How do I use FOIL method with a binomial cubed ? Our teacher said we need to use FOIL method.?
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
71477
6x^2 + dx - 12 would need to factor into two binomials.  If you know that one is (2x - 3), then the first term of the other would have to be 3x, in order to make the 6x^2 term.  In order to get -12 as the constant term, we would multiply -3 x +4.  \n\nSo the other factor is (3x + 4)\n\nUsing the FOIL method for multiplying two binomials, we get\n(3x + 4)(2x - 3) \n= 6x^2 - 9x + 8x - 12\n= 6x^2 - x - 12\nSo d must be -1.
889348
As the previous poster said, you don't have enough information.  If you have n-sided polyhedra, are the sides numbered 1 to n?  Is the expected value in reference to a throw of the dice?  If so, are we talking about two dice?\n\nFor two dice with the sides numbered 1 to n, the answer is n + 1.
694740
Is this for the sum of the first n cubes formula? I can tell you what your problem is right now: that two should be _inside_ the brackets. Otherwise, this formula doesn't work.\n\nNow, using the correct formula:\n(k(k+1)/2)² + (k+1)³\nWe start by actually multiplying out the stuff in the parentheses:\n((k²+k)/2)² + (k+1)³\n(k^4 + 2k³ + k²)/4 + (k³ + 3k² + 3k + 1)\nNow we give these fractions a common denominator:\n(k^4 + 2k³ + k²)/4 + (4k³ + 12k² + 12k + 4)/4\nAnd add:\n(k^4 + 6k³ + 13k² + 12k + 4)/4\nFactoring the polynomial:\n(k+1)²(k+2)²/4\n((k+1)(k+2)/2)²\nThus completing the proof.
449249
ABOUT THE BINOMIAL THEOREM I AM TEEMING WITH A LOT OF NEWS\n\nWITH MANY CHEERFUL FACTS ABOUT THE SQUARE ON THE HYPOTENEUSE\n\n(W. S. Gilbert, I am The Very Model of a Modern Major General)\n\nI am assuming we are dealing with what we know is a PRIMITIVE Pythagorean triple ie one like (3,4,5) or (5,12,13) or (8,15,17) or (20,21,29) in which there is no common factor between the three sides, NOT one like (6,8,10) which is divisible by 2.\n\nThere are tables published of the smallest triples, in ascending order of hypooteneuse, so you may be able to look it up and not need to calculate it. \n\nhttp://www.math.rutgers.edu/~erowland/tripleslist-long.html\n\nHowever, two points.  \n\nNot every integer can be a hypoteneuse. There are gaps in the table eg after (3,4,5) the next smallesr hypoteneuse is 13 (in (5, 12, 13). There are no primitive triples with a hypoteneuse of 6, 7, 8, 9, 10, 11 or 12. And therefore if you were given a hypoteneuse of 12, there are no integer solutions to it.\n\nSecondly, there are cases where a number is hypoteneuse to more than one triple. 65 is the lowest such number and 85 is the next.\n\n(16, 63, 65) and (33, 56, 65)\n(13, 84, 85) and (36, 77, 85)\n\nThese are known as siblings. Both give correct solutions and you can't say one is right and the other is wrong. Here are a couple of special cases, with more than two solutions:\n\n(47, 1104, 1105)\n(744, 817, 1105)\n(576, 943, 1105) \n(264, 1073, 1105)\n\nThere are EIGHT siblings with a hypoteneuse of 32045, SIXTEEN siblings with a hypoteneuse of 1185665, THIRTY-TWO siblings with a hypoteneuse of 48612265, AND SIXTY FOUR siblings with a hypoteneuse of 2576450045.\n\nHere are rthe eight siblings with a hypoteneuse of 32045:\n\n(2277, 31964, 32045) \n(8283, 30956, 32045)\n(17253, 27004, 32045) \n(21093, 24124, 32045)\n(22244, 23067, 32045)\n(15916, 27813, 32045)\n(6764, 31323 , 32045)\n(716, 32037, 32045)\n\nLacking access to tables, for small numbers, you may be able to work it out, by trial and error, using the following properties\n\n(a) the hypoteneuse must be the biggest side and must be an odd number (to be a primitive triple)\n\n(b) one of the legs will be an odd number and the other an even number\n\n(c) As a result, the difference between the two small sides will always be an odd number. In fact it is either a prime number (eg 1, 7 and 17 feature, but 11, 13 and 19 do not) or a power of a prime number (49 is the lowest such).\n\n(d) the sum of the two legs must be bigger than the hypoteneuse\n\n(e) one and only one of the legs must be divisible by 3\n\n(f)  one and only one of the legs must be divisible by 4\n\n(g)  one and only one side (which might be the hypoteneuse) must be divisible by 5\n\n(h) follows from (e) (f) and (g)\n\n(h) the product of the three sides is always divisible by 60\n\nThus if you knew the hypoteneuse and could see it was not divisible by 5, then it follows the legs must, between them, be divisible BY 3 4 AND 5\n\nSo if you knew the hypoteneuse is 61, it seems obvious to try one of the legs as 60 and see if that works (it does, (11, 60, 61) is a triple. \n\nIn fact you can work out the third member is 11 by adding 60 and 61 to see if it is a perfect square (it is 121) and taking the square root of it, if so. (another property of triples whose two largest sides differ by 1).\n\nGiven that 17 is a hypoteneuse and knowing one of the other sides must be 3 x 4 or 3 x 5 (4 x 5 would be larger than 17) you have a choice of (12, x, 17) where x = 5 or 15 (has to be odd as 12 is even) or (y, 15, 17) where y is 4, 8, 12 or 16)\n\nIt turns out to be (8, 15, 17)\n\nAgain, knowing a property of all triples where the hypoteneuse and the larger leg differ by 2, there is a shortcut, that leads to 8 ... take the average of the two longer sides, multiply by 4 and take the square root. (thus 16 => 64 => 8)\n\nThis holds true for \n(12, 35, 37)\n(16, 63, 65)\n(20, 99, 101) etc\n\nSo it is possible to make intelligent guesses at the
677899
yes we did it in science class.  I forget how exactly but you fold the paper and then start cutting tiny rows but everyother row you change directions and then open it up when you are done and you get a loop of paper like a hoop and then you can step through it.  E-mail me for more details at fairy_dust_xo@yahoo.com  Hope this helps!
977457
1 to 64k.. so there exists about 40 cubes (1^3, 2^3...  39^3, 40^3)..  and sum of 4 cubes. that's like. "40 choose 4" (combinatorics) so that comes out to be about 91390 combinations of 4 cubes and then you just find the cube roots of each. each should have its own cube root. and 91 thousand isn't too bad for an average computer to handle. i suggest you program a computer to do it for you. it wouldn't be hard, just basic programming language.\n\nif you cant do that, then i suggest you try to narrow down what you're looking for...
926476
You could start here and write a program:\nhttp://www.math.nmsu.edu/~breakingaway/Lessons/spiral_K_1/spiral_K.html
884978
There are numerous solving strategies... too long to explain here.  Most people work a layer at a time (top third, middle third and bottom third).  There are other strategies that align the corners then do the middle squares.\n\nI would just do a search of the web and you'll find lots of information on solving the cube, speed cubing, blind cubing, etc.
712544
ok so you know 15 can only be broken down to 5*3, -5*-3, 15*1 and -15*-1\ntherefore when you break it up\n(x+3)(x+5) using FOIL (I assume you know how to do that) k would equal 8. For (x-3)(x-5) k would be -8. Do the same for the set of (x+15)(x+1) and (x-15)(x-1)
220197
The sum or difference of 2 cubes can always be factored the way everyone is telling you.  It does not have to be set equal to something to factor it.  For the difference of cubes, a^3-b^3, the first set of parentheses is (a-b) and you will use this to find what goes in the second set.  Square the first (a^2), multiply the 2 together and change the sign (+ab), square the last (b^2), so a^3-b^3 = (a-b)*(a^2 + ab + b^2).\n\nTo address some of your questions in the details - you cannot solve for both variables with only 1 equation.  You will need some other information - another equation, the value of a or b, etc. to solve for both, even if you set it equal to 0.\n\nWhen a=2, b=4 and a^3-b^3=p, you CAN still factor it into (a-b)(a^2+ab+b^2)=p, but it's not necessary.  Since you are given values for a and b, it's much easier to just plug them in and solve for p.  Let's try it both ways, though.\n\na^3-b^3 = p; a=2,b=4\n(2)^3 - (4)^3 = p\n8-64 = p\np = -56\n~~~~~~~~~~~~~~~~~~~~\n(a-b)(a^2 + ab + b^2) = p; a=2, b=4\n(2-4)(2^2 + 2*4 + 4^2)=p\n(-2)(4+8+16) = p\n-2(28)=p\np= -56
845986
I dont know what you mean by the first half, but (x+2)(x+1) is simple.\n\n\nUse the FOIL method.\n\nFront\nOutside\nInside\nLast\n\nso\n\nFront\nX*X=X^2\n\nOutside\nX*1=X\n\nInside\n2*X=2X\n\nLast\n2*1=2\n\nSum them up\nX^2 + X + 2X + 2\n\nSimplify\nX^2 + 3X + 2
892755
FOIL stands for First term, Outer term, Inner term, & Last term. So, take for example, (X+3) (X+5).\n We're multiplying these terms, so multiply:\n                           \n                              First, \n                         X * X= X^2\n                           \n                            Outer,\n                         X * 5= 5X\n                         \n                           Inner,\n                         X *3= 3X\n  \n                           Last,\n                         3* 5 =15\n             \n                Now combine the terms\n                     X^2+5X+3X+15\n              \n              Now combine LIKE terms (5X+3X=8X)\n                       X^2+8X+15\n \n\n                And your done =)
707089
Fold the circle in half\n\nStart to fold the half in half again, but only crease it near the 90 degree angle formed; this identifies the center of the circle.  Open it back up to where it's only folded in half once.\n\nFold the half into thirds with each of the 2 folds running through the center of the circle\n\nFold once more in half, again with the fold running through the center of the circle.  You Now have something close to a skinny triangle\n\nHolding the skinny triangle pointing up, cut from just above the bottom right corner to about midway of the opposite long side.  \n\nBTW, you don't have to actually start with a circle.  You can just start with a regular sheet of paper if you adjust the cut for the correct results. Mine came out nice!
172667
You use FOIL (First, Outside, Inside, Last).\n\nF - 2x*3x=6x^2\nO - 2x*-4=-8x\nI - 5*3x=15x\nL - 5*-4=-20\n\nExpanded. 6x^2-8x+15x-20\nSimplify by joining like items - (-8x+15x=7x)\n\nFinal answer. 6x^2+7x-20
900433
Discounting rotations and mirror images, there is only one way to make a magic square using the digits 1 through 9.  \n\nFirst, we can prove the 5 has to go in the middle:\n\nLet the numbers in the top row from left to right be a, b, and c.  Let the numbers in the middle row from left to right be d, e, and f.  (Note e is the middle box.)  Let the numbers in the bottom row from left to right be g, h, and i.     \n\nWe know a + b + c = 15 and g + h + i = 15\n\nWe also know a + e + i = 15, b + e + h = 15, and c + e + g =15.  Therefore, (a + e + i) + (b + e + h) + (c + e + g) = 15 * 3 = 45.  We can rearrange this to (a + b + c) + (g + h + i) + 3e = 45.  Since a + b + c = 15 and g + h + i = 15, that reduces to 15 + 15 + 3e = 45, or 3e = 15.  Therefore e (the middle box) must equal 5.   \n\nKnowing the middle box is 5, try 9 in a corner and you'll notice a problem - the 6, 7 and 8 can't all be placed in the square without the row or column with the 9 exceeding 15.  So, the 9 has to occupy a side box.  Since we are discounting rotations and mirror images, we can arbitrarily select box b for the 9.  This forces the placement of the 1 in box h.   \n\nWith the 9 and 1 now placed, the 8 and 6 have to go in the row with the 1.  So, we'll put the 8 in box g and the 6 in box i.  (We still haven't done anything to create a second square if rotations and mirror images are discounted.)  That forces the placement of the final four numbers.
444843
It's a perfect square trinomial.  Take the square root of the first term and the square root of the last.  If you multiply them and then double it, and it equals the middle term, then it's a perfect square trinomial.\n\nIn this case, it factors into (x/3 + 6)^2.  Using FOIL, you'll see that it expands into your original expression.