# Where am I going wrong when removing brackets from (s+1)(s+5)(s-3)

One of my text book exercises is to remove the brackets from this expression:

$(s+1)(s+5)(s-3)$

The I've tried a number of times and the result I keep getting is:

$s^3 -3s^2 -13s -15$

However, my textbook says the answer is:

$s^2 -3s^2 -13s -15$

I keep getting the same answer, but I'm reluctant to think the text book is wrong.

Here's how I come to that result:

$(s+1)(s+5)(s-3)$

$= s((s+1)(s+5)) - 3((s+1)(s+5))$

$= s(s(s+1) + 5(s+1)) - 3(s(s+1) + 5(s+1))$

$= s(s^2 + s + 5s + 5) - 3(s^2 + s + 5s + 5)$

$= s^3 + s^2 + 5s^2 + 5s - 3s^2 -3s -15s -15$

Then re-arrange that to:

$= s^3 +6s^2 - 3s^2 + 5s - 3s -15s -15$

$s^3 +3s^2 -13s -15$

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thank you! that's good to hear, think I'm starting to get a grasp on this daunting subject! –  bot_bot Mar 16 '12 at 13:26
(Just a small comment on language - you remove parentheses from an expression (it is a thing) and not a statement (which is a verb). :) –  AD. Mar 16 '12 at 13:30
thanks, updated. –  bot_bot Mar 16 '12 at 13:33
Incidentally, you have mismatched parentheses in the second line of your argument. –  David Mitra Mar 16 '12 at 13:38
fixed, thank you. –  bot_bot Mar 16 '12 at 13:43

Neither of these results is correct; the correct result is $s^3+3s^2-13s-15$. The error in your calculation is where you replace $6s^2-3s^2$ by $-3s^2$.

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Isn't that the result I have? sorry I see now –  bot_bot Mar 16 '12 at 13:31
@mal Almost, as joriki say - you have $1\cdot s^2 + s\cdot 5\cdot s + s^2\cdot(-3)= 3s^2$... –  AD. Mar 16 '12 at 13:33
@mal: Perhaps there's a delay in displaying your edits? In the version of the page that I'm being shown, you have $-3s^2$ instead of $+3s^2$. –  joriki Mar 16 '12 at 13:34
Thank you both. –  bot_bot Mar 16 '12 at 13:35

The error is, in fact, in your textbook: The product $(s+1)(s+5)(s-3)$ will, after expansion, have a term $s^3$ in it (as in your result). Since it is missing in the textbook, the answer given there is wrong.

On the other hand, your derivation looks fine to me, so I would say that you are right and the textbook has a typo in it.

Edit: In fact, there is an error in your calculation; see the other answer. Still, the textbook answer is wrong.

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Thanks for your help –  bot_bot Mar 16 '12 at 14:17

$1$. The textbook answer obviously has a typo. It is highly implausible that anyone would intend $s^2 -3s^2 -13s -15$ as a final answer.

$2.$ While you were doing the calculation, you had the intermediate expression $s(s^2 + s + 5s + 5) - 3(s^2 + s + 5s + 5)$, and then multiplied without changing the $s^2+s+5s+5$ automatically to $s^2+6s+5$. The more terms you have, the more work you need to do. Each step up in complexity introduces new opportunities for error.

$3.$ I would have as a matter of strategy preferred to find $(s+5)(s-3)$ first, since multiplying by $s+1$ is more pleasant than multiplying by $s-3$.

$4.$ Any calculation involves the possibility of computational error, or error in writing down the result of the computation. It is not a bad idea to make a final plausibility scan, like evaluating the original expression and the expanded one at some nice number, such as $x=1$.

$5.$ You have undoubtedly acquired visual shortcuts for expanding things like $(x+a)(x+b)$, and even $(px+a)(qx+b)$. Note that $(x+a)(x+b)(x+c)=x^3+(a+b+c)x^2+(ab+bc+ca)x+abc$.

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