why  p+2p^2-1=0 is = to (2p-1) (p+1)?
820036
2p^2+2p-p-1\n=2p(p+1)-1(p+1)\n=(p+1)(2p-1)\nthat is why!!!!!
1086798
(I am assuming you mean that neither a nor b are 0).  Since p is prime, you are really working in the multiplicative group G of nonzero elements of the field Zp, the integers modulo p.  Thus, for b^x = a (mod p) to have a solution,  the integer a must be in the subgroup generated by b.\n\nIt is well known that G is cyclic.  Thus, for b^x = a (mod p) to have a solution for any integer a<p, then b must be a generator of G.\n\nIf there is a solution, there will be infinitely many solutions.  This is because b^(p-1) = 1 (mod p) for any nonzero integer b<p (the order of G is p-1).  Hence, if b^x = a (mod p) and n is any multiple of p-1, then b^(nx) = a (mod p).  However, the solution is unique modulo the order of the element b.
800348
b^2 - y^2 - 2yp - p^2\n=b^2-(y^2+2yp+p^2)\n=b^2-(y+p)^2\n=[b+y+p][b-(y+p)]\n=(b+y+p)(b-y-p)
901836
I disagree with Peter because when I solved for p in both equations the resulting expressions are different.\n\nHere are my steps and my solutions:\n\nIn the second equation, multiply both sides by (p + 1)\np = r (p + 1)\n\nDistribute r through:\np = rp + r\n\nSubtract rp:\np - rp = r\n\nFactor out a p:\np(1 - r) = r\n\nDivide both sides by (1 - r):\np = r / (1 - r)  \n\nIn the first equation, multiply both sides by r:\np = r(r - 1)\n\nThese are different...
797453
you set the bottom of the equation equal to 0. So x^2 - x = 0, which leads to two cases. The first one is x = 0. The second one is x-1 = 0, or x = 1.
25136
1 = sqrt(1) = sqrt(-1*-1) = sqrt(-1)*sqrt(-1)=i*i = i^2 = -1\nSo 1 = -1.  Add 1 to both sides and 2 = 0.  Divide both sides by 2 and 1 = 0.
220199
The problem factors to\n\n(x - 2)(x^2 + 1)(x^2 + 1) or (x - 2)(x^2 + 1)^2\n\nx = 2, i, -i\n\nthe only real zero is 2.
287019
1 + 1 = 2\none plus one = oneone
516632
Let P=A\nwhere P = 2x + 2y and A = xy\nthen\n2x+2y = xy\nMove xy to the left side and 2x to the right side\n2y-xy = -2x\nFactor a y out\ny(2-x) = -2x\nDivide by 2-x\ny = -2x/(2-x)\nThis is positive whenever x is greater than 2.\nAnswer: x = any number greater than 2 and y = -2x/(2-x)
909020
Recall the distributive property: a(b+c) = ab +ac\nUse this on the second part of your equation where a = -2, b = p and c = 1.\n-2(p+1) = -2p + (-2)(1) = -2p -2\n\nSo 3p - 2(p+1) = 3p - 2p - 2 = p - 2
200150
for all of those, all u do is set once variable at a time equal to zero. for instance..\n\n-x+0=2\n-x=2  <--- multiply x and 2 by -1\nx=2\n\nthen..\n\n0+y=2\ny=2
201685
If y is a root of this and b is in F_p, then (y+b)^p -(y+b)=y^p +b^p -y-b=a, so y+b is another root. Hence the complete list of roots is y, y+1, ....y+p-1.\n\nThus we have two cases, the first where the polynomial splits in F_p (a=0) and the other where we get a degree p extension, so F_(p^2).\n\nNotice that we are using the algebra student's dream (x+y)^p=x^p +y^p which is justified because we are working over a field of characteristic p.
163896
yes the previous equation mentioned would "prove" it, but the problem occurs as you're dividing by zero.  you can also make 1=2, with a similar process, involving the division of zero.  if you don't get what i mean, the previous explanation has this:\n\n(x-y)(x+y)=y(x-y)\nx+y=y\n\nwhat happens at this step?  you are dividing by (x-y).  since x=y is given, then (x-y)=0, thus you are dividing by 0, which is impossible in math.
692175
( x^2 -x +2 )( x^2 + x + 2 )\n\n=>\n a= -1\n b= +2\n c= +1\n d= +2
840090
P1V1 = nRT = P2V2\nP1/P2 = V2/V1
1027274
2x^2 - 2y^2 = 2(x^2 -y^2) = 2(x-y)(x+y)\n-2x^2 + 2xy = -2x(x-y)\nx^2 + xy = x(x+y)\n2(x-y)(x+y)/[-2x(x -y)]/[x(x+y)]\n= 2(x-y)(x+y)/[-2x(x-y)(x)(x+y)]\n= -1/x^2
747036
0.12 \nP(A or B) = P(A) + P(B) - P(A and B) = P(A) + P(B) - P(A)P(B|A) = 0.1 + 0.1 - 0.1(0.8)