Factor this expression as a difference of squares. a^2 + 4a +4 -b^2
(a+2)^2-b^2 (a+2-b)(a+2+b)
brackett come to chat and i will show you the fractions
(a+2)^2-b^2 =(a+2+b)(a+2-b) use the identity a^2-b^2=(a+b)(a-b)
think there is a mistake here \[a^2+4a+4=(a+2)^2\] giving \[(a+2)^2-b^2\] get \[(a+2+b)(a+2-b)\] yes?
i find this to be correct sat . i can't find any mistake :)
i'm in chat now sorry it -took a mniute
\[a\div b\]
\[\sum \frac{1}{2^n}\]
\[\sum_{k=1}^\infty (\frac{3}{4})^k\]
\[\dbinom{n}{k}=\frac{n!}{k!(n-k)!}\]
how d'u remember so much stuff? :O too good !
here to learn it basically
i typed what i wrote in chat so you can see it
\[\int_1^{\infty}\frac{1}{x^{\frac{2}{3}}}dx\]
i'm copy pasting everything in my notepad...will learn them soon :)
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