how do I do polynomials in Alegbra 1 and the foil method?
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
142158
Use the FOIL method, since this is really (6x+5)(6x+5)\n\nFirst terms (6x)(6x)=36x^2 +\nOutside terms (6x)(5)=30x +\nInside terms (5)(6x)=30x +\nLast terms (5)(5)=25\n\nFor a total of  36x^2 + 60x + 25
584142
To simplify radical rational expressions, multiply top and botom by the conjugate of the bottom.  This is found by switching the operation from + to - or - to +.  This will eliminate the radical on the bottom becaise of the way "FOIL" works.  Here, multiply by sqrt(3)-2 over sqrt(3)-2:\n\n\nTop: (sqrt(3)-5)(sqrt(3)-2)\n= 3 - 7sqrt(3) + 10 \n= 13 - 7sqrt(3)\n\nBottom: (sqrt(3)+2)(sqrt(3)-2) \n= 3 - 2sqrt(3) + 2 sqrt(3) - 4\n= 3 - 4\n= -1\n\nAnswer: (13 - 7sqrt(3))/-1, or -13 + 7sqrt(3)
10610
X^2-9X+18\n\nShort and simple just factor it FOIL
168995
plynomial expression could be factorized as the product of \n(x-r) where r is a root (a zero) of the polynomial soyou must get the roots first\nMany numerical methods exists
589334
http://aleph0.clarku.edu/~djoyce/mathhist/mathhist.html
638676
1.)\nstep one: if they are zeroes, use opp. \nexample....\n (x-2) (x+9) (x-7)          if you solve for x it should be opposite of given zeroes.\n\nstep two: FOIL the first two...\nyou'll get   x^ +9x -2x -18 which simplifies to\n\n(x^2 + 7x -18) (x-7) \n\nstep three: FOIL these two together...\nyou'll get     x^3 +7x^2 - 18x -7x^2 -49x + 126\n\nwhich simplifies to  \n   x^3 - 67x +126\n\n\n2.) I don't know about the second one... I could solve it if I spent some time, but start with...\n (x - 4i) (x+5) (x + 2i)\nthan just do what I did in the problem above, factor the first two, than factor your answer with the last one... and remember that i^2 is equal to 1...
661206
For the vertical method, you want to multiply out the problem one term at a time, and line the terms up over each other according to what variables are in them.  Like this:\n\n(2r-3s)*(4r^2+6rs+9s^2)\n\n(+2r)*(4r^2+6rs+9s^2) = 8r^3 +12r^2s +18rs^2\n(-3s)*(4r^2+6rs+9s^2) = ........-12r^2s -18rs^2 -27s^3\n.............................. = 8r^3....................-27s^3\n.............................. = 8r^3-27s^3\n\nI had to put periods in to make it space out right.\n\nSo just line the terms up over each other so it's easy to see which ones are the same type, and in this case, they cancel each other out.\n\nThe answer here is 8r^3-27s^3.\n\nGood luck!
964615
Please type smith's algorithm for polynomial matrices at yahoo search you get links hope fully one of them should provide you with a solution.
387063
X=[-(30y-433)]/1232\nY=[-(1232x-433)]/30
638682
Domain ~ x values\nRange ~ y values\n\nYou have to work the function and figure them out.
483089
First write your equation using hyperbolic function:\n\ny = a cosh(x/a)\n\nDifferentiate, dy/dx = sinh(x/a)\n1 + (dy/dx)^2 = 1 + sinh^2(x/a) = cosh^2(x/a)\nSQRT (1 + (dy/dx)^2 ) = cosh(x/a)\n\nIntegrate from x = 0 to x = a:\n\nAnswer = [asinh(x/a)]@x=a  - [asinh(x/a)]@x=0\n       = asinh1 - asinh0\n       = asinh1\n       = a(e - (1/e))/2\n\nRefer to the following website for details on the catenary
882676
sure it does\n\nx^2-8x+4x-32\n\nx^2-4x-32\n\nRemember FOIL\nFirst Outer Inner Last
1079811
It *is* awfully hard to write algebra expressions on Yahoo Answers, isn't it? Oh well, we do what we can.\n\nAddressing your first question, let's write it out like a long division problem.  You'll have to imagine the horizontal bar above the larger polynomial, because I can't format it properly here:\n\nx^2 + 3x - 2  )  x^4 + x^3 - 5x^2 + 13x - 6\n\nOur first step is to look at the first term in each polynomial, and divide the smaller into the larger.  x^4 / x^2 = x^2, so the first term we write in our quotient (above the bar) is x^2.  You can write it directly above the x^4, if you wish, or above the 5x^2 term, your preference.\n\nNow, we're going to multiply that "partial quotient" by the entire divisor (x^2 + 3x - 2), which gives us x^4 + 3x^3 - 2x^2.  Write that "partial product" below the first three terms of the original dividend (x^4 + x^3 - 5x^2).  Our next step is to subtract the partial product: (x^4 + x^3 - 5x^2) - (x^4 + 3x^3 - 2x^2) = -2x^3 - 3x^2.  We then drop down the next term from the dividend, the 13x term, and repeat:\n\n-2x^3 / x^2 = -2x (add that term to our quotient)\n\n-2x * (x^2 + 3x - 2) = -2x^3 - 6x^2 + 4x\n\n(-2x^3 - 3x^2 + 13x) - (-2x^3 - 6x^2 + 4x) = 3x^2 + 9x\n\nFinally, we drop down the last term from the dividend, -6, and repeat once more.  I'm sorry that I can't write this out the way your math book can... I really hope Yahoo Answers creates some sort of way of inputting formulas, or at least monospacing text entry!\n\n3x^2 / x^2 = 3 (so we add a +3 term to our quotient)\n\n3 * (x^2 + 3x - 2) = 3x^2 + 9x - 6\n\n(3x^2 + 9x - 6) - (3x^2 + 9x - 6) = 0... no remainder!\n\nSo our final quotient is x^2 - 2x + 3, with no remainder.\n===\nNow, when doing the 2nd problem, you'll want to write the larger polynomial, the dividend, as x^4 - 3x^3 + 0x^2 + 6x - 5.  We incorporate that extra 0x^2 term so that we never find ourselves trying to subtract unlike terms.\n\nSorry there's no better way to write this out on Yahoo Answers.  Hope that helped anyway!
175327
Try this, maybe?\n\nhttp://mathforum.org/library/drmath/drmath.elem.html
72362
I forgot how to do those, but it's some sort of trig identity isn't it?  Like\n\nsqrt (1-tan [x])\n\nProbably not though.
950045
Correct formula should be as follows:\n\nE = 1/2*I*w^2