Multiplying out brackets efficiently

OK so I've managed them...

[MATH]{f}+{g}({f}+{h})[/MATH] The term (f) is not included as discussed.

[MATH]{g}\times{f}={f}{g}+{g}\times{h}={g}{h}[/MATH] The solution to the expression is [MATH]{f}+{fg}+{gh}[/MATH]
I have three others of the same type that I also made the same errors with, but have now corrected them. The final expression also tripped me up because I changed the sides of the terms from the right side to the left side, and the reason I did this was the coursework book said in previous examples of expressions that if a student wanted to, then the terms of the expression could be moved from the right side to the left side of the expression, hence I did and I could not get the correct answer, and now I understand why...

I have the expression;

[MATH]({d}-{c}){c}-{c^2}[/MATH]
From here I can change the sides if I don't feel comfortable with the layout as the coursework advised. So I did and in this example you cannot change the sides and here is why...

[MATH]{c^2}-{c}({d}-{c})={-cd}+{c^2}[/MATH] So I'd end up with [MATH]{c^2}{-cd}+{c^2}[/MATH]
But if I keep the same layout I then get;

[MATH]({d}-{c}){c}-{c^2}={-c^2}+{dc}-{c^2}={dc}-{2c^2}[/MATH]
This lot is not as easy as it looks to me.

Sorry I made another error in the above. My misunderstanding again. [MATH]{c^2}-{c}[/MATH] should have been written as [MATH]{c}-{c^2}[/MATH]
 
Probability said:
[MATH]{g}\times{f}={f}{g}+{g}\times{h}={g}{h}[/MATH]
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The above is very disrespectful to the equal sign.

The second expression - can you explain what you mean by "changing sides"? Your final solution is correct, but the first attempt is very confusing. You can move the terms around, but you have to keep their signs. (-c2) has to keep the minus sign if you move it.
 
Mathematically I meant change the order. Here is the example;

[MATH]{c^2}-{c}({d}-{c})=-{cd}+{c^2}[/MATH] The result would then be;

[MATH]-{cd}+{c^2}+{c^2}=-{cd}+{2c^2}[/MATH]
So it seems in this example that I can't change the order if I prefer as I was advised in the books strategy. Hence;

[MATH]({d}-{c}){c}-{c^2}={-c^2}+{dc}{-c^2}={dc}-{2c^2}[/MATH] Now I have two queries here, 1 I was also advised in previous strategies to follow the order of the alphabet in arranging the order of letters, so in this example it multiplies out as [MATH]{cd}[/MATH] and not [MATH]{dc}[/MATH] but the solution shows it that way, and the second point I've forgotten at the moment!

Of course you are correct I should not have written; [MATH]{g}\times{f}={fg}+{g}\times{h}={gh}[/MATH] I see I got that completely wrong, hence I should have said something like;

[MATH]{g}\times{f}={fg}[/MATH] and [MATH]{g}\times{h}={gh}[/MATH] therefore this implies [MATH]{fg}+{gh}[/MATH]
 
Probability said:
Mathematically I meant change the order. Here is the example;

[MATH]{c^2}-{c}({d}-{c})=-{cd}+{c^2}[/MATH] The result would then be;

[MATH]-{cd}+{c^2}+{c^2}=-{cd}+{2c^2}[/MATH]
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I don't understand what you are doing here. The first line is wrong. The result there is -cd+2c2
 
lev888 said:
I don't understand what you are doing here. The first line is wrong. The result there is -cd+2c2
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I've just broke it up into two steps that is all. But you have just reminded me about my second point I forgot about earlier!

[MATH]{c^2}-{c}({d}-{c})=-{cd}+{c^2}-{c^2}[/MATH] This last term where I have changed the order by moving [MATH]+{c^2}[/MATH] from the left hand side of the expression to the right hand side of the expression is where I was getting confused probably over the sign convention, but I've got it now. If [MATH]{c^2}[/MATH] is addition on the left hand side of the expression and then I move it to the right hand side of the expression, then the [MATH]{c^2}[/MATH] becomes [MATH]{-c^2}[/MATH] I understand it now.

But the point I was trying to make was that I can't do this;

[MATH]{c^2}-{c}({d}-{c})[/MATH] from this [MATH]({d}-{c}){c}-{c^2}[/MATH] just because I prefer to change the order as the book previously advised. The answers are completely different.
 
Probability said:
I've just broke it up into two steps that is all. But you have just reminded me about my second point I forgot about earlier!

[MATH]{c^2}-{c}({d}-{c})=-{cd}+{c^2}-{c^2}[/MATH] This last term where I have changed the order by moving [MATH]+{c^2}[/MATH] from the left hand side of the expression to the right hand side of the expression is where I was getting confused probably over the sign convention, but I've got it now. If [MATH]{c^2}[/MATH] is addition on the left hand side of the expression and then I move it to the right hand side of the expression, then the [MATH]{c^2}[/MATH] becomes [MATH]{-c^2}[/MATH] I understand it now.

But the point I was trying to make was that I can't do this;

[MATH]{c^2}-{c}({d}-{c})[/MATH] from this [MATH]({d}-{c}){c}-{c^2}[/MATH] just because I prefer to change the order as the book previously advised. The answers are completely different.
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I think you are confusing moving terms around in an expression (signs DO NOT CHANGE) and moving terms to the other side if the equal sign in an equation (signs change).
 
Probability said:
I've just broke it up into two steps that is all. But you have just reminded me about my second point I forgot about earlier!

[MATH]{c^2}-{c}({d}-{c})=-{cd}+{c^2}-{c^2}[/MATH] This last term where I have changed the order by moving [MATH]+{c^2}[/MATH] from the left hand side of the expression to the right hand side of the expression is where I was getting confused probably over the sign convention, but I've got it now. If [MATH]{c^2}[/MATH] is addition on the left hand side of the expression and then I move it to the right hand side of the expression, then the [MATH]{c^2}[/MATH] becomes [MATH]{-c^2}[/MATH] I understand it now.

But the point I was trying to make was that I can't do this;

[MATH]{c^2}-{c}({d}-{c})[/MATH] from this [MATH]({d}-{c}){c}-{c^2}[/MATH] just because I prefer to change the order as the book previously advised. The answers are completely different.
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If you want to change the order of operation like that - you have to remember that a negative sign is actually multiplication by "-1"

so

[MATH]{c^2}-{c}({d}-{c})[/MATH] becomes

= [MATH]{c^2}+{c} * (-1) * ({d}-{c})[/MATH] ........... now you can change order "blindly"

= [MATH]{c} * (-1) * ({d}-{c}) + {c^2}[/MATH]
= [MATH]{c} * ({c}-{d}) + {c^2}[/MATH]
= [MATH]{c} * {c} - {c} * {d} + {c^2}[/MATH]
= [MATH]{c^2} - {c} * {d} + {c^2}[/MATH]
= [MATH]{2c^2} - {c} * {d}[/MATH]
Above, you "changed order" (correctly) and did not change "answer". But as you saw, if you write "every step" explicitly, the process becomes excruciatingly long - but avoids mistakes.
 
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I suspect that you try to generalize too soon.

f + g(f + h). Yes, in the expression

f + g(f + h), the f in blue is not operated on by anything else, but the f in parentheses is operated on by g. The point is that f takes on different roles in different parts of the expression. You can be a father in one aspect of your social responsibilities and an employee in a different aspect of your social responsibilities. Math notation can imply many subtleties.

[MATH]f + g(f + h) = f + fg + gh = f(1 + g) + gh.[/MATH]
I for one do not find one expression simpler than the other, but they are very different ways to view the same thing. People who believe that math is mechanical miss one of its great points: what is the same can have multiple looks. In fact, what is ANY equation except two different ways to describe the exact same number.
 
Thank you for all your input and efforts to help me along my journey trying to learn this subject. However in this instance I feel that (all my fault) the interpretation of what I'm trying to explain is being lost along the way somehow?

I'll try and keep it simple. I have the expression; [MATH]({d}-{c}){c}-{c^2}[/MATH] This is exactly how the author presented the expression.

I was previously advised by the author that I could change the order if I preferred, however what I'm trying to explain here is that sometimes you CAN'T change the order because the layout might not suit the individual person doing the expressions!

If I now change the order of the expression; [MATH]{-c^2}+{c}({d}-{c})=({cd}-{c^2}-{c^2})={cd}-{2c^2}[/MATH] This solution is not in the same order as the authors, which is [MATH]{dc}-{2c^2}[/MATH]
The author of the material advised previously to write the algebra in alphabetical order, but the author has not followed their own rules here!

So is the solution I've calculated correct or incorrect when compared to the authors?

Now if we take a step back and look at changing the order and putting numbers in as previously advised by JeffM to check the solution, then we can see clearly that the order here in this expression CANNOT be changed, hence;

[MATH]{dc}-{2c^2}={2}\times{3}-{2}\times{3^2}={-12}[/MATH] and

[MATH]{-dc}+{2c^2}={-2}\times{3}+{2}\times{3^2}=12[/MATH]
The solutions to each method is different. It seems that there are a few ways to tackle this problem, but maybe best to learn the method shown by the author, provided the authors material is correct in itself!
 
Probability said:
I'll try and keep it simple. I have the expression; [MATH]({d}-{c}){c}-{c^2}[/MATH] This is exactly how the author presented the expression.

I was previously advised by the author that I could change the order if I preferred, however what I'm trying to explain here is that sometimes you CAN'T change the order because the layout might not suit the individual person doing the expressions!

If I now change the order of the expression; [MATH]{-c^2}+{c}({d}-{c})=({cd}-{c^2}-{c^2})={cd}-{2c^2}[/MATH] This solution is not in the same order as the authors, which is [MATH]{dc}-{2c^2}[/MATH]
Click to expand...

Multiplication is commutative, therefore, cd = dc above. So the solution IS the same. The order in multiplication does not matter.

Probability said:
The author of the material advised previously to write the algebra in alphabetical order, but the author has not followed their own rules here!

So is the solution I've calculated correct or incorrect when compared to the authors?
Click to expand...
Whenever you refer to something the author wrote, it would help if you included the complete quotation. Write algebra in alphabetical order? What?
Your solution is correct.

Probability said:
Now if we take a step back and look at changing the order and putting numbers in as previously advised by JeffM to check the solution, then we can see clearly that the order here in this expression CANNOT be changed, hence;

[MATH]{dc}-{2c^2}={2}\times{3}-{2}\times{3^2}={-12}[/MATH] and

[MATH]{-dc}+{2c^2}={-2}\times{3}+{2}\times{3^2}=12[/MATH]
The solutions to each method is different. It seems that there are a few ways to tackle this problem, but maybe best to learn the method shown by the author, provided the authors material is correct in itself!
Click to expand...

This is extremely confusing to me. Are we talking about the same expression as in the beginning of your post? [MATH]{dc}-{2c^2}[/MATH]?
Do you see that [MATH]{-dc}+{2c^2}[/MATH] is different? Why in the first part of your post changing the order kept signs of the terms unchanged, but here they changed? Of course if you change signs and plug in numbers the result will be different.
 
Possibly the thing you are missing is that the author's answer is not the only valid answer. You should never compare your answer to someone else's and conclude that you are wrong because the order is different. Your answer of [MATH]{cd}-{2c^2}[/MATH]is equivalent to theirs of [MATH]{dc}-{2c^2}[/MATH]. (Possibly the reason they wrote it that way is that they are thinking of the result as a polynomial in c, so they put c last. Or it may just be some quirk of their thinking.)

There is no rule that expressions must be written in alphabetical order. That is a common recommendation, simply because (a) it makes it more likely that your answer will match others', so you can more quickly check that you agree, and (b) it helps you compare parts of your own work, and gives you a habit that frees you from making lots of trivial decisions about order.
 
I think I'll leave it now at this point. I've learned a lot about order of operations and what can happen when order is changed. I can see that I'm also creating some confusion in what I trying to explain, however the main point to be brought out of this discussion being that I've now understood what can happen when I change the order of an expression. Something I am now very mindful about. Thank you all who have helped me out in this thread, this is very much appreciated.
 
Probability said:
I think I'll leave it now at this point. I've learned a lot about order of operations and what can happen when order is changed. I can see that I'm also creating some confusion in what I trying to explain, however the main point to be brought out of this discussion being that I've now understood what can happen when I change the order of an expression. Something I am now very mindful about. Thank you all who have helped me out in this thread, this is very much appreciated.
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If you can't explain your point maybe it's an indication that you don't understand the topic fully.
 
lev888 said:
Multiplication is commutative, therefore, cd = dc above. So the solution IS the same. The order in multiplication does not matter.


Whenever you refer to something the author wrote, it would help if you included the complete quotation. Write algebra in alphabetical order? What?
Your solution is correct.



This is extremely confusing to me. Are we talking about the same expression as in the beginning of your post? [MATH]{dc}-{2c^2}[/MATH]?
Do you see that [MATH]{-dc}+{2c^2}[/MATH] is different? Why in the first part of your post changing the order kept signs of the terms unchanged, but here they changed? Of course if you change signs and plug in numbers the result will be different.
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Yes we are talking about the same expression. In my limited understanding of the expression I changed the order. This is what I did.

[MATH]({d}-{c}){c}-{c^2}[/MATH] This is the original expression. In previous expressions like this I was advised that I could change the order. I wrote;

[MATH]{c^2}-{c}({d}-{c})[/MATH] Initially I changed the sign but later I changed the sign again of [MATH]{c}[/MATH] on the left hand side making it addition. I was not sure whether that practice was allowed or not. Reading through this thread will show that I gained two different solutions, (1) showed (-dc) while the other showed (dc). They are not the same, hence the discussion about changing the order. In conclusion I've learned that it is better to just leave the expressions as presented.
 
Probability said:
Yes we are talking about the same expression. In my limited understanding of the expression I changed the order. This is what I did.

[MATH]({d}-{c}){c}-{c^2}[/MATH] This is the original expression. In previous expressions like this I was advised that I could change the order. I wrote;

[MATH]{c^2}-{c}({d}-{c})[/MATH] Initially I changed the sign but later I changed the sign again of [MATH]{c}[/MATH] on the left hand side making it addition. I was not sure whether that practice was allowed or not. Reading through this thread will show that I gained two different solutions, (1) showed (-dc) while the other showed (dc). They are not the same, hence the discussion about changing the order. In conclusion I've learned that it is better to just leave the expressions as presented.
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No, I wouldn't say it's better to leave every expression as it is (though it may often be better not to change something without a good reason). What's best is to know what changes are valid -- that is, what you can do that doesn't change the value of an expression. You can change the order of certain things only, such as terms in an addition.

In this example, you have something of the form A - B, and hopefully you've learned that you can't change it to B - A, as for example 3 - 2 and 2 - 3 are not equal. What you can do is to change it to -B + A, if there is some reason to do so. You can informally think of this as dragging the negative sign along with the B when you move it; this is because A - B means the same thing are A + -B. Since addition (but not subtraction) is commutative, you can now change the order and make it -B + A.

Once you are familiar with what you can do, and what you can't do, it becomes easy to do the right things. The way to become familiar with these things is to practice a lot, and always ask yourself what change you just made, and whether it is justified by any facts that you know. If you can't explain why it can be done, don't do it.
 
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