(a) simplify the product (x + y)(x^2 - xy + y^2)\n(b) simplify the product (x + y)(x^4 - x^3y + x^2y^2 - xy^3 + y^4)\n(c) based on your answers to part (a) and (b), write a general formula. use "2n" to represent a general even integer and "2n + 1" to represent a general odd integer, and use "..." for missing terms.\n\nI only need (c). Thank youuu for your help!!
199400
So, for a) & b) the two things that you should notice are:\n1 - The exponents inside the second polynomial alternate, meaning that as the powers of the x variable decrease, the powers of the y variable increase, both by increments of one.\n2 - The answer will only contain the variables x and y raised to the same power, which is one more than the leading exponent on the x variable.  \n\nSince (x + y)(x^2 - xy + y^2) = x^3 + y^3\n and  (x + y)(x^4 - x^3y + x^2y^2 - xy^3 + y^4) = x^5 + y^5,\nthen for any given polynomial the pattern is written out...\n(x + y)(y^2n - x^(2n-1)y + x^(2n-2)y^2 + ... + x^2y^(2n-2) - xy^(2n-1) + y^2n = x^(2n+1) + y^(2n+1)
608172
4(a+b)^2-9(a+b)^2\n\nStep 1. 4(a^2 + b^2) - 9(a^2 + b^2) Distribute the 4 and 9\nStep 2. 4a^2 + 4b^2 - 9a^2 - 9b^2 Combine like terms\nStep 3. -3a^2 - 3b^2 You get the sum\nStep 4. -3(a^2 + b^2) Remove common factors\nStep 5. -3(a + b)(a - b) Factor even more to get final answer (if you multiply this, you should get the sum in step 3)
177377
(4x + 5y)(2y^2)\n8y^3 + 10y^3\n\na.)\n(4x + 2y^2)(5y)\n20xy + 10y^3\n\nb.)\n(4x) + (5y)(2y^2)\n4x + 10y^3\n\nc.)\n4x + (5y + 2y^2)\n4x + 5y + 2y^2\n\nd.)\n(2y^2)(4x + 5y)\n8xy^2 + 10y^3\n\nSo the answer is d.) (2y^2)(4x + 5y)\n\nI just did the math out.
563733
# 1    \n(x+1)*(x+6)=x2 +x+6x+6=x2 +7x +6\n#2\n(y-6)*(y-5)= y2-5y-6y+30= y2-11y+30\n#3\n(a-2) * (a+10)=a2 +10a-2a -20= a2  +8a -20\n#4\n(x-1) *(x+1)= x2 -x+x -1= x2 - 1
71368
I believe it simplifies to n - w.\n\nHere's the expression you wrote: (a-y/w+n * w^2-n^2/y-a).\n\nHere's what I think it should look like:\n\n((a-y)/(w+n)) * ((w^2-n^2)/(y-a))\n\nBecause:\n\n((a-y)/(w+n)) * ((w^2-n^2)/(y-a))\n\n= ((a-y)/(w+n)) * ((w+n)(w-n))/(y-a)\n= ((-1)(y-a)/(w+n)) * ((w+n)(w-n))/(y-a)\n\nThe (w+n)'s cancel out, as do the (y-a)'s, so you're left with this:\n\n= (-1)(w-n)\n= (n-w)
411922
1)x+y-z=0.........(eqn. 1)\n2x-y+z=3...........(eqn. 2)\n4x+2y-2z=-2       (eqn.3)\nMultiplying eqn.1 by by 2 and subtracting it from eqn.2,we get\n-3y+3z=3...........(eqn. 4)\nMultiplying eqn.2 by 2 and subtracting it from eqn.3,we get\n4y-3z=-4...........(eqn 5)\nAdding eqn, 4 and 5 we get,y=7\nPutting the value of y in eqn 4 .we get\n-21+3z=3 or,3z=24  or z=8\nputting the value of y and z in eqn.1,we get\nx+7-8=0 \n=>x-1=0 or x=1\nTherefore,x=1.y=7 and z=8  ans\n2)x+2y=9....(eqn 1)\n 2x-3y=4     (eqn.2)\nmutiplying eqn.1 by 2 and subtracting it from eqn 2,we get\n-7y=-14 or y=2\nPutting the value of y in eqn 1,we get.\nx+4=9 or x=5\nTherefore x=5 and y=2 ans.
531774
the first one:\n\n4x^4y - 6x^3y^2 + 2x^2y^3\n\nassuming you forgot the "-" after the first term in the parenthesis\n\nand also assuming the original equation was:\n\n2x^2y ( 2x^2 - 3xy + y^2)
525993
**************\nQuestion 1\n**************\n\n2x - 4y = 8\nx - 2y = 4\n\nx - 2y = 4\nMultiply both sides of this equation by -2\n(-2)(x - 2y) = -2*4\n-2x + 4y = -8\n\nNow we will add this equation to the other\n  2x - 4y =  8\n-2x + 4y = -8\n____________\n0 + 0 = 0\n0 = 0\n\nThere are an infinite number of solutions to this problem because both sides of the equal sign end up the same.\n\n**************\nQuestion 2\n**************\nWe can add these together right away because the x and -x will cancel each other out.\n-x - y =  4\n x - y = -8\n_________\n0 -2y = -4\n\nNow we will solve for y.\n-2y = -4\nDivide each side by -2\ny = 2\nNow we plug that into one of the equations to get x\n x - y = -8\n x - 2 = -8\nAdd 2 to both sides\nx = -6\n\nSo, x = -6 and y = 2
294641
here are the general steps you want to take when approaching these math problems\n\n1.) check for the greatest common factor\n2.) write out your parenthesis\n3.) since it's a squared problem, fill in the x's\n4.) write out your + & - signs\n5.) list the multiplication factors for the last term\n6.) plug in factors that will lead to the correct product\n\nwe'll start with #1 to get a better idea\n\n1.) -x^2+5x+24\n\n1. there is no greatest common factor\n2. write out your parenthesis\n\n(    )(    )\n3. fill in the x's\n\n( x   )(x   )\n\n4. fill in + or - signs\n\n-since we're dealing with -x^2 we know that one of the x's need to be + and the other one -\n\n(x   )(-x   )\n\n-since our last term is positive we know that either both the #'s are + or both of them are -\n\n(x-  )(-x-  )\n\nor\n\n(x+  )(-x+  )\n\n\n\n5.  list the multiplication factors for the last term\n\n24\n\n1*24=24\n2*12=24\n3*8=24\n4*6=24\n\n6. plug in factors that will lead to the correct product\n\n(x-8)(-x-3)=-x^2+5x+24\n\n**now, with a little more practice, you'll be able to figure these out faster...here are the answers to the rest of the problems\n\n2. (x-12)(x-12)\n3. (x-9)(x-12)\n4. 2(x^2-8x+16)=2(x-4)(x-4)\n5. 3(x^2-6x+72)=3(x-12)(x+6)
197253
((y^2 - 16)/(y^2 + 4y))/((y^2 - 7y + 12)/(y^2 - 6y + 9))\n\n(((y - 4)(y + 4))/(y(y + 4)) / (((y - 4)(y - 3))/((y - 3)(y - 3))\n\n((y - 4)/y) / ((y - 4)/(y - 3))\n\n((y - 4)/y) * ((y - 3)/(y - 4))\n\n((y - 4)(y - 3))/(y(y - 3))\n\nANS : (y - 3)/y
382070
I presume you mean\n\n[(6x^2) / (4x^2y-12xy)] * [(x^2+x-12) / (3x^2+12x)]\n\nThere seems to be something wrong. Are you sure about those "y" terms?
476810
It's obvious that 3 can be factored out of each term immediately.  That leaves us with: \n\n3(2x^2+xy-3y^2), \n\nwhich can be factored much more easily than the original expression.\n\n(2x^2+xy-3y^2) factors into (x-y)(2x+3y).  \n\nSo we are left with the following complete factorization for the expression:\n\n3(x-y)(2x+3y)
227314
We have 3 products of single digit numbers (0-9) that all equate to the same value (call it x):\n\n1) abc = x\n2) bge = x\n3) def = x\n\nWe can't have one of the digits be 0 because its product would be 0.  To make the other products zero, one of their digits would also have to be zero, but no letter appears in all 3 equations.\n\nSimilarly, 5 cannot be one of the digits because that would require a five in all the other equations.  No other digit is divisible by 5 so there is no way to get a multiplier of 5 except from another 5.\n\nLikewise 7 cannot be one of the digits for the same reasons.  There are no other digits that are divisible by 7.\n\nThat leaves us with the set {1,2,3,4,6,8,9} for the seven letters a,b,c,d,e,f,g.  So we know the values of abcdefg, just need to figure out what equates to what.\n\nLet's think about the combinations for abc and def that will multiply to the same product:\n\nThe only combinations that work are:\n2x3x6 = 36\n1x4x9 = 36\n\nOr:\n3x4x6 = 72\n1x8x9 = 72\n\nIf we try to work with the first product (36), there is no way to form another product bge that is also 36.  Try it for yourself.\n\nHow else can we make 72 though?  Only one way.\n2x4x9 = 72\n\nThe one new digit used is 2.  So we are done.  We don't know the exact assignments of abc and def, but we do know that g must equal 2.\n\nThe products are:\n3x4x6 = 72\n1x8x9 = 72\n2x4x9 = 72\n\na is one of {1,3,6,8}\nb is one of {4,9}\nc is one of {1,3,6,8}\nd is one of {1,3,6,8}\ne is one of {4,9}\nf is one of {1,3,6,8}\ng is 2\n\nThe answer is g is 2.
523329
LCM of 2, x+1, x+2 is the product of the terms.  Express everything fractions with the LCM as the denominator.  Then solve.\n\n4(x+2) / 2(x+1)(x+2) + 6(x+1) / 2(x+1)(x+2) = 7(x+1)(x+2) / 2(x+1)(x+2)\n\nThen multiply everything by the denominator (LCM):\n4(x+2) + 6(x+1) = 7(x+1)(x + 2)\n\nMultiply out:\n4x + 8 + 6x + 6 = 7(x+1)(x+2)\n10x + 14 = 7(x+1)(x+2)\n10x + 14 = 7(x² + 3x + 2)\n10x + 14 = 7x² + 21x + 14\nPut it all on one side:\n7x² + 11x = 0\nx(7x + 11) = 0\n\nx = 0\nor\n7x + 11 = 0\n7x = -11\nx = -11/7\n\nSo x = 0 or x = -11/7
278040
(6y^2-3y)(y+7)-->(2y)(y-1)(y+7)\nif that helps
187835
y-y^1=m(x-x^2)\ny-8=2(x-(-2)\ny-8=2(x+2)\n A is the correct answer.\ny-8=2x+4\ny=2x+4+8\ny=2x+12 >> y form\n(transpose)\n-2x+y=12\n(if the variable x is negative make it positive)\n-(-2x+y=12) \n2x-y=-12 >> standard form
136946
This is the best I could come up with:\n\n(a)  (x^2-4)(x^4+4x^2+16)\n      (x-2)(x+2)(x^4+4x^2+16)\n\n(b)  (x^3-8)(x^3+8)\n     (x-2)(x^2+2x+4)(x+2)(x^2-2x+4)\n\n(c)  If you take out the factors that are the same, what's left should be equal hence:\n     x^4+4x^2+16=\n(x^2+2x+4)(x^2-2x+4)\n\n(d)  x^(a^2)+2ax^a+4a^2=\n(x^a+ax+2a)(x^a-ax+2a)