACTLY the same.

Really, factoring by grouping is kind of magic. You just have to see it or figure it out some other way that will TELL you how to see it.

Anyway, are you sure we have the right problem statement on that first one? Maybe there should be an \(\displaystyle x^{2}\) in there somewhere?

 

[masters]

Re: factoring by grouping

1. 8x - 4x - 6x + 6
(8x -4x) - (6x +6)
x(8-4) - 6(x+6)

I got this far, but I don't even know if this is the correct way to complete the problem. Please help with figuring this out factoring by grouping.

2. x(x-1) - 3(x-1)

This is the same type of factoring by grouping. I just need help solving both of these problems.

chasann29

Edit: I was a little bit slow on this one, TK

I think, maybe, there's something wrong with your first expression. Did you leave out an exponent or something? If you factor that (for whatever reason), you end up with x(8-4-6)+6 which simplifies to -2x+6.

The second one is almost complete. Factor out (x-1) from both terms and you get (x-1)(x-3).

 

[jonboy]

Re: factoring by grouping

factoring by grouping is just pulling out a gcf, something i'm sure you've learned before this method.

so if you have: \(\displaystyle 3(x + 1) - 2x(x + 1)\)

What's the gcf? (x + 1)

So pull it out: \(\displaystyle (x + 1)(3 - 2x)\)

 

[chasann29]

Re: factoring by grouping

No, the problem 1 is the right way, so should I ask about this problem?

For prolem #2, are you telling me the answer is (x-3)(x-1), or do I need to factor this out more?

The next problem goes with this problem.
Solve for x - use answer to #2

x(x-1) - 3(x-1) =0

For this problem, would I just plug in (x-3)(x-1) into the x?

 

[tkhunny]

Re: factoring by grouping

This is not encouraging. You don't seem to have an idea WHY factoring might be useful.

1) Pulling out the GCF is part of the magic I was talking about. First, you must get it into a form where there IS a GCF. There is no clear way to do that.

2) The point of factoring is to exploit this property: If A*B = 0, then either A = 0 or B = 0. Try it out. See if you can get zero (0) without one of them actually being zero (0).

Thus, if (x-3)(x-1) = 0, then either x-3 = 0 or x-1 = 0. Solve those two for 'x' and you should be done.

Note: I still don't beleive the first one is correct. That may be the way it appears on your sheet or in your book, but I'm thinking "typo". As you have it:

8x - 4x - 6x + 6 = 0

-2x + 6 = 0

x = 3

There is no need for factoring. The solution is simple. As presented, this problem makes no sense in a "factoring" section.