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!
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!