Find a ploynomial that has the given zeros and degree. (do not leave in factored form-multiply out)?
831999
y = (x+3)(x-1)(x-2) = x^3 - 7x + 6
629310
nything that doesn't involve either zero or U
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^_^
889891
~..-8,-6,-4,-2,0,2,4,6,8
617264
Since each term has at least x in it, divide out the x by distributive property:\n\nx(x^2 - 4xy + 4y^2)\n\nThen factor the rest by FOIL:\n\nx(x-2y)(x-2y)
445985
if by this you mean\n\n((3x^-2 * y)^-2)/((4xy^-2)^-1)\n\nthis becomes\n\n((3^-2) * x^(-2 * -2) * y^-2)/((4^-1) * x^-1 * y^(-2 * -1))\n\n((1/9)x^4 * y^-2)/((1/4) * x^-1 * y^2)\n\n((1/9)/(1/4)) * x^(4 - (-1)) * y^(-2 - 2)\n\n((1/9)*(4/1)) * x^(4 + 1) * y^(-2 + (-2))\n\n(4/9)x^5y^-4\n\nwithout using negative exponents, the answer would be\n\n(4x^5)/(9y^4)
296877
It has a solution, indeed. It actually has two solutions, -3 and 7. To find it out, set each pair of nomials, (e.g. (x + 3)) to zero. So it would be (x+3)=0. Then solve for x. Do the same for the other nomial, and there you have it.
1177677
shouldn't it be x^2??? if it IS x-squared, then sure, you could use the factorization method. \n\nit goes like this:  (6x + 1) (x - 1) = 0\n\ntherefore, x = -1/6 , 1
383816
!!, !!!!!!!! are "multifactorials". They are described in the first attached article.\n\nBriefly,\nn!! = 1 for n=0 or 1,\nn!! = n * (n-2)!! for n>=2.\n\nn!!...! (k "!" signes) = 1, for n<k\nn!!...! (k "!" signes) = n * (n-k)!!...! (k "!" signes)\n\n(2n)!! = n! * 2^n\n(kn)!!...! (k "!" signes) = n! * k^n\n\nSo, we have\n1) 6! = 1*2*3*4*5*6 = 720\n2) 6!! = 2*4*6 = 48\n3) 0! = 1\n4) 5!! = 1*3*5 = 15\n(5!!)!! = 15!! = 1*3*5*7*9*11*13*15 = 2027025\n\nOther factorials are too huge to compute precisely, but we can evaluate them using Stirling formula (second article)\nln n! ~= n*(ln n) - n + (ln(2*pi*n))/2 + 1/(12*n)\n\n5) 5! = 120\n(5!)! = 120! =~ e^(120*ln(120) - 120 + (ln(2*pi*120))/2 + 1/(12*120)) =~ e^457,81 =~ 6.69*10^198\n6) 1234!!!!!!!! = 1234 * 1233 * 1232!!!!!!!! =\n= 1521522 * (154*8)!!!!!!!! =\n= 1521522 * 154! * 8^154 = e^972.43 =~ 2.1*10^422\n\nThus, 6!! - 6! + (5!)! - 1234!!!!!!!! - 0! + (5!!)!! =\n=~ 48 - 720 + 6.69*10^198 - 2.1*10^422 - 1 + 2027025 =~ -2.1*10^422\n\nAnswer: about -2.1*10^422
178452
y^2 + 3y + 6 -4y^2 - 6y simplifies to:\n\n-3y^2 - 3y + 6\n\nIt's a second-degree trinomial expression.
511948
There are an infinite number of ordered pairs that would satisfy this equation. If you want to find a few of them, assign a value to x and then solve for y.
782446
x:-    0** 2** 4\ny:-  -4** 0** 4\n\ni hope that this helps
911592
Factorize!\n4x (x^2-9) = 0\n4x (x+3) (x-3)\n\nyour solution is \nX = 0\nx= -3\nx= 3
78119
I guess you must factorize them.\nAll the polynomials you wrote are to be factorized by partial factorization.\nI think you know the general rule.\nA(x)P(x)+A(x)Q(x) + B(x)P(x) + B(x)Q(x) = \n---> factorize between the 1°-2° and 3°-4° <---\nA(x)[P(x) + Q(x)] + B(x)[P(x) + Q(x)]\nYou get the same binomial term [P(x) + Q(x)] which is a common factor. Factorize:\n[P(x) + Q(x)]*[A(x)+B(x)]\n\nFor example, in the first:\nA()=a\nB()=b\nP(x)=2x\nQ(y)=3y\nYou get \na(2x+3y)+b(2x+3y) = (2x+3y)(a+b)\n\nIn all the others you must rearrange the terms to use the above steps:\nin xy+32+8x+4y the repeating patterns are 8x + xy and 32 + 4y ....\n\nThink about it and good work
799253
Polynomial is a general term for any expression, including monomial, binomial, trinomial, etc.  Monomials have just one term with no + or -, such as x or 3x or 5x^6.  Binomials are two monomials added or subtracted, such as 3x + 5 or 5x^2 - 6x.  The degree of a polynomial is the highest exponent of any of its variables, assuming there is only one variable.  So for x^3 + 8x^2 + 3x - 4, the degree is 3.
778763
1) (a - b)(x - y)\n2) (x^9 + y^3)(x^9 - y^3)\n3) (6x + 5)(x + 8)\n4) (x - 9)(x - 7)
973495
Replace the y in the first equation with its value from the second:x+9.  You get x^2 + (x+9) = 9\n\nSubtract 9 from both sides to get x^2 + x = 0\n\nFactor out an x:  x(x+1) = 0\n\nSo x is 0 or x is -1\n\nPlug these into the original equations to get y.  \ny= 0 + 9; y=9\n\ny = -1 + 9; y=8\n\nSo the complete answers are (0,9) and (-1,8)Check them in the other equation: 0^2 + 0 + 9 = 9 check\n(-1)^2 + 8 = 9; 1+8=9 check