# Expanding/Factoring Q.

• Jan 17th 2006, 03:03 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!

• Jan 17th 2006, 03:33 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!

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.
• Jan 17th 2006, 03:35 PM
dud
I suppose you mean to expand this, there would be no x in the first order.

$
(x-7)(x+7) \Leftrightarrow (x+7)(x-7) \Leftrightarrow (a+b)(a-b)
$

This last one there is a theorem/rule which I don't know what is called in English.
$
(a+b)(a-b) \Leftrightarrow a^2 - b^2
$

Which gives.
$
(x-7)(x+7) \Leftrightarrow x^2 - 7^2 \Leftrightarrow x^2 - 49
$

There you have it, no x of first degree! :)