"Adding, Subtracting,and Multiplying Polynomials?"?
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.
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.
751280
multiplication\n\n\nDoug
208764
This is actually elementary math. If you remember the order of operation, it helps. \n\nStep 1: Do the part of the problem in the parenthesis.\n  Example: (4/3)= 1.3333333333 (you get where this is going)\n\nStep 2: Take that answer and multiply it by the named multiplier.\n  Example: 1.33333 * 12= 16\n\nStep 3: Take the result of that transaction and devide it by the named devisor. For this method don't use the fraction way, and don't use a calculator. Do a basic devision problem like you were taught in the 3rd grade using the "bracket."\n  Example: 16/13 = 1  (13 goes into 16 1 time)\n\nStep 4: Subtract the 13 from the 16 and it will give you your remainder.\n  Example: 16/13 = 1   16-13= 3\n\nThe answer when written correctly isn't with a decimal but would look like this- 1 R3.\n\nAlso for a little note for this- Order of operation:\nP (parenthesis)\nE (exponents)\nM (multiply)\nD (devide)\nA (addition)\nS (subtraction)\n\nHope this helps!
1109323
The others are correct:  3x is one such example.\n\nIf you want to be pedantic about it, there are an infinite number of functions that fulfill this.  Just add any constant:\ny=3x+c.
890645
Wow, there is a lot here!  I'll start with some tips for solving algebra problems.  The first thing you should always remember is to play the equal game.  Whenever you add, subtract, multiply, or divide, you should do the same to the other side.  The other tip is to use the order of operations when simplifying.  Remember the sentence: Please excuse my dear Aunt Sally.  The initial letters of each word stands for the order of operations.  Do the Parenthesis first, followed by the Exponents, then Multiplication or Division, and Addition and/or subtraction next.  Finally, you should always simplify first using the order of operations before you start shifting numbers from one side of the = to the other.  \n\nI'm not going to work all of these problems, but I'll show you a few to get you started.\n\n1)\n3x-7 = 2   Add 7 to both sides\n3x = 9   Divide by 3\nx = 3\n\n4)\n6x - 3 = 2x - (4-3)/3   Simplify the stuff inside the parenthesis first\n6x - 3 = 2x - 1/3  Add 3 to both sides\n6x = 2x - 5/3  Subtract 2x from both sides\n4x = -5/3  Multiply both sides by 3\n12x = -5  Divide both sides by 12\nx = -12/5\n\nI hope this helps.  Good luck!
913484
are you talking about a polynomial :\na + bx +cx^2 + dx^3 +...\n\nbecause those a,b,c,d... do not appear any where on the graph... you have to look at the intercepts and the extreama, and the inflection points to get any good idea about the shape of the graph
54176
The instruction is to write a polynomial function with the zeros 0, 1, and 1+i. Please keep in mind that when a zero consists radicals or imaginary numbers, the conjugate of that zero is also a zero. For exmaple, here 1+i is a zero. Then 1-i must also be a zero.\n\nThis problem is like working backwards. Instead of factoring the polynomial, you are asked to do it backwards, starting with the zeros. Since 0, 1, and 1+i and 1-i are zeros, you would expect the last step of the factoring to look something like this:\n\nx(x-1)[x-(1+i)][x-(1-i)]\n\nNow simplify this expression a bit. It becomes:\n\nx^2-x[(x-1-i)(x-1+i)]\n\nSimplify this expression even further and you get:\n\nx^4-3x^3+4x^2-2x
415757
I hate to tell you, but I don't think there is an algebraic method for arriving at a value for x. Sometimes, this can be the case, in which case all you can do is find a numeric approximation, here around x=1.0587...\n\nI'll see if another user can prove me wrong, but I'm going to go with there's no way to solve this problem using analytic methods.
555590
Just multiply:\n(X/3)(3/4 x) -(x/3)(3/5) + (3/4)(3/4x) - (3/4)(3/5)\n\nand simplify from there
917477
where else?\n\nhttp://www.factoring-polynomials.com/
1137023
(2x-9)(2x+9)\n\na guess... its been a long time since I did that stuff
78119
I guess you must factorize them.\nAll the polynomials you wrote are to be factorized by partial factorization.\nI think you know the general rule.\nA(x)P(x)+A(x)Q(x) + B(x)P(x) + B(x)Q(x) = \n---> factorize between the 1°-2° and 3°-4° <---\nA(x)[P(x) + Q(x)] + B(x)[P(x) + Q(x)]\nYou get the same binomial term [P(x) + Q(x)] which is a common factor. Factorize:\n[P(x) + Q(x)]*[A(x)+B(x)]\n\nFor example, in the first:\nA()=a\nB()=b\nP(x)=2x\nQ(y)=3y\nYou get \na(2x+3y)+b(2x+3y) = (2x+3y)(a+b)\n\nIn all the others you must rearrange the terms to use the above steps:\nin xy+32+8x+4y the repeating patterns are 8x + xy and 32 + 4y ....\n\nThink about it and good work
913334
x^2 + 2x + 7 Remainder 10\n\ni can add the steps if you don't think i did it by hand
806827
depends what your trying to sum from the polynomial... what are you trying to get, theres to many thing just to say sum of given polynomials
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!
32760
You can use the AC Method to answer this question:\n\nFirst, rewrite the expression so it's in the form Ax^2 + Bx + C where A,B, and C are constants (numbers)\n\n16x^2 + 8x - 15\n\nThere's no common factors we could take out first, so...\n\nMultiply A x C, which in this case is 16 x -15 or -240.  Then look for factors of this number that add up to B.  You're looking for two numbers that fill in the blanks below:\n__ x __ = -240\n__ + __ = 8\n\nPrime facorization will help give you some ideas, or make an ordered list of numbers that multiply to -240.  \n240 and -1\n120 and -2\n80 and -3\n60 and -4\n48 and -5\n40 and -6\n30 and -8\n24 and -10\n20 and -12\n\nSee how the last pair of numbers satisfies the conditions?\n\nRewrite your polynomial, substituting the 2 values in for B to have 4 terms:\n\n16x^2 + 20x - 12x -15\n\nThen factor by grouping\n(16x^2 - 12x) + (20x - 15)\n4x (4x - 3) + 5 (4x - 3)\n(4x + 5)(4x - 3)\n\nThis method works for most polynomials of the form Ax^2 + By + C!  \n\nYou can check your answer of (4x + 5)(4x - 3) by using the FOIL method and see that it does work.\n\nGood luck!