Expanding expressions

[Bagpipes]

Hello,

I've been trying to learn to expand expressions. I've found a particular tutorial with an example that I can't seem to understand, even though I know it's supposed to be very simple.

My understanding is that you multiply each term from within the brackets by the expression outside them. The example given is:

6g + 2g (3g + 7) = 6g + 2g x 3g + 2g x 7g (this is then simplified but that's not relevant to my question).

If 6g + 2g is the expression to multiply the bracketed terms by, then what's happening to the '6g' part when multiplying the '7'? Ie, how come it's not 6g + 2g x 3g + 6g + 2g x 7?

why does the '2g' get multiplied by both terms from within the brackets, but not '6g'?

I'd really appreciate it if anybody can explain what I'm getting wrong to me.

Thanks!

 

[Deleted member 4993]

Hello,

I've been trying to learn to expand expressions. I've found a particular tutorial with an example that I can't seem to understand, even though I know it's supposed to be very simple.

My understanding is that you multiply each term from within the brackets by the expression outside them. The example given is:

6g + 2g (3g + 7) = 6g + 2g x 3g + 2g x 7g (this is then simplified but that's not relevant to my question).

If 6g + 2g is the expression to multiply the bracketed terms by, then what's happening to the '6g' part when multiplying the '7'? Ie, how come it's not 6g + 2g x 3g + 6g + 2g x 7?

why does the '2g' get multiplied by both terms from within the brackets, but not '6g'?

I'd really appreciate it if anybody can explain what I'm getting wrong to me.

Thanks!

Following PEMDAS rule - you need to multiply first. Multiply 2g first with the (3g + 7) term - to get 6g² + 14g

Now you have 6g + 6g² + 14g

That is the RULE we follow - there is no "why" to it. We can only answer "How".

 

[lookagain]

My understanding is that you multiply each term from within the brackets by the expression outside them. The example given is:

6g + 2g (3g + 7) = 6g + 2g x 3g + 2g x 7g \(\displaystyle \ \ \ \)This last term should be 7. It might be a typo as it shows up as a 7 in a later step.

If 6g + 2g is the expression to multiply the bracketed terms by, then what's happening to the '6g' part when multiplying the '7'?

Please do not use "x" for multiplication as it looks like the x-variable. Pairs of parentheses or the asterisk can be good
symbols for that when typing that in the forum for clarity. Also, you should place the "2g" right up against the
"(3g + 7)" binomial as in "2g(3g + 7)." With the spacing in what you typed that may have led to some of your confusion
in what was actually intended to be multiplied together.

6g + 2g(3g + 7) = 6g + (2g)(3g) + (2g)(7) is one of the clearer ways.

Also, 6g + 2g(3g + 7) = 6g + 2g*3g + 2g*7 is another way out of the other total ways to be relatively clearer.

 

[Bagpipes]

Thank you. I'm new to this, sorry about the errors.

Think I've got it now.

Thanks again.

 

[Steven G]

Why are you not concerned about multiplying the 6g by 3g??

Let's move away from variables for a moment.

If you have 3 + 2(4+5), the 3 is not to be multiplied by the 4 and 5

3 + 2(4+5) = 3 + 2*4 + 2*5 = 3 + 8 + 10 = 21

To see that this is correct note that (4+5) = 9 so 2(4+5) = 2*9 = 18. Then 3 + 2(4+5) = 3 + 18 = 21.

Now if you had (3+2)(4+5) you will get (3+2)(4+5) = 3*4 + 3*5 + 2*4 + 2*5 = 12 + 15 + 8 + 10 = 45.
Note that (3+2)(4+5) = (5)*(9) = 45

The brackets make a big difference

 

[Bagpipes]

Why are you not concerned about multiplying the 6g by 3g??

Let's move away from variables for a moment.

If you have 3 + 2(4+5), the 3 is not to be multiplied by the 4 and 5

3 + 2(4+5) = 3 + 2*4 + 2*5 = 3 + 8 + 10 = 21

To see that this is correct note that (4+5) = 9 so 2(4+5) = 2*9 = 13. Then 3 + 2(4+5) = 3 + 18 = 21.

Now if you had (3+2)(4+5) you will get (3+2)(4+5) = 3*4 + 3*5 + 2*4 + 2*5 = 12 + 15 + 8 + 10 = 45.
Note that (3+2)(4+5) = (5)*(9) = 45

The brackets make a big difference

Thanks, yes - this is the mistake I was making. I misunderstood the way the brackets work and assumed the whole expression outside the brackets must be multiplied by the terms inside it. It seems so obvious now…

thanks!