Guest Sep 4, 2017

Hello! That's quite an interesting expression you got there of \((a-7)^2-2(a-7)(a+7)+a+7\). Let's simplify it the we can.


\((a-7)^2-2(a-7)(a+7)+a+7\) First, let's deal with \((a-7)^2\) by knowing the expansion of a binomial squared. In general, it is \((x-y)^2=x^2-2xy+y^2\).
\(a^2-14a+49-2(a-7)(a+7)+a+7\) Now, let's expand \((a-7)(a+7)\) by using the following rule again of \((x+y)(x-y)=x^2-y^2\).
\(-2(a-7)(a+7)=-2(a^2-49)\) Distribute the -2 to both terms in the parentheses.
\(a^2-14a+49-2a^2+98+a+7\) Let's rearrange the equation such that all the terms with the same degree are adjacent.
\(-2a^2+a^2-14a+a+98+49+7\) Now, simplify.


Therefore, \((a-7)^2-2(a-7)(a+7)+a+7=-a^2-13a+154\).

TheXSquaredFactor  Sep 4, 2017

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