The answer to the question is: (2p+1)(2p+81) for Equation 4p^2+164p+81
P.S ^ is exponent
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p2 + 2pq + q2 = 1q2 + 2pq + (p2 - 1) = 0q = 1/2 [ -2p plus or minus sqrt( 4p2 - 4p2 + 4 ) ]q = -1 - pq = 1 - p
It is 820p2 - 4p - 32.
8p3 + 1 = (2p + 1)(4p2 - 2p + 1)
120 four letter permutations if you don't allow more than one 'o' in the four letterarrangement.209 four letter permutations if you allow two, three and all four 'o'.1.- Let set A = {t,l,r,m,}, and set B = {o,o,o,o}.2.- From set A, the number of 4 letter permutations is 4P4 = 24.3.- 3 letters from set A give 4P3 = 24, and one 'o' can take 4 different positions in theword. That gives us 24x4 = 96 four letter permutations.4.- In total, 24 + 96 = 120 different four letter permutations.5.- If the other three 'o' are allowed to play, then you have 2 letters from set A thatgive 4P2 = 12 permutations and two 'o' can take 4C2 = 6 position's, giving 12x6 = 72four letter permutations.6.- One letter from set A we have 4P1 = 4, each one can take 4 different positions, therest of the spaces taken by three 'o' gives 4x4 = 16 different permutations.7.- The four 'o' make only one permutation.8.- So now we get 72 + 16 + 1 = 89 more arrangements adding to a total of 89 + 120 = 209 different 4 letter arrangements made from the letters of the word toolroom.[ nCr = n!/((n-r)!∙r!); nPr = n!/(n-r)! ]