multiplying some polynomials.?
814548
no the correct answer will be\n(1/2)y^-5v^-12
87719
x^2 + 9x + 10x + 90\n\n(x + 9)(x + 10)
245684
To do this multiply each term in the first () by each term in the second () , a process often called FOIL.  When multiplying, multiply the numbers in front but add the exponents.  So 2n^3 times 6n^2 is 12n^5.  Be careful of the signs- you can do the rest.
983225
remember the rule nx^n-1\n2x^2-1\n \nSo in this case it is 2x
919174
first multiply the first two factors and multiply the product by the third factor\n\nTo multiply something complicated like (4x + 6)(5x - 3)(15 - x), just do FOIL on two of the binomials and then distribute the answer onto the remaining binomial
38209
actually polynomial functions make me happy...they are actually pretty easy.  You evaluate each term (thats the stuff separated by the addition (+) or subtraction (-) sign) individually.  So if you had sin(x) -x^3 + 3x\n\nthen you have the following\n\ncos(x) -3x^2 +3\n\nhope this helps!!
498556
Add and subtract only those monomials in a polynomial that has the same exponent.
918948
When you multiply a polynomial, the degrees are added\n\nex: x^2 * x^5 = x^7\n\nbut... if they are in parenthesis, you multiply the degrees\n\nex: (x^3)^6 = x^18\n\nanswer #1: 6\n\nbut you also have to take the coefficient into account. That must be multiplied as well...\n\nex: 3x^2 * 2x^4 = 6x^6\n\nanswer #2: x^3 - 41x^2 + 22x + 3\n\nThe only two things that can be combined are 8x and 14x = 22x\n\nI don't know what you need to do with #2, solve for x or just add? The adding rule for polynomials is that monomials with different degrees cannot be combined.
169512
In multiplying you have to multiply the numerators and demoninators and simlify. and in dividing you have to find the resiprocal of the second # by switching the numerators and demoninators and then you multiply the #s. Peace!
761571
1.(4x+1)(16x^2-4x+1)\n\n2.(1-5x)(1+5x+25x^2)\n\n3.(2xy^2+3)(4x^2y^4-6xy^2+9)\n\n4.b^3(a^3-b^3)\nb^3(a-b)(a^2+ab+b^2)\n\n5.(3x-2y)(9x^2+6xy+4y^2)
119559
its (x+2)(3x+1) now thats rite lol
971390
{3XY^2M/1}-{2M/[X^-1Y^-1]^2\n+{[4X^2Y^2]/M^-1}-\n(3X^(3Y)^(3M)^-1/(XYM)^-2
18182
2x^2 + 5x + 20x + 50\n2x^2 + 25x + 50
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
652953
I got:\n1. 6x^2 - 36x +48\n2. 6x + 22
475786
Represent these binomials in Vector Notation to do the calculations. The Vector Space will be in 7 dimensions, So choose a 7 dim vestor to represent each polynomial.\n\nLet V1: 9x^5 -2x^3 +6x^2 + 3 \nLet V2: 5x^4 - 7x^2 + 6 \nLet V3: 3x^6 - 5x^5 -x^2\n\nMatix Vector Form for Vector 1=V1 etc..\nWe are in 7 space, so express our vector with (&) and seven\ndigits. Let V=V1+V2+V3\nV1= (0, 9, 0, -2, 6, 0, 3)\nV2= (0, 0, 5, 0, -7, 0, 6)\nV3= (3,-5, 0, 0, -1, 0, 0)\n\nV = (3, 4, 5, -2,-2, 0, 9)\nIn algebraic representation\nAns is:\n3x^6+ 4x^5 +5X^4 -2x^3 -2x^2 +9\n\n---\nVectors in 2 space, ex.\nV1=(x1, x2)\nV2=(y1, y2)\nV1+V2=(x1+y1, x2+y2)\nThat is what I did for the 7 space vector above. I expressed the 6 th degree binmial with a 7 dim vector,\nto make calcualting the answer faster. On paper, you could line up all of your equations so that each binomial eqs line up. All x^6 lined up under all x^6, all x^5 lined up under x^5,...etc. But using vectors makes this quicker and faster to calculate the binomials.\n\nMy sources: Two MS's in Mathematics that I received many, many years ago.
945218
[(x^2)^2]^2\n\n(x^2 -4x +4)^2\n\nmultiply      x^2  (x^2 -4x +4)^  = x^4 -4x^3 +4x^2\n\nmultiply     -4x (x^2 -4x +4)^    = -4x^3 +16x^2 -16x\n\nmultiply     4  (x^2 -4x +4)^      =  4x^2 -16x +16\n__________________________________________\n\nthe answer should be                 x^4 -8x^3 +24x^2 -32x +16\n\ni hope that helped