Multiplying out brackets efficiently

Probability

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In a activity I was asked to multiply out the brackets. It seems I did not go far enough from what I thought was correct.

\(\displaystyle {2}(\frac{1}{2}{A^2}+\frac{3}{2})=\)

I thought;

\(\displaystyle {2}(\frac{1}{2}{A^2}+\frac{3}{2})={2}\frac{1}{2}{A^2}+{2}\frac{3}{2}\)

Which I thought was multiplying out the brackets. The answer given is \(\displaystyle ({A^2}+{3})\)

My idea now seem to be based around how am I going to cancel down the fractions to end up with \(\displaystyle {+}{3}?\)

So the first thought I had was to convert the mixed fractions to proper fractions (top heavy);

\(\displaystyle \frac{5}{2}{A^2}+\frac{7}{2}=\)

From this point on-wards using trial and error I can end up with \(\displaystyle {A^2}-\frac{3}{5}\) which is obviously incorrect.

If I were to cancel down \(\displaystyle \frac{5}{2}{A^2}={A^2}\) but I can't do the same to \(\displaystyle \frac{7}{2}\) so from this point I'm stuck!!
 

lev888

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You are confusing 2 multiplied by 3/2 with a mixed number 2 and 3/2.
You wrote yourself that you multiplied, right? 2 times 3/2 = 2*(3/2). This is NOT 2 and 3/2.
2*(3/2) can be simplified: (2*3)/2 = (2/2)*3 = 1*3 = 3.
 

Probability

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lev888 I did not see that at all, thank you so much for that. I must now ask a question about this to gain a deeper understanding of the procedure.

So I multiply out the brackets and I end up with \(\displaystyle {2}\frac{1}{2}{A^2}+{2}\frac{3}{2}\)

I've look at those as mixed fractions and never thought of that arrangement for the product rule multiplying the numerators before the denominators!

So to perform this operation is mathematically correct then;

\(\displaystyle {2}\frac{1}{2}{A^2}={2}\times{1}=\frac{2}{2}={A^2}\:and\:{2}\times{3}=\frac{6}{2}={3}\:hence\:{A^2}+{3}\)

I did not see that method at all.
 

Subhotosh Khan

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lev888 I did not see that at all, thank you so much for that. I must now ask a question about this to gain a deeper understanding of the procedure.

So I multiply out the brackets and I end up with \(\displaystyle {2}\frac{1}{2}{A^2}+{2}\frac{3}{2}\)

I've look at those as mixed fractions and never thought of that arrangement for the product rule multiplying the numerators before the denominators!

So to perform this operation is mathematically correct then;

\(\displaystyle {2}\frac{1}{2}{A^2}={2}\times{1}=\frac{2}{2}={A^2}\:and\:{2}\times{3}=\frac{6}{2}={3}\:hence\:{A^2}+{3}\)

I did not see that method at all.
Your answer is correct but your use of "=" sign is WRONG. It should be:

\(\displaystyle {2} * \frac{1}{2}*{A^2}={A^2}\:and\:{2}\times\frac{3}{2}=\frac{6}{2}={3}\)

\(\displaystyle \:hence \ it\ is\ reduced\ to \)

\(\displaystyle \:{A^2}+{3}\)
 

Probability

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Your answer is correct but your use of "=" sign is WRONG. It should be:

\(\displaystyle {2} * \frac{1}{2}*{A^2}={A^2}\:and\:{2}\times\frac{3}{2}=\frac{6}{2}={3}\)

\(\displaystyle \:hence \ it\ is\ reduced\ to \)

\(\displaystyle \:{A^2}+{3}\)
Every day is a new learning curve for me. Thank you.
 

Subhotosh Khan

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Every day is a new learning curve for me. Thank you.
It is not a new "curve" - it is a new height!

You climb a bit - and you can see further and "clearer" !
 

Probability

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Otis

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\(\displaystyle {2}(\frac{1}{2}{A^2}+\frac{3}{2})\)
I think of all quantities in terms of fractions, when multiplying.
\[\left( \frac{2}{1} \right) \left( \frac{A^2}{2} + \frac{3}{2} \right)\]
\[\left( \frac{2}{1} \right) \left( \frac{A^2}{2} \right) + \left( \frac{2}{1} \right) \left( \frac{3}{2} \right)\]
Those 2s are common factors, so they cancel, leaving A2+3.
\[\left( \frac{\cancel{2}}{1} \right) \left( \frac{A^2}{\cancel{2}} \right) + \left( \frac{\cancel{2}}{1} \right) \left( \frac{3}{\cancel{2}} \right)\]

😎
 

topsquark

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\(\displaystyle {2}\frac{1}{2}{A^2}={2}\times{1}=\frac{2}{2}={A^2}\:and\:{2}\times{3}=\frac{6}{2}={3}\:hence\:{A^2}+{3}\)
I've notice this mistake from you before. I have found many students that will try to solve a problem using this "notation"
\(\displaystyle 4 \times (3 + 2) / 7 = 4 \times 5 = 20/7\)

With a little guesswork an instructor can figure out what you are trying to say. But it is not at all clear. When I was teaching I took points off if I had to guess what my student was trying to write. (I warned them ahead of time and they did it for the first exam anyway. They learned fast after that.)

My advice: Do these problems single step by single step using a line for each step.

-Dan
 

Probability

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I've notice this mistake from you before. I have found many students that will try to solve a problem using this "notation"
\(\displaystyle 4 \times (3 + 2) / 7 = 4 \times 5 = 20/7\)

With a little guesswork an instructor can figure out what you are trying to say. But it is not at all clear. When I was teaching I took points off if I had to guess what my student was trying to write. (I warned them ahead of time and they did it for the first exam anyway. They learned fast after that.)

My advice: Do these problems single step by single step using a line for each step.

-Dan
Dan it's not always that easy when you are learning the subject from first principles. So much thinking going on trying to move from one step to the next, and then I forget about other parts of the math like when an = is actually = to the next part of the math I'm trying to solve. So much going on sometimes that innocent mistakes do occur as well.
 

topsquark

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Dan it's not always that easy when you are learning the subject from first principles. So much thinking going on trying to move from one step to the next, and then I forget about other parts of the math like when an = is actually = to the next part of the math I'm trying to solve. So much going on sometimes that innocent mistakes do occur as well.
I understand it's difficult. But writing a clear solution is also something you need to learn. I struggle with this in my own studies. But one of the practical principles of Mathematics (and all the "Sciences") is to be able to convey your meaning clearly.

-Dan
 

Probability

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I understand it's difficult. But writing a clear solution is also something you need to learn. I struggle with this in my own studies. But one of the practical principles of Mathematics (and all the "Sciences") is to be able to convey your meaning clearly.

-Dan
I'll do my best thanks
 

Probability

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Just a second opinion on this expression please. I'm ask to multiply out the brackets and simplify where possible.

I have two examples which I have carried out using the same mathematical techniques, yet according to the solutions one is correct and the other I'm to understand I'm wrong, but I can't see where I've gone wrong!

This one is correct;

\(\displaystyle {x}-{y}({x}+{2y})\)

I concluded;

\(\displaystyle {x^2}+{xy}-{2y^2}\)

This expression I carried out using exactly the same mathematical techniques but the authors answer \(\displaystyle {f}+{fg}+{gh}\) I can't see to achieve it.

The expression is;

\(\displaystyle {f}+{g}({f}+{h})={f^2}+{fh}+{fg}+{gh}\)

Where am I going wrong!
 

lev888

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Just a second opinion on this expression please. I'm ask to multiply out the brackets and simplify where possible.

I have two examples which I have carried out using the same mathematical techniques, yet according to the solutions one is correct and the other I'm to understand I'm wrong, but I can't see where I've gone wrong!

This one is correct;

\(\displaystyle {x}-{y}({x}+{2y})\)

I concluded;

\(\displaystyle {x^2}+{xy}-{2y^2}\)

This expression I carried out using exactly the same mathematical techniques but the authors answer \(\displaystyle {f}+{fg}+{gh}\) I can't see to achieve it.

The expression is;

\(\displaystyle {f}+{g}({f}+{h})={f^2}+{fh}+{fg}+{gh}\)

Where am I going wrong!
1. The first x is not involved in any multiplication, leave it alone. The 2 terms in parentheses are multiplied by (-y). Note, that in the result the signs are different, why?
2. As above, the first f is not involved in any multiplication, leave it alone.
 

Probability

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1. The first x is not involved in any multiplication, leave it alone. The 2 terms in parentheses are multiplied by (-y). Note, that in the result the signs are different, why?
2. As above, the first f is not involved in any multiplication, leave it alone.
Thanks. In the first expression the first\(\displaystyle {x}\) must be involved otherwise how does the solution end up with \(\displaystyle {x^2}\) and the author seems to agree with the method based on the answer!

In the second example it looks like the author has not properly worked out the brackets fully, hence not given a complete solution. I've also just seen this with three other examples. I'm sure part of a term cannot just be ignored as if it were not present!!
 

lev888

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Thanks. In the first expression the first\(\displaystyle {x}\) must be involved otherwise how does the solution end up with \(\displaystyle {x^2}\) and the author seems to agree with the method based on the answer!

In the second example it looks like the author has not properly worked out the brackets fully, hence not given a complete solution. I've also just seen this with three other examples. I'm sure part of a term cannot just be ignored as if it were not present!!
1. Well, I can't guess what the problem should be in order to have x2 in the answer. As written, your solution is not correct. Do you see that x is sitting there, all alone, watching as (-y) multiplies the terms in parentheses? It would object vociferously if it found out about your intentions to multiply it by another x, that magically appeared out of nowhere. And, as I wrote, there is a problem with signs.
2. The author's answer is correct. First we have f, then g multiplied by the terms in parentheses.
 

Dr.Peterson

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Just a second opinion on this expression please. I'm ask to multiply out the brackets and simplify where possible.

I have two examples which I have carried out using the same mathematical techniques, yet according to the solutions one is correct and the other I'm to understand I'm wrong, but I can't see where I've gone wrong!

This one is correct;

\(\displaystyle {x}-{y}({x}+{2y})\)

I concluded;

\(\displaystyle {x^2}+{xy}-{2y^2}\)

This expression I carried out using exactly the same mathematical techniques but the authors answer \(\displaystyle {f}+{fg}+{gh}\) I can't see to achieve it.

The expression is;

\(\displaystyle {f}+{g}({f}+{h})={f^2}+{fh}+{fg}+{gh}\)

Where am I going wrong!
Can you show images of the two problems and the author's solutions, so we can be sure we are interpreting them correctly?

If they are as you show, then in your work you are forgetting the order of operations. For example, f+g(f+h) means that you have f alone, and then you are multiplying g times (f + h), which can be distributed to get f + gf + gh. You are taking it as if it were (f+g)(g+h), which is the product of two factors, (f+g) and (g+h). These are entirely different.
 

Jomo

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I've notice this mistake from you before. I have found many students that will try to solve a problem using this "notation"
\(\displaystyle 4 \times (3 + 2) / 7 = 4 \times 5 = 20/7\)

With a little guesswork an instructor can figure out what you are trying to say. But it is not at all clear. When I was teaching I took points off if I had to guess what my student was trying to write. (I warned them ahead of time and they did it for the first exam anyway. They learned fast after that.)

My advice: Do these problems single step by single step using a line for each step.

-Dan
I wanted my students to use equal signs (and valid ones). For the 1st test I took off 1 point for every missing equals sign (but not in my grade book!) and on top of the paper along with their grade (lowered because of missing signs) I noted how many points they lost for missing equal signs. Some students lost as many as 15 points. I told them this time those mistakes are free, but next time I will takes those point off for real. They all started using equal signs.

As far as what you did, I do it all the time. Equals signs must be valid! The students who solve 7+7+3+5 +3 by writing 7+7+3+5 +3 = 14 + 3 = 17 + 5 = 22 + 3 = 25 just hate me as all I grade are equal signs and only that last one is valid.
 

Probability

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1. Well, I can't guess what the problem should be in order to have x2 in the answer. As written, your solution is not correct. Do you see that x is sitting there, all alone, watching as (-y) multiplies the terms in parentheses? It would object vociferously if it found out about your intentions to multiply it by another x, that magically appeared out of nowhere. And, as I wrote, there is a problem with signs.
2. The author's answer is correct. First we have f, then g multiplied by the terms in parentheses.
lev888, see my reply to Dr.Peterson's post number 17. Thank you for your explanation.
 

Probability

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Can you show images of the two problems and the author's solutions, so we can be sure we are interpreting them correctly?

If they are as you show, then in your work you are forgetting the order of operations. For example, f+g(f+h) means that you have f alone, and then you are multiplying g times (f + h), which can be distributed to get f + gf + gh. You are taking it as if it were (f+g)(g+h), which is the product of two factors, (f+g) and (g+h). These are entirely different.
Looking at the written expression in the activity, which was the first question, I was asked to multiply out the brackets and simplify where possible.

This is the expression;

\(\displaystyle {f}+{g}({f}+{h})\)

I can see now my misunderstanding. The \(\displaystyle {f}\)is a term on its own and although I knew this the \(\displaystyle +\) belongs to the \(\displaystyle {g}\)Looking at the activity I read it as \(\displaystyle {f}\) to be multiplied by \(\displaystyle ({f}+{h})\). The example shown to me before the activity did not have exactly the same type of expression, but showed an expression with two terms, e.g. \(\displaystyle {x}({y}+{1})+{2y}({y}+{3})\) when this is multiplied out the solution will show;

\(\displaystyle {xy}+{x}+{2y^2}+{6y}\)

In my expression above \(\displaystyle {f}+{g}\) I can see now that \(\displaystyle {f}\) is not a term in \(\displaystyle {g}({f}+{h})\)

\(\displaystyle {f}\)is a term on its own and I did not recognise that!
 
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