What is completely factored answer to (x-y)(x-y)?
600427
It is already factored, you can write it more concisely as (x-y)^2.\n\nThe EXPANDED form is x^2-2xy+y^2.
87985
The expression was not set equal to anything.\n\nIf you're doing simple factoring in algebra class, the \nsimplest factors would be (b+4) (b+3).\n\nIf the expression was set to equal 0, solve for b, then your answer set would include -4 and -3.\n\nHope this helps!
818222
1.) (x^2y)/(xy) = ((x)(x)(y))/(x)(y) = (x/x)(y/y)(x) = (1)(1)(x) = x\n\n2.) (3(x-2)^2)/(6(x-2)^4) = ((3)(x-2)(x-2))/((3)(2)(x-2)(x-2)(x-2)^2)\n    *Now what can be "1" \n\n3.) There may be a type-o in the denom having the "+" sign, but if such the case, the expression is at it's simplest form.\n\n4.) Now for this think about factoring out a number and I think you'll see something that can be in its simplest form.
917321
x+y=2\n2x-y=0\n\nx=2-y\n2(2-y)-y=0\n\nX=2-y     \n4-3y=0   \n\nx=2-4/3\ny=4/3
445887
you have to take out an x because there are x's that are there that are common to both terms. x(1-x^4) then -x(x^4-1)\nso factoring is easier. Then -x(x^2+1)(x^2-1). Then it becomes -x(x^2+1)(x+1)(x-1). That's the answer.
60538
if [ -7x + x^2 =0 ]\nthe you will get y = 7/x
685420
x-y = 4\nx = 4 + y\nx-4 = y\n\nSo all pairs of the for (x, x-4).
421674
y = (x mod 5) + 1.\n\nThough that does seem a bit like cheating :)
914223
difference of squares rule:\n\nx^2 - y^2 \n\n(x-y)(x+y)
1124422
[(y^2 + y )/(y+ 2)] / [(y^2 + 3y)/(y^2-4)]\n= [(y^2 + y )/(y+ 2)] times (y^2-4)/(y^2+3y)\nFactor everything\ny^2 + y = y(y+1)\ny^2 + 3y = y(y+3)\ny^2 - 4 = (y+2)(y-2)\nCancel anything on the top with anything on the bottom\nAnswer: (y+1)(y-2)/(y+3)
1068964
simplify expression: -2(y+3)+6(y+1)\n\nsimply distribute then add.
861275
x(y-1) = 2y + 3\n\nxy - x = 2y + 3\nxy - 2y = x + 3\n(x - 2) = (x + 3) / y\n y = (x + 3) / (x -2)
532215
Your question is to factor in the following form y=a(x-p)^2+q for the question y=x^2-8x+8. Okay, let's roll.\n\ny=x^2-8x+8\n = (x-4)^2 -16 +8   [using (a-b)^2 = a^2 -2ab +b^2]\n = (x-4)^2 -8\n\nTherefore, in the form  y=a(x-p)^2+q, a=1, p=4 and q= -8.
859744
(x + 4y)(x - 4y)\nI answered 2 of your questions. Do the rest of the questions yourself.
982146
Just add and substract terms in both sides.\n\nSince \ny-x^2+8x\nis equal to 7, \nyou can add x^2 and substract 8x to both sides.\n(y-x^2+8x) +x^2-8x = y\n7 +x^2-8x = x^2-8x+7\n\nso y=x^2-8x+7\n\nby the way, if you factorize in terms of x then you'd get\ny=(x-1)(x-7)
549551
The way this is written, there is nothing to solve because it's not an equation.\n\nTo use the zero product property, you want to factor an expression that equals zero. Then you conclude that if a product equals zero, one of its factors must be zero. To factor the expression in your problem, combine any like terms to simplify, then factor the simplified expression.\n\nHope that helps.
1032854
given the zeroes of y:\nif x = -4, 0 or 2, then y = 0.  Therefore it passes through\n(-4, 0)\n(0,0)\n(2,0)\n\nIn other words -4, 0 and 2 are x-intercepts of the polynomial.\n\nThe polynomial (3rd degree) can be expressed as:\ny = x³ + ax² + bx + c\nIt can also be expressed as\ny = (x - x1)(x - x2)(x - x3) \nwhere n is a constant.\n\nWhen you substitute x1 to the expression\ny = (x1 - x1)(x1 - x2)(x1 - x3)\ny = (0)(x1 - x2)(x1 - x3)\ny = 0\n\nTherefore, x1 is a zero\n\nThe same goes with x2 and x3. Therefore you can put the zeroes into our expression\n\ny = (x - (-4))(x - 0)(x - 2)\ny = (x + 4)(x)(x - 2)\ny = x(x + 4)(x - 2)\nor\ny = x(x² + 2x - 8)\ny = x³ + 2x² - 8x\n\nThat is your 3rd degree polynomial! Hope this helps^_^