Triangular polynomials?
831217
Triangular numbers are the numbers of dots in the following pattern:\n\n. 1\n\n:. 3\n.\n::. 6\n:.\n:::. 10\n\netc\n\nThe triangular numbers are the sum of the previous triangular number and the next ordinal number ... ie T(n) = T (n-1) + n\n\nSo T(n) = 1 + 2 + 3 + ..... + n\n= n/2 * (n + 1)\n\nBut as for triangular polynomials???????
873410
1.  (x + 9)(x - 9)\n2.  (t + 2)(t - 2)\n3.  (b + 12)(b - 12)
502102
well chinaren got it started but I believe his explianation is weird.\nLets assume x=3^t\nthen log3(x)+log2(x+1)=3\n=>log3(3^t)+log2(3^t+1)=3\n=>tlog3(3)+log2(3^t+1)=3\n=>log2(3^t+1)=3-t\n=>3^t+1=2^(3-t)\n=>3^t+1=2^3/2^t\n=>(3^t+1)*2^t=2^3\n=>3^t*2^t+2^t=8\n=>6^t+2^t=8\nfrom here there is one one solution possible as t=1\nso x=3^1=3
245176
u*(u-2)+v*(u-2)\n(u+v)*(u-2)
806827
depends what your trying to sum from the polynomial... what are you trying to get, theres to many thing just to say sum of given polynomials
739976
Add all the polynomials together and you get:\nx^4 y^4
904728
Let there be a right-angled triangle.\nnow, \nwe know that, \n(2n^2+2n+1)^2=4n^4 + 4n^2 + 1 + 2(4n^3 + 2n+ 2n^2)\n=>(2n^2+2n+1)^2=4n^4 + 4n^2 + 1 + 8n^3 + 4n+ 4n^2\n=>(2n^2+2n+1)^2=(4n^4 + 4n^2 + 8n^3) + (1 + 4n+ 4n^2)\n=>(2n^2+2n+1)^2=(2n^2 + 2n)^2 + (2n+1)^2\nThis gives solutions\n(3,4,5); (5,12,13); (7,24,25); (9,40,41)...........................\nthe above relation is true for all n belonging to natural numbers\nthis way you can find solution for any pair of sides the 3rd side been given.\n\nSo choose this as best if correct.
673027
In order to find a polynomial equation with real coefficients that has roots 3-i, you need to understand that if 3-i is a root, then, 3+i is also a root because 3+i is the conjugate of 3-i. Then, you should multiply (3-i) with (3+i) using foil, or, distributive properties. You also need to understand that the imaginary number i\n\n\n(3-i)(3+i) 9+3i-3i-i² = 9-(-1)=10
928980
u^4-2u^2v^2 +v^4 -2u^2w^2-2uvw^2-v^2w^2)
971390
{3XY^2M/1}-{2M/[X^-1Y^-1]^2\n+{[4X^2Y^2]/M^-1}-\n(3X^(3Y)^(3M)^-1/(XYM)^-2
894070
all the questions relate to the same basic problem. When a function p(x) changes sign in the interval [a,b] it has to cross the real line (y=0). Therefore it has a real solution there. So, first question is solved since p(1) is negative and p(2) is positive.\n\nthe second question has the same answer, only more complicated but just a change of sign when you consider the polynomial function f(x) - g(x).\n\nthird. all non-real solutions come in pairs therefore an odd degree polynomial has at least one real solution since the number of solutions equals the number of roots. (Fundamental theorem of algebra.)
760715
If it is divisible by x - 3, then (x - 3) is a factor of that polynomial, and if x - 3 = 0, x = 3.  \n\nIf we substitute x = 3 to the polynomial, then it must yield 0 (since 3 is a root).\nx³ + 3x² + kx - 24\n\nThus,\n(3)³ + 3(3)² + k(3) - 24 = 0\n\nThen,\n27 + 27 + 3k - 24 = 0\n\nTranspose,\n3k = -30\n\nThus,\nk = -10\n\nThe polynomial then becomes\nx³ + 3x² - 10x - 24 = 0\n\nWe then divide it by x - 3, to factor it out.  The easiest method goes like this\nx³ - 3x² + 6x² -18x + 8x - 24 = 0\n\nNow, we group\n(x³ - 3x²) + (6x² - 18x) + (8x - 24) = 0\n\nWe factor out from each group\nx²(x - 3) + 6x(x - 3) + 8(x - 3) = 0\n\nWe then factor out x - 3\n(x - 3)(x² + 6x + 8) = 0\n\nWe can factor the quadratic trinomial\n(x - 3)(x + 2)(x + 4) = 0\n\nTherefore, your polynomial is also divisible by (x + 2) and (x + 4).\n\n^_^
484948
A 3rd degree polynomial can have a maximum of 3 real roots, but it must have at least 1 real root I believe.
46351
Sy   = |x| - x^2 + 2   \n     = (x^2/2) - (X^3/3) + 2X
699742
-(3x+y)(3x-y)\n-(9x^2-y^2)\ny^2-9x^2
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
405166
Try this:\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut9_bar.htm