Algebra: Dividing Polynomials: How to solve: (5+10y^2+5y^4)/(-5+y)?
605605
It's hard to format using this system.  It's pretty much just like long division with regular numbers.  First you need to rearrange them so the coefficients of x are decreasing.  Also, in the numerator, fill in any missing powers of x with a 0.\n\n5x^4 + 0x^3 + 10x^2 + 0x + 5\n\nNow you will divide by x-5\n\nx-5| 5x^4 + 0x^3 + 10x^2 + 0x + 5\n\nJust ignore the -5 and look at the x.  One x goes into 5x^4 how many times?  ***5x^3***.  So that is the first part that goes on top of the long division sign.  Now, same as with numbers, multiply this by the x-5 and put it underneath and subtract: \n5x^3 * (x-5) = 5x^4 - 25x^3.  So subtracting this from 5x^4 + 0x^3 + 10x^2 + 0x + 5 you get  25x^3 + 10x^2 + 0x + 5. (Note the 25x^3 is positive because we subtracted a negative number).\n\nNow again, x goes into 25x^3 how many times?  ***25x^2***.  Multiply 25x^2 * (x-5) = 25x^3 - 125x^2.  Subtract and you are left with 135x^2 + 0x + 5.\n\nx goes into 135x^2 ***135x*** times.  135x * (x-5) = 135x^2 - 675x.  Subtract and get 675x + 5.\n\nx goes into 675x ***675*** times.  675 * (x-5) = 675x - 3375.  Subtract: 3380. This is the remainder, since x does not go into 3380 at all, and so it goes over x-5 (just like with plain numbers).\n\nSo then the final answer is to gather up all the numbers across the top of the long division bar, and add on the remainder:\n\n5x^3 + 25x^2 + 135x + 675 + 3380/(x-5)
629654
x^2+y^2-2y-5=0\n\nx^2 + y^2-2y + 1 - 1 -5 =0\n(x-0)^2 + (y-1)^2 = 6\n\nCenter (0, 1) radius sqrt(6)
930601
put -2y=5-3x call it equation {1}\ncall 4x-5y=86 {2}\n\nmultiply all of the terms i {1} by 5\nmultiply all ot the terms in {2) by 2\nthis gives -10y=25-15x {3} &  8x-10y=172 {4}\nyou've got -10y in terms of x in {3}, so just put that into {4}  8x+25-15x=172, -7x=157, \nx = -172/7 sub in to {1} to get y \nman that was ablast from the past, haven't done that for years.
141598
If we break down 25x^2, we get 5x.\n\nIf we break down 49y^2, we get 7y.\n\nThe resulting equation is:\n\n(5x + 7y)(5x - 7y)\n\nSolve it from there.
17351
The equation to solve looks like:\n\n(x+10)/(y+5)=3x/y\n\nwhere x is the numerator and y is the denominator.\n\nIf you cross multiply the expression, it yields:\n\n-15x+10y=2xy\n\nThere is no single answer to that problem. Any (x,y) pair that satisfies the equation above works. (y cannot be 0 or -5). x=2 indeed yields y = 5. x=1 yields y=15/8 etc...
576742
ANSWER: y = 2; x = -1/5\n\nSOLUTION:\n\n1) First solve for y.  You can do this by adding the two equations together:\n\n5x - 2y = -5\ny - 5x = 3\n\nNotice that you have a positive 5x on the first equation, and a minus 5x on the second equation, so they both cancel out and you can re-write the equation as follows:\n\n- 2y = -5\n   y = 3\n\nAdd up these 2 equations and you get: \n\n- 1y = -2   or y =  2\n\n2) Then solve for x by replacing y with 2.  You can use any of the 2 equations from the problem.\n\n5x - 2(2) = -5\n\n5x - 4 = -5\n\n5x = -5 + 4\n\n5x = -1\n\nx = -1/5
1177326
1) q^3/(5p^8)\n\n2) 1/(-2y)^4 = 1/[(-2)^4(y)^4]\n= 1/(16y^4)
828966
This is exactly what I am doing in math right now:\nso with the remainder theorem you plug d+4 into the equation so it becomes -4\nso\n(-4)^4+2(-4)^2+5(-4)-10\n256+32-20-10=258
401465
(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.
902478
wow its been a long time.. i use a calculator to do that now.. but here goes one thing i remember...\n\nlets say we have a polynomial like this\n\nx^2 -3x-10  and we need to factor it..\n\nwe take a look at what factors of -10 we could use that would add up to -3  in this case there are two way to get -10 (-2*5) or (2*-5)  if we add the two sets together one gives us -3 and the other gives us 3 so we can throw out the -2*5 terms.\n\nSo now we lay out the factors.... we need two x's to get x^2 so we need two x's in the first terms (remember FOIL?)\n\nlay the terms out so they give us the correct values\n\n(x+2)(x-5)  now we can use FOIL (firs, outer, inner, last) to check ourselves...\nX*X = x^2\nX*-5 = -5X\n2*x =2x\n2*-5=-10\n\nso we get [x^2] +[-5x] +[2x] +[-10]\nor x^2 -5x+2x-10\ncombine like terms now\n-5x+2x =-3x\n\nwhich gives us\nx^2-3x-10\n\nhope that helps
30589
2(5)-5(5)+2(5)=\n10-25+10\n20-25\n-5
847519
here's how you solve:\n\nstep 1 - distribute:\n\n5*1/8*y + 64*1/8 = 1/4*20+1/4*2*y\n5/8y + 8 = 5 + 1/2y\n\nstep 2 - combine like terms:\n\n5/8y - 1/2y = 5 - 8\n5/8y - 4/8y = -3\n1/8y = -3\n\nstep 3 - solve for y:\n\n1/8 * 8 * y = -3 * 8\ny = -24
959550
x-5y=4, therefore x=4+5y and you plug that into the other problem to get 4(4+5y)+8y=-5, 16+20y+8y=-5, 28y=-21, y=-21/28 or -3/4, then plug -3/4 into one of the equations for y and solve for x, although I'm afraid it's been to long for me to remember the answer to dependant, independant, etc.
287717
-4y^+8y+3y^2-6\n-4y(y^2-2)+3(y^2-2)\n(-4y+3)(y^2-2)\n\ndoes that make sense?
807151
you can factor by using this:\n(y + 4)( y + 4)=0\n\nnow you have\ny + 4 = 0\n\nsubtract 4 from each side:\ny + 4 = 0 \n   - 4    -4\nyou'll end up with:\ny = -4\n\nPS please award best answer
395515
6x+2y=10\nsubract 6x fro both sides\n2y=-2x+10\ndivide both sides by 2\ny=-x+5
226701
For the second equation, use x=-1+y or x = y-1, then substitute that for x in the first equation, and solve.\n\n5(y-1) - 2y = 10\n\nYou should be able to take it from there.  You really should do your own, or you won't do well on the test.