Expanding/Factoring Q.

Printable View

January 17th 2006, 02:03 PM
Constance

Expanding/Factoring Q.

Alright. I do not know the answer to this question, and need to know.

If you were to expand (x-7)(x+7) there would be no x-term. Explain this!

Please someone help!
January 17th 2006, 02:33 PM
Jameson

Quote:

Originally Posted by Constance

Alright. I do not know the answer to this question, and need to know.

If you were to expand (x-7)(x+7) there would be no x-term. Explain this!

Please someone help!

There is an x term! :cool: FOIL. First, Outer, Inner Last.

This gives us $x*x+7*x-7*x-49=x^2-49$ . This is a special expansion called the difference of squares.
January 17th 2006, 02:35 PM
dud

I suppose you mean to expand this, there would be no x in the first order.

$<br /> (x-7)(x+7) \Leftrightarrow (x+7)(x-7) \Leftrightarrow (a+b)(a-b)<br />$
This last one there is a theorem/rule which I don't know what is called in English.
$<br /> (a+b)(a-b) \Leftrightarrow a^2 - b^2<br />$
Which gives.
$<br /> (x-7)(x+7) \Leftrightarrow x^2 - 7^2 \Leftrightarrow x^2 - 49<br />$
There you have it, no x of first degree! :)

All times are GMT -8. The time now is 01:05 PM.

Copyright © 2005-2013 Math Help Forum. All rights reserved.

Copyright © 2005-2013 Math Help Forum. All rights reserved.