how do you multipy polyomials?
340490
(ax^2 + bx + c)(dx^2 + ex + f)\n=ax^2(dx^2 + ex + f) + bx(dx^2 + ex + f) + c(dx^2 + ex + f)\n=adx^4 + aex^3 + afx^2 + bdx^3 + bex^2 + bfx + cdx^2 + cex + f\n=adx^4 + (ae + bd)x^3 + (af + be + cd)x^2 + (bf + ce)x + f
439000
Since you cannot factor any binomial to form a monomial, the only factoring you can do is by "common monomial factoring". After that, cancel like factors, but if there are no common factors, that means you cannot factor it anymore.\nExample:\n\n(3x - 3)/(2x - 2)\nyou can get common monomial factors for each:\n3(x - 1)/2(x - 1)\ncancel the common factors\n3/2\n\n(4x² - 6x)/(2x + 10)\n= 2x(2x - 3)/2(x + 5)\n= x(2x - 3)/(x + 5)\n\n(x + 3)/(x + 5)\n---> this cannot be simplified anymore\n\n^_^
673673
Use the multinomial coefficient, which is an extension of the binomial coefficient.\n\n10!/(2!2!2!2!2!) = 113,400\n\nBasically it works like this: if your have people A, B, C, ..., J, then there are 10! ways to arrange them.  One arrangement of these 10 people with 2 people in five groups looks like this\n\nAB|CD|EF|GH|IJ\n\nwhere the bars are separating the groups.  Now another arrangement of these people looks like this\n\nBA|CD|EF|GH|IJ\n\nbut everyone is still in the same group.  So is this one\n\nBA|DC|EF|GH|IJ\n\nand this one\n\nBA|DC|FE|GH|IJ\n\nand so on.  It turns out there are 2!2!2!2!2! = 32 ways to rearrange this without changing the groups, since each pair has 2! ways of being arranged.\n\nSo since there will be fewer ways to make 5 groups of 2 than there are ways to arrange the 10 people, you come up with \n10!/2!2!2!2!2! ways to put 10 people into 5 groups of 2.\n\n10C2 and 10C5 would not be correct.  10C5 is the number of ways to select 2 groups of 5, not 5 groups with 2.  10C2 is the number of ways to select one group with 2 people and one group with 8 people.
504285
It has been a long time, here are suggestions:\n1. 5x^3 + 11x^2 + 3x + 1  \n2. 4x^3 - 13x^2 - 8x + 7 \n3. 2x^4 - x^3\n4. 21x^2 + 26 x + 8
1007787
distribute and then combine like terms
221530
Add them together and divide by 5.\n\nWow. I could seriously have you people do my homework for me. Every night. I'll type out the problems and see how you all do. :P JP
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.
1166133
Multiply both sides by R+r\nI(R+r) = E\nDistribute the I\nIR + Rr = E\nSubtract IR form each side\nRr = E - IR\nDivide both sides by R\nr = (E-IR)/R
1136055
Here's how we can solve the problem logically, without doing all the multiplication.\n\nFirst of all, the coefficients aren't relevant to the problem; we don't care what the terms are specifically, just how many of them there are.\n\nLet's create two new temporary variables: let a = x + 5y, and b = 3z + w.  That reduces the problem, for the moment, to (a + b)^6.\n\nNow, any time you take a binomial to a power -- such as (a+b)^n -- there will be n+1 terms in the final expansion: (a+b)^2 has 3 terms, (a+b)^3 has 4 terms, etc.\n\nSo in (a+b)^6 there will be seven terms, which will have various coefficients but generally be of the forms a^6, a^5b, a^4b^2, a^3b^3, a^2b^2, ab^5, and b^6.  Note that, in each term, the sum of the powers is 6.\n\nNow, if we reinstate the original variables, substituting back in for a and b, we can consider each of those seven terms.\n\na^6 becomes (x+5y)^6, which will have 7 terms.\na^5b becomes (x+5y)^5 (3z+w), which will have 6 * 2 = 12 terms.\na^4b^2 becomes (x+5y)^4 (3z+w)^2, which will have 5 *3 = 15 terms.\na^3b^3 becomes (x+5y)^3 (3z+w)^3, which will have 4 * 4 = 16 terms.\na^2b^4 becomes (x+5y)^2 (3z+w)^4, which will have 3 *5 = 15 terms.\nab^5 becomes (x+5y) (3z+w)^5, which will have 2 *6 = 12 terms.\nAnd, finally, b^6 will have 7 terms.\n\n7+12+15+16+15+12+7 = 84.\n\nHope that made sense.
905726
three sizes\nthree crusts\nten toppings\n\nLook in your math text under "combinations and permutations." You can do this!
548405
please dont hate math\nshow interest\nthen you do well\n\nFOIL\nFirst-Out-Inner-Last\n\nThis method is used for multiplying two binomials.\nsay (a+b)(x+y)\n\n( a + b ) (x + y)\n\nFirst------first terms in the two binomials. they are to be multiplied. so ax\nthen Outer a and y. so ay\nInner-----bx\nLast-----last terms in the binomials. so 'by'\nthen add all the terms.\nax+ay+bx+by\n\nyou may consult me
263042
If they are all the same size die (same number of faces), take the number of faces of one die and raise it to the power of the number of dice. If the dice are of different sizes, multiply the number of their faces together.
920579
use synthetic  divsion method \ni think this link would help\nhttp://www.purplemath.com/modules/synthdiv.htm
534374
Ex. (6,2) \n\n6 times 5 over 2 times 1\n6 x 5/2 x 1= 30/10 = 3/1\n\nEx. (7,4)\n\n7 times 6 times 5 times 4 over 4 times 3 times 2 times 1\n7 x 6 x 5 x 4/4 x 3 x 2 x 1= 840/12=70\n\nGet it?
773850
Tip:  If the sum of the coefficients of the polynom are zero, then x =1 is a zero of the polynom.\n\nx = 1\nIf you need more roots do the synthetic divisiona and then use the quadratic formula.
7772
You can.  Use the mapping (x, y, z) -> x.  Of course, that is a many-to-one mapping, and kind of useless.
1098310
There are lots of them.  The easy way is to multiply 5 times 6 times 7.  This number is divisible by 5, 6, and 7.  So -- just add two to it.\n\nThis is 212.\n\nYou want another?  How about 2102?  210000000002?\n422? 420002?\n\nI got a million of 'em.