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This must be a stupid question with an obvious answer that is hidden from me. No one I have found even mentions a conflict.

lets say I want to find the square $(2 + 3)^2 = 2^2 + 2\times2\times3 + 3^2$

Why do I not simplify first? Parenthesis first?

Edit

To clarify for future views. The main confusion was why is the above problem always (in my limited experience) solved one way in situation A and another way in situation B...Z without ever being noted of why the method used is preferred for the given situation (again in my limited experience).

Thanks for the clarification. I think I get it now. In some cases it is arbitrary because the order is divorced from the output.

In some cases you have an variable that you cannot add or subtract to simplify so you must use the distributive property.

I suppose also you may not simplify to show the trinomial pattern of the output produced by the distributive property (in a pedagogical situation).

mLovelady85
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    sorry I don't understand your question - could you clarify exactly what you're asking – mrnovice Apr 17 '17 at 23:03
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    What do you mean? You can simplify first if you want to. $(2+3)^2 = 5^2 = 25$. Or you don't have to. $(2+3)^2 = 2^2 + 223 + 3^2 = 4+ 12 + 9 = 25$. You can do whatever you want. Who the heck is telling you otherwise? – fleablood Apr 17 '17 at 23:11
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    Are you familiar with the 'FOIL' method, silly though it is? – The Count Apr 17 '17 at 23:13
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    Only question I wanna raise is Why so many upvotes? – The Dead Legend Apr 17 '17 at 23:16
  • This depends on the algebra. In some algebras e.g. "freshman's dream" you can just lift down the $.^2$ on the individual terms inside the parentheses, but not for the real or complex numbers. – mathreadler Apr 18 '17 at 08:02
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    You can look at $(2+3)^2 = 5^2 = (2+3)(2+3) = 2^2 + 2\times2\times3 + 3^2 = 25$. In this case I think when you can simplify what's in parentheses, in this case a sum of integers, it is quite clear that it makes sense to add 2+3 5, then square $5$ to get $5^2 = 25$. The expansion at the end is useful when variables are involved. E.g. $(x+3)^2.$ There's nothing within the parentheses that we can evaluate/simplify, since $x$ is variable. This is when we might need to expand $$(x+3)^2 = (x+3)(x+3) = x^2 + 3x + 3x + 3^2 = x^2 + 6x + 9$$ – amWhy Apr 18 '17 at 13:40
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    @UddeshyaSingh There are gangs here. They pad their reputation by up-voting each other. By amassing points, they can offer plethora bounties on questions and get others to do their research for them, and in turn, add the others' names to the papers in order to pad their vitae. It's now possible to parley MS points into tenure, if you just have a few like-minded friends. – B. Goddard Apr 18 '17 at 14:28
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    @UddeshyaSingh I wonder why so few upvotes (8 at this moment cancelled by 5 downvotes) for a question interesting enough to have attracted several answers, one with 31 upvotes. And why haven't the downvoters commented with reasons? – Ethan Bolker Apr 18 '17 at 19:19
  • @EthanBolker I find it more odd that the answers have gained so many up votes. – mrnovice Apr 18 '17 at 20:16
  • @Ethan , Idk about answers. As for the question, I find it silly DMAS problem. As for the answer. M not really impressed because OP's query is solved within first few lines. The rest seem to be out of place. – The Dead Legend Apr 18 '17 at 22:44

5 Answers5

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It seems you confuse the precedence of operations, which describe how to interpret an expression (and thus which transformations are actually valid), with the order in which you apply valid transformations to an expression.

For your specific formula, $(2+3)^2$, the precedence rules say that the parentheses override the implicit precedences that otherwise would have been in effect. That is, the precedence rules say that the expression means that the sum $2+3$ is taken to the second power (i.e. squared), as opposed to $2+3^2$ which would mean that $2$ and the square of $3$ would be added.

However the precedence rules do not say that you need to calculate the term in parentheses first (although in this special case, it's the most economic choice). You have two valid transformations that you can apply (well, actually there are many more, but the others would only make the problem more complicated):

  • You can do the sum first: It's a subexpression $2+3$ which you can replace by $5$, because $2+3=5$. Doing so, you'll get $(2+3)^2 = 5^2$. Then you can go on by using the fact that $5^2=25$ to finally arrive at the result $(2+3)^2 = 25$.

  • You can do the square first: Since it is the square of a sum, you can use the binomial formula to obtain $(2+3)^2 = 2^2 + 2\cdot 2\cdot 3 + 3^2$. Again, you can continue to find $2^2+2\cdot 2\cdot 3 + 3^2 = 4 + 12 + 9 = 25$. The result is the same again, as it of course has to be.

However what you cannot do is to first replace $3^2=9$ because the original expression does not contain the subexpression $3^2$. Of course after applying the binomial formula, a subexpression of that form appears, which you then can replace by $9$.

Similarly, in $2+4\cdot 5$, the precedence rules say that this is the same as $2+(4\cdot 5)$, and therefore there's a subexpression $4\cdot 5$ which you can replace with $20$, but there is no subexpression $2+4$ which could be replaced by $6$. However you could, in principle, first write $2=2\cdot 1$ and $4=2\cdot 2$, and then use the distributive law to get $2\cdot 1 + 2\cdot 2\cdot 5 = 2\cdot (1+2\cdot 5) = 2\cdot 11 = 22$, and again, you get, necessarily, the same result as when simplifying in precedence order, $2+4\cdot 5 = 2 + 20 = 22$. However if you got the precedence wrong and started by replacing the non-subexpression $2+4$ with $6$, you'd arrive at $6\cdot 5 = 30\ne 22$, which is the wrong result.

Note that while in the cases above, applying the operations in precedence order is the most efficient, this is not always the case. Indeed, you can even have cases where first going to a more complicated expression simplifies the complete calculation. As an example, consider the expression $999^2$. While you can directly calculate it as $999\cdot 999$ using the standard multiplication algorithm, it is much easier to first split it into a difference and apply the binomial formula: $$999^2 = (1000-1)^2 = 1000^2 - 2\cdot 1000\cdot 1 + 1^2 = 1\,000\,000 - 2000 + 1 = 998\,001$$ This is possible because each single step is a valid operation; note how the replacement of $999$ by $1000-1$ requires the addition of parentheses due to the precedence rules (we replace the expression $999$ by the expression $1000-1$, but without the parentheses we'd get $1000-1^2$ which doesn't even contain the subexpression $1000-1$). Also note that the first step would be futile if we then proceeded by doing the addition in parentheses first, as we'd just arrive at the starting point again.

celtschk
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    It doesn't help that precedence is almost always taught as "order of operations", in terms of what operations to do first, and then you're lucky if anyone ever explicitly tells you later that it's really just an argument grouping convention. – user2357112 Apr 18 '17 at 07:39
  • Nice and verbose answer. – Gewure Apr 18 '17 at 10:39
  • @user2357112 Your comment is the point of my answer to this question. – Ethan Bolker Apr 18 '17 at 13:56
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    One valid transformation you might add: $(2+3)^2=(2+3)\times(2+3)$. (This essentially leads to the binomial formula, but it is certainly a useful rule to try in a computation.) – Joonas Ilmavirta Apr 18 '17 at 18:08
  • in C and C++ programming, operator precedence is specified but order of operations are undefined and that caused a lot of confusions – phuclv Apr 19 '17 at 01:28
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As a general rule, you can do anything you want in any order as long as each step is a valid operation.

Replacing $2+3$ by $5$ is certainly fine.

Expanding an expression $(a+b)^2$ as $a^2+2ab+b^2$ is also correct.

Do whichever you want.

Zev Chonoles
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    Indeed, you sometimes even want to split single numbers into sums. For example, $999^2$ is much easier to calculate if you use $999=1000-1$ and use the binomial formula for $(1000-1)^2$. – celtschk Apr 17 '17 at 23:27
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    @amWhy: Of course replacing $5+6\cdot 7$ by $11\cdot 7$ is not a valid operation. – celtschk Apr 17 '17 at 23:30
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    @amWhy: At least to me, this particular student seems more confused due to the expectation that there is a magic recipe order, and would instead benefit from a more conceptual understanding. – Zev Chonoles Apr 17 '17 at 23:33
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    @amWhy: Precedence of operations matters. Precedence rules say that $5+6\cdot 7$ means $5+(6\cdot 7)$, and especially there is not any subexpression $5+6$ that could be replaced by $11$. Note that precedence and order of operations are unrelated. For example, explicit parentheses override implicit precedence rules, but there's no requirement that in the expression $(2+3)\cdot 5\cdot 7$ you start by calculating $2+3$; you could just as well start by first calculating $5\cdot 7$, or by using the distributive law to transform it into $2\cdot 5\cdot 7+3\cdot 5\cdot 7$. – celtschk Apr 17 '17 at 23:42
  • @amWhy: OK, I've now created an answer out of my comments. – celtschk Apr 18 '17 at 00:29
  • @amWhy I see this answer as clarifying that the way one is taught to do things is not always the way they have to be done. As a tutor the number one thing I bring to my students is the concept that the exact techniques shown by their teachers are not the only valid techniques. To me, this question is asking why we don't have the option of using another valid approach, and Zev is clarifying that we can in fact use any valid approach. Too many teachers imply or even explicitly state that only their method is acceptable. To me Zev is granting ownership of math to the asker. – Todd Wilcox Apr 18 '17 at 06:13
  • @Todd, I agree with everything you say! What you say, but only , as expressed in celtschk answer and comments, and given in your perspective above, I never disagreed with any of it. My issue with Zev's answer only about its flippant tone, not explaining anything he claims. celtschk is the one that actually elucidated his own thoughts, Zev never bothered to elaborate. So be careful when addressing someone you assume disagrees with you. I don't. When you ass-u-me what what you have no clue about, you make an a*s out of u and me. – amWhy Apr 18 '17 at 21:36
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This is a very good question.

There are some standard conventions for order of operations - for example, when you see $2 + 3 \times 4$ you do the multiplication first. This convention is not really a part of arithmetic, it's just an agreed upon way to write the calculation that avoids having to write all the pairs of parentheses.

If we didn't have the convention you would have to write exactly what you meant: $2 + (3 \times 4)$. You can always do that anyway - you don't have to rely on the convention. Use as many "extra" parentheses as you wish. (Sometimes that's a good idea when you're writing a computer program and aren't sure the computer is following the same conventions you are.)

Anything surrounded by a pair of parentheses is evaluated as an independent calculation. Think of the parentheses as a circle surrounding the expression, just cut off so it doesn't interfere with the lines above and below it on the page.

In this particular question you are probably confused by (the horrible ugly) PEMDAS, which puts the P before the E. So in $(2+3)^2$ you evaluate the sum in the parentheses first. And you need those parentheses, because PEMDAS says $2+3^2$ is $2+3\times3 = 11$.

But when you rewrite $(2+3)^2$ as $(2+3) \times (2+3)$ there is no longer an exponent. You already know the answer is $5 \times 5$. But there's another way to get there that involves real mathematics, not just a convention. That's the distributive law that tells you how multiplication and addition combine: $$ a \times (b + c) = (a \times b) + (a \times c). $$ (I wrote that out with all the parentheses - didn't rely on PEMDAS.) Using the distributive law twice (and the commutative law too): $$ \begin{align} (2+3) \times (2+3) & = (2+3) \times 2 + (2+3) \times 3 \\ & = 2 \times 2 + 3 \times 2 + 2 \times 3 + 3 \times 3 \\ & = \text{and so on} \end{align} $$

(Here I relied on the "multiplication before addition" convention and left out the extra parentheses).

You may recognize that calculation as (the horrible ugly abbreviation) FOIL - which is just a mnemonic for the distributive law twice.

Ethan Bolker
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If I'm understanding your question correctly, then note:

$(2+3)^2 = (5)^2=25 $

$(2+3)^2 = 2^2+2\cdot 2\cdot 3+3^2=4+12+9 = 25$

You can expand the brackets, or simplify the contents of the brackets and then square it and obtain the same answer via both methods, if that's what you mean.

mrnovice
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By the rules of precedence, we compute $$ (2+3)^2 = 5^2 = 25.$$ By the same rules, we compute $$2^2+2\cdot 2\cdot 3+3^2=4+12+9=25. $$ The very fact that both computations produce the same result justifies us to write down the interesting fact $$ (2+3)^2=2^2+2\cdot 2\cdot 3+3^2.$$

Hagen von Eitzen
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