how do i factorise this sum:p(to the power 2)+p?
594239
ok so you have p^2 + p, you can factor out p from both parts of the expression, it will end up like this: p(p + 1).  Hope that works.
333863
Diagonalisation is the answer here.  Not sure how this didn't turn up in a web search, as I found plenty of resources for it, but the idea is that you find a matrix that diagonalises your initial matrix, and this allows powers to be done trivially.\n\nAssuming you're aiming to get a matrix A to the power n:\n\nFind P such that:\nP^-1 A P = D, where D is diagonal.  Then:\nA = (P D P^-1) and:\nA^n = (P D P^-1)(P D P^-1)(P D P^-1)....\n\nThe P^-1 P terms cancel and you're left with:\nA^n = P D^n P^-1\n\n... and this is much easier to calculate as D^n is trivial :)\n\nA good reference is in the sources.  Hope that helps!
294579
First multiply p times the second polynomial.\np(p^4 - 2p^3 + 3) = p^5 - 2p^4 + 3p\nthen multiply 2 times the second polynomial.\n2(p^4 - 2p^3 + 3) = 2p^4 - 4p^3 + 6\nNow add these two together.\n\n(p^5 - 2p^4 + 3p) + (2p^4 - 4p^3 + 6)\n= p^5 - 4p^3 + 3p + 6
875476
(a + b - c)(a + b + c)\n((a+b)+c)((a+b)-c)\n(a+b)^2-c^2\na^2+2ab+b^2-c^2
1012911
P = 2 L + 2 W\nL = 2 W + 8\n2W + (4 W + 16) = 64\nW = 8
761108
p= 2L + 2 W \n2w=p-2l\nw=(p-2l)/2
408254
p^5 +2p^4 -2p^3 + 3\n\nNote: P times P^4 is P^5, b/c when you multiply exponents, you add the exponential numbers.  Since P is really P^1, 1+4 = 5, giving you P^5.
92762
A = P + Prt\n\n==> P(1 + rt)\n\n==> P = A/(1 + rt)\n\n<done>
283645
if p is a factor of both, than p should also be a factor of the difference.\n\nn + 10 - (n + 3)\n\nThe resulting integer only has one factor greater than 1.  That's your answer!\n\n(I'll let you figure out the rest, don't want to spoil it for you.)
919174
first multiply the first two factors and multiply the product by the third factor\n\nTo multiply something complicated like (4x + 6)(5x - 3)(15 - x), just do FOIL on two of the binomials and then distribute the answer onto the remaining binomial
773695
Look in your text book under "factoring polynomials"... or you could try this web site that explains this fairly thoroughly...\n\nhttp://faculty.ed.umuc.edu/~swalsh/Math%20Articles/Factoring.html
533127
2*2*2*5*5 or 2to the third power times 5 to the second power
868234
P = 2W + 2L\nReplace W with 10 and L with 19\nP = 2(10) + 2(19)\nP = 20 + 38\nP = 58 in
728458
Here is how you do it (I'll use ^ for the power sign).\n- Divide by P:    (1+r/n)^n = F/P\n- Take the 1/n the power:     (1+r/n) = (F/P) ^ (1/n)\n- Solve for r:   r= n* ((F/P) ^ (1/n) -1 )\n\nIn this case F=20000, P = 12600, n=12 therefore r = 0.47 or 47%\n\nSeeing this answer I wonder if I didn't misinterpret your notation, but with the method given, you can solve the question for yourself.
895900
This is essentially a maths question.\n\nP1 = P2V2 / V1\nV1 = P2V2 / P1\nP2 = P1V1 / V2\nV2 = P1V1 / P2
945604
1.adding-2f-7\n-6f-2f+7-7<2f-2f-9-7\n-8f<-16\ndividing by -8\nx>2\n\n2.5/8g-4>2/3g+6\nadding 4\n5/8g>2/3g+10\nmultiplying by 24\n15g>16g+240\nadding -16g\n-g>240\ndividing by -1\ng<-240\n\n3.sum incomplete\n\n4.-2 1/9d+3/5<-3 2/9d-4 2/5\n-19/9d+3/5<-29/9d-22/5\nmultiplying by45\n-95d+27<-145d-198\nadding 145d\n50d+27<-198\nadding -27\n50d<-225\ndividng by 50\nd<-5 1/2
56508
first, do 2 to the 12th power (exponents take precedent over multiplication) which is 4096. Then multiply that by 86 and get 352256.  It's quite simple, really.