P^4- 81= ( p^2-9)(p^2+9) = (p+3)(p-3)(p^2+9); Why can't you further factor (P^2+9)? By factoring by grouping P^2+0P+9, I get (P+3)(P-3), So my complete answer is (P+3)(P-3)(P+3)(P-3).

Question

phi · Answer

notice that (p+3)(p-3) = p^2 - 9  which is different from p^2 + 9
in other words,  p^2 + 9 does not factor

anonymous · Answer

I see it now. I couldn't for the life of me figure it out. Thank you so much!!

anonymous · Answer

if we consider p^2+9 as quadratic polynomial
then discriminent=0^2-4*1*9=-36 which is negative so can not be factored for real numbers.

anonymous · Answer

I haven't gotten into quadratic equations just yet, it's coming up in the next section though. Wish me luck!

phi · Answer

luck doesn't work with math (unfortunately).  But hard work and concentration do.

anonymous · Answer

I know, I've been dedicated to my math class and am among the top students in my class. :)

anonymous · Answer

Thanks again, Phi! :D