(1+2x^3)^8\n\nsry i added to many x's
122241
In general the binomial expansion of\n(a+b) to the 8th degree is:\n\na^8+8a^7b+28a^6b^2+56a^5b^3+\n\n70a^4b^4+56a^3b5+28a^2b^6+\n\n8ab^7+b^8.\n\nHere your "a" term is 1 and your "b" term is 2x^3.\nDo the substitution and the rest is tedious arithmetic.
669034
8x^2 -5 -3x^2 + 7\n8x^2 - 3x^2 -5 + 7\n5x^2 + 2
631390
8/(x+3) = (1+1)/x\nCross Multiply, so:\n8x=(1+1)(x+3)\n8x=2x+6\n8x-2x=2x+6-2x\n6x=6\nx=1
647368
I hope you are trying to say x+2 = (2/3)(2x-1)..if it is correct then\n\n3(x+2) = 2(2x-1) --> 3x+6 = 4x - 2 --> 6+2 = 4x-3x --> 8 = x
404469
7x^2+16x over (x-2)(x+2)(x+3)
259562
Um...\n\n2(-X) + 3(-X) + X + 8\n\n-2X - 3X + X + 8\n\n-5X + X + 8\n\n-4X + 8\n\n8 - 4X
353525
x^2+6x+8/x^2-4-21 / x^2-4/x^2-7x\n=x^2 + 6x +(8/x^2) - 4 - (21/x^2) - (4/x^2) - 7x\n=x^2 + 6x -4 -7x +(8/x^2) - (21/x^2) - (4/x^2) \n=x^2 + 6x -4 -7x +[8-21-4]/x^2\n=x^2 + 6x -4 -7x -17/x^2\n=(x^2)*[x^2 - x -4  -17]/x^2\n=(x^2)*(x^2) - (x^2)*(x) -4*(x^2)  -17]/x^2\n=(x^4) - (x^3) - 4(x^2)x -17]/x^2
253991
Case 1 : 2x+10<4\n   \n-> 2x+10-4<0\n-> 2x + 6 <0\n-> 2(x+3)<0\n-> x < -3\n\nCase 2 : 2x+10>-4\n\n-> 2x + 14>0\n-> 2(x+7)>0\n-> x > -7\n\nNow we combine the results of the 2 cases\n\ni.e, x<-3 & x>-7\n\nthis will come out to be -7 < x < -3
365645
simple!!!\nthe answer is : 12x^2\nthe rule of addition intergers with exponents is simple.\nyou just add the whole number (that is 4+8) and then follow the integers (that is x^2)... get my point?\ni hope i have helped you with your assignment....
460869
1) 4-3x<_8+x\n4+8<_x+3x\n12<_4x\n3<_x or x in the interval [3,plus infinity)\n\n2) 3x-3>-x+1\n3x+x>1+3\n4x>4\nx>1 or x in the interval (1, plus infinity)\n\n3) 3(x-1)_<4x\n3x-3<_4x\n-3<_x or x in the interval [-3, plus infinity)\n\n4)1/2x<x+1\n-1<x/2\n-2<x or x in the interval (-2, plus infinity)\n\n5)-1/2x>1-x\nx/2>1\nx>2 or x in the interval (2, plus infinity)
382422
You left out the negative sign in front of the 4 on the last term and the x^4 cancels in the first term\nAnswer: 8 + x^-2 + 9x^-4 or you could write it like this\n8 + 1/x^2 + 9/x^4
674280
7x+27 over (x+9)(x-9)
314729
Here's the best I can do to help with the first one.\n   __x_+9___\nx-2)X^2+7x-2\n    x^2-2x\n    ------\n        9x-2\n        9x-18\n        -----\n           16\nFollow the example for the rest
528285
You can use the long way of expanding out each term and adding:\n(x+1)(x+1) = x² + 2x + 1\n(x+1)(x+1)(x+1) = (x + 1)(x² + 2x + 1) \n= x(x² + 2x + 1) + 1(x² + 2x + 1)\n= x^3 + 2x² + x + x² + 2x + 1\n= x^3 + 3x² + 3x + 1\n\nBut there is a much easier method to solve this called "binomial expansion".  First you need to know Pascal's triangle.  For (x+1)^3 you need row 3 (the first row is row 0).\n\nPascal's triangle:\nRow 0: ..... 1 \nRow 1: .... 1 1\nRow 2: ....1 2 1\nRow 3: ...1 3 3 1 <---\nRow 4: ..1 4 6 4 1\nRow 5: 1 5 10 10 5 1\n\nNow you need to make all the combinations of x and 1 in descending order.  First you need x^3, then x^2, then x, then 1.\n\nNow put each number (1, 3, 3, 1) from that row of the Pascal's triangle in front of each term.\n\nSo (x + 1)^3 = x^3 + 3x² + 3x + 1\n\nYou can use it to do even higher powers easily... for example:\n(x + 1)^7 = x^7 + 7x^6 + 21x^5 + 35x^4 + 35x^3 + 21x²  + 7x + 1\n\nI think that is the " special formula" you were looking for.  With it you can quickly see:\n(x + 1)^3 = x^3 + 3x² + 3x + 1
69467
2x-1(xsquared-x+10)squared
333751
[(2x^3+2x^2+3x+3)/(x^2-9)]*\n[(x^2+2x-15)/(4x^2+6)]\n\nFactor each expression individually.\n\n(2x^3+2x^2+3x+3) = (x+1)(2x^2+3)\n\n(x^2-9) = (x+3)(x-3)\n\n(x^2+2x-15) = (x+5)(x-3)\n\n(4x^2+6) = 2(2x^2+3)\n\nSubstituting these in, our expression becomes\n{[(x+1)(2x^2+3)]/[(x+3)(x-3)]}\n*{[(x+5)(x-3)]/[2(2x^2+3)]}\n\nCombining these into one larger expression, we get\n[(x+1)(2x^2+3)(x+5)(x-3)]/\n[2(x+3)(x-3)(2x^2+3)]\n\nCancel the quantities common to the numerator and denominator and we get our final answer\n[(x+1)(x+5)]/[2(x+3)]\n\nGood luck.
369098
2x^3+x^2+2x+1\n\nWhat I look for in the beginning is similar constant numbers.  I notice that the x^3 has a 2 before it as does that x.  At the same time I notice that the x^2 and the 1 have the 1 in common.  I also see that both sets are x^2 apart from each other.\n\nI will try and start there.\n\n(2x^3 + 2x) +(x^2 + 1)\n2x(x^2 + 1) + 1(x^2 + 1)\n\nThis then equals\n(2x + 1)(x^2 + 1)