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!

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- Jan 17th 2006, 02:03 PMConstanceExpanding/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! - Jan 17th 2006, 02:33 PMJamesonQuote:

Originally Posted by**Constance**

This gives us $\displaystyle x*x+7*x-7*x-49=x^2-49$. This is a special expansion called the difference of squares. - Jan 17th 2006, 02:35 PMdud
I suppose you mean to expand this, there would be no x in the first order.

$\displaystyle

(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.

$\displaystyle

(a+b)(a-b) \Leftrightarrow a^2 - b^2

$

Which gives.

$\displaystyle

(x-7)(x+7) \Leftrightarrow x^2 - 7^2 \Leftrightarrow x^2 - 49

$

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