how do you expand a polynomial? Any suggestions?
96075
you are talking about "Pascalles (not sure about spelling) Triangle." It goes 1,11,121,1331 and so on.
942305
Okay. You have a few things that you have to consider.\n\nNegative exponents: moves it to the denominator\nFractional exponents: the denominator tells you what the root is, the numerator tells you what the root is rasied to\n\nSo: (x+1)^-2/3 rewritten would be:\n\n1/[cuberoot(x+1)^2]\n\nwhen x is seven you have\n1/[cuberoot(8)^2]\n1/[2^2]\n1/4
703053
so the Q is basically asking for two binomials which, if multiplied will end up as a polinomial...\nAs a matter of fact any two binomials can form a polynomial.\n(2X+3) (x-7)...this is an e.g.
941571
1) f(x) = A(x-1)(x-2)(x-4) for some A. Substitute in 0 to get -16 = A*-1*-2*-4, so A = 2. So 2(x-1)(x-2)(x-4).\n2) Whenever you have a polynomial with real coefficients and a+bi is a root, you also have a-bi is a root. So you know the 3 roots are 4, 3+i, 3-i. Thus f(x) = (x-4)(x-(3+i))(x-(3-i)). Multiplying those last two together, you get x^2 - x(-3-i-3+i) + (3+i)(3-i) = x^2 - 6x + (9-i^2) = x^2-6x+10. So f(x) = (x-4)(x^2-6x+10).\n3) I don't think you wrote the question down properly. Expanding the left hand side gives 3x^3 + (-a-27)x^2 + (9a+42)x - 14a. You need a=5 for the first part to work, but that doesn't give what you wrote.
594239
ok so you have p^2 + p, you can factor out p from both parts of the expression, it will end up like this: p(p + 1).  Hope that works.
484948
A 3rd degree polynomial can have a maximum of 3 real roots, but it must have at least 1 real root I believe.
854560
(x+2)(x²-2x+4) is the correct answer. It expands to x³ + 8.\n\n(x+2)³, while tempting, is NOT the correct answer; its expansion is x³ + 6x² +12x + 8, NOT x³ + 8.
208169
Factoring, I guess, but it really depends on the type of polynomial question. Can you give a sample?
891987
The first step is to apply the Rational Roots Test to the polynomial to get a list of values that might possibly be solutions to the polynomial equation. You can follow this up with Descartes' Rule of Signs, if you like, to narrow down which possible zeroes might be best to check. Of course, if you've got a graphing calculator, it's a good idea to do a quick graph, since x-intercepts of the graph are the same as zeroes of the equation. Seeing where the graph looks like it crosses the axis can quickly narrow down your list of possible zeroes.\n\nOnce you've found a value you want to test, you use synthetic division to see if you can get a zero remainder. If you get a zero remainder, you've not only found a zero, but you've also reduced your polynomial by one degree. Remember that synthetic division is, among other things, division, so checking if x = a is a solution is the same as dividing out the linear factor x – a. This means that you should not return to the original polynomial for your next computation (for finding the other zeroes); you should instead work with the output of the synthetic division. It's smaller, so it's easier to work with.\n\nYou should expect to get some complicated solutions (that is, solutions containing square roots or complex numbers, or both). You will find these zeroes by applying the Quadratic Formula for the final quadratic factor.\n\nFind all the zeroes of the following polynomial: 2x5 + 3x4 – 30x3 – 57x2 – 2x + 24 \nFirst, I'll apply the Rational Roots Test--\n\nWait. Actually, the first thing I'll do is check to see if x = 1 or x = –1 is a root, because these are the simplest roots to test for. This isn't an "official" first step, but it can often be a timesaver, because you can just look at the powers and the numbers. When x = 1, the polynomial evaluates as 2 + 3 – 30 – 57 – 2 + 24 = –60, so x = 1 isn't a root. But when x = –1, I get –2 + 3 + 30 – 57 + 2 + 24 = 0, so x = –1  is a root, and I can do the synthetic division with x = –1 right now.\n\n\n\nThis leaves me with the smaller polynomial 2x4 + x3 – 31x2 – 26x + 24. (Since I've divided out the factor x + 1, I've reduced the degree of the polynomial by 1. That's how I know this is a degree-four polynomial.) Now I'll apply the Rational Roots Test to get a list of values to look at:
221826
1) \nxy - x + 4y - 4\nx(y-1) + 4(y-1)\n(x+4)(y-1)\n\n2)note: x^3 means x cubed; x^2 means x squared\nx^3 -6x^2 + 2x - 12\nx^2(x-6) + 2(x-6)\n(x^2 + 2)(x-6)\n\n3)\n6tx + r - 3t - 2xr\n3t(2x - 1) - r(2x - 1)\n(3t - r)(2x - 1)\n\n4)Are you sure it wasn't -x^2 - 2x - xy - 2y?\n-x^2 - 2x - xy - 2y\n-x(x + 2) - y(x + 2)\n(-x - y)(x + 2)\n\nTo "factor by grouping," you must split the original polynomial that you are given into two different binomials by "grouping" them together.  This is done with perenthesis in the problems above.  Then, you factor out the "greatest common factor" from each of the binomials.  After factoring, the two quantities still within the perenthesis should be the same.  You could call these quantities the "common binomial."  Lastly, you factor out the common binomial to get the final answer.
381493
It's set up like a long division problem.  Let's say you have x^2 - 6x - 27 divided by x + 3.  You'd set up like this:\n\nx + 3 ) x^2 - 6x - 27\n\nFor polynomial division, just focus on the variables and the exponents.  To start this one, look at how many times x goes into x^2: x times.  So x would go on top of the dividing symbol, and then you'd multiply the value on top by the value on the outside and subtract, just like a long division problem.  So you'd end up with\n\nx^2 - 6x - 27\n-(x^2 + 3x)\n----------\n-9x - 27\n\nRemember that, just like long division, you're subtracting, so the problems becomes x^2 - 6x - 27 - x^2 - 3x if you wrote it out this way.  The x^2 terms cancel out, and you'd have to bring down the -27.\n\nNow repeat the process - how many times does x go into -9x?  -9 times.  So the -9 goes on top (making your top x - 9) and your problem look like this:\n\nx^2 - 6x - 27\n-(x^2 + 3x)\n----------\n-9x - 27\n-(-9x - 27)\n----------\n0\n\nSince you have no remainder, your answer is x - 9.  And if you're ever unsure, you can always use FOIL to see if you get your original trinomial:\n\n(x + 3)(x - 9) = x^2 - 9x + 3x - 27\n= x^2 - 6x - 27
961741
Im not going to prove it completely but I'll send you on your way..\n\nLook at a cubic, ax^3 + bx^2 + cx + d \nwe want to know for what value of x is this cubic = 0.\nLet's say r is a root.\nSet up the synthetic division:\n\nr || ..... a ........... b .............. c .................. d\n........................ ar .......... ar^2 + br ......... ar^3 + br^2 + cr\n───────────── ────────── ──────────\n......,....a......... ar+b ..... ar^2 + br + c......|| ar^3 + br^2 + cr + d\n\nin synthetic division, we say that if this last term,\nar^3 + br^2 + cr + d\nequals zero, then r is a root.\nNotice that the term is just the polynomial evaluated at x=r\n\nAnother way to look at it is:\nf(x) = ax^3 + bx^2 + cx + d\n= x( ax^2 + bx + c ) + d\n= x( x (ax + b) + c ) + d\n= x( x ( x (a) + b ) + c ) + d\nor,\ntake a,\nmultiply that by x, add b\nmultiply that by x, add c\nmultiply that by x, add d\nif that total is 0, then x is a root.\nThis is just what you do in synthetic division.
908083
Tried to do it algebraically, that was a booger.\n\nI plugged that into mathematica and it gave me an error normally generated when the inverse of a function is needed to represent a solution and the solution variable occurs in more than one argument of that function.\n\nThat one might not have a solution?
913334
x^2 + 2x + 7 Remainder 10\n\ni can add the steps if you don't think i did it by hand
241036
I'm assuming you're looking for information on factoring polynomials since it's a College Algebra courese.  \nThis website is very helpful, with the steps for factoring laid out, and it also provides websites you can use if you need more information.  Good luck!\n\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm
294579
First multiply p times the second polynomial.\np(p^4 - 2p^3 + 3) = p^5 - 2p^4 + 3p\nthen multiply 2 times the second polynomial.\n2(p^4 - 2p^3 + 3) = 2p^4 - 4p^3 + 6\nNow add these two together.\n\n(p^5 - 2p^4 + 3p) + (2p^4 - 4p^3 + 6)\n= p^5 - 4p^3 + 3p + 6
562954
You simply divide each term in the polynomial by that integer.. For your example here it goes..\n\n(3x^2+8x-2) divided by 2 \n\n> {3(x^2)/2} + {8x/2} - {2/2}\n\n>> Answer: 1.5 x^2 + 4x -1    or   (3/2)x^2 + 4x -1