# Calculate "(2a+8b)(4a-2b)"

15 replies to this topic

### #1-Alex-

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Posted 22 November 2012 - 08:24

Hi all,

I'm interested in becoming a pilot and was just reading some information online at a pilot programme. They give a sample entrance exam, and on it is this question (question 9):

Quote

Calculate:
(2a+8b)(4a-2b)

Is this supposed to read (2a+8b)+(4a-2b)=0, or am I missing something?

I'm not looking for the answer to this question, rather, how to calculate it; for my own knowledge.

Thanks!

### #2XerXis

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Posted 22 November 2012 - 08:32

(2a+8b)(4a-2b)=8a^2 - 4ab + 32ab - 16b^2=8a^2 + 28ab - 16b^2

probably just to see if you know your distribution rule

### #3abecedarian paradoxious

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Posted 22 November 2012 - 08:37

It would be read as "(2a+9b)*(4a-2b)" with no absolute answer: you would be solving for one variable in terms of the other: as in "a=something b" or "b=something a"

Edited by abecedarian paradoxious, 22 November 2012 - 08:42.

### #4+Neo003

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Posted 22 November 2012 - 08:37

&#045;Alex&#045;, on 22 November 2012 - 08:24, said:

Hi all,

I'm interested in becoming a pilot and was just reading some information online at a pilot programme. They give a sample entrance exam, and on it is this question (question 9):

Is this supposed to read (2a+8b)+(4a-2b)=0, or am I missing something?

I'm not looking for the answer to this question, rather, how to calculate it; for my own knowledge.

Thanks!

6a-6b
a-b

### #5Xahid

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Posted 22 November 2012 - 08:37

-Alex-, on 22 November 2012 - 08:24, said:

Is this supposed to read (2a+8b)+(4a-2b)=0, or am I missing something?

No, I believe it means (2a+8b)x(4a-2b)=0
Since brackets () mean multiplication so its not necessary to use 'x' specifically.

### #6OP-Alex-

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Posted 22 November 2012 - 08:43

I was going through the exam slowly, working out each answer in my head... I just decided to scroll through it fully, and saw that the answers are at the end. It says:

Quote

8a2+28ab-16b2

So XerXis was correct.

### #7+ozzy76

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Posted 22 November 2012 - 08:43

FOIL (First Outside Inside Last)

this takes me back

### #8+GreenMartian

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Posted 22 November 2012 - 08:44

Xahid, on 22 November 2012 - 08:37, said:

.. its not necessary to use 'x' specifically.
Exactly. The same way that everyone knows 2b means 2xb.
a(2b) = a times 2 times b.

### #9abecedarian paradoxious

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Posted 22 November 2012 - 08:55

-Alex-, on 22 November 2012 - 08:43, said:

I was going through the exam slowly, working out each answer in my head... I just decided to scroll through it fully, and saw that the answers are at the end. It says:

So XerXis was correct.

Every time I had ?'s like that in school we had to solve for each variable.... Oh well, good thing I'm not a pilot.

### #10XerXis

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Posted 22 November 2012 - 08:59

abecedarian paradoxious, on 22 November 2012 - 08:55, said:

Every time I had ?'s like that in school we had to solve for each variable.... Oh well, good thing I'm not a pilot.

That would be true if there was an equation, however it wasn't there. Also it's impossible to solve an equation with two variables. It wouldn't make any sense

### #11Axel

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Posted 22 November 2012 - 09:28

I'm a pilot and as far as I can help with that question would be to interpret it as (2a+8b)*(4a-2b). Any further than that is fairly lost on me! I don't remember my entrance exams being like that!

### #12mercenary

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Posted 22 November 2012 - 10:08

http://www.wolframal...b%29%284a-2b%29

That should help you understand

### #13nik louch

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Posted 22 November 2012 - 10:12

Why is this ridiculously simple thing being posted again?

It's been solved and debated to death less than 3 months ago.

### #14OP-Alex-

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Posted 22 November 2012 - 13:06

nik louch, on 22 November 2012 - 10:12, said:

Why is this ridiculously simple thing being posted again?

It's been solved and debated to death less than 3 months ago.

1) I didn't see the thread 3 months ago
2) Good luck searching the forums for a random equation
3) What's the big deal?

### #15+Xinok

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Posted 22 November 2012 - 16:47