How do you factor polynomials on the TI-83 Plus calculator?
342302
Below is the manual. I could not find it maybe you can.\n\nGood Luck I will still look and update with anything I can find.
259003
Use the quadratic formula.  It works everytime!\nx = (-b plus or minus sprt(b^2 - 4ac))/2\nwhere\nax^2 + bx + c = 0\nSome of these will factor, however not all of them will.
175207
Next time use ^2 instead of typing out "to the 2". Anyway, write it in standard form, so that the variables are in descending order.\n\n14w^3 + 49w^2 - 8w - 28\n\nYou can't factor this. If it was 7w instead of 8w, then you could factor out a 7 by dividing all the coefficients by 7 and pulling it outside the parentheses. Alternatively, you could factor the first two terms separately from the second two:\n\n7w^2(2w + 7) - 4(2w + 7)\n\nNow it's in a great form for factoring. Group them.\n\n(2w+7)(7w^2-4)\n\n---\n\nDo the same thing for the second example.\n\n44t^3 - 4t^2 + 66t - 6\n\nFor this one, you can factor out a 2.\n\n2(22t^3 - 2t^2 + 33t - 3)\n\nOr, again, you can factor the terms separately:\n\n4t^2(11t - 1) + 6(11t - 1)\n\nGroup.\n\n(11t-1)(4t^2+6)\n\n\n\nHope this helped.
26326
A procedure for finding the quadratic factors for the complex conjugate roots  of a polynomial P(x) with real coefficients.\n\n[x-(a+ib)][x-(a-ib)]=\nx^2+2ax+(a^2+b^2)=\nx^2+Bx+C.
781910
not possible to factor easily ... try the quadratic formula.
811639
Calculator Library help?I don't know!
882816
You can't do that on a calculator. What you can do is use the calculator to get the graph drawn then use your algebra to figure out where to shade. \nLooking at the graph, you know that it's a parabola with the vertex (4, -3). test the point (0,0) outside of the parabola.....\nsince you get\n0 is greater than or equal to 0 - 0 + 13\nsince that is a false mathematical statement, you know that you shade inside of the parabola....\nlets test a point inside/on the parabola and see, like (4, -3)\n- 3 is greater than or equal to 16 - 32 + 13\n-3 is greater than or equal to - 3\nthat is true so you now know to shade inside your parabola\nthat's how you do problems like that w/a calculator\nsince your calculator can do all of the operations, it's really easy
684474
You have to factor each part first, then cross off (cancel) identical factors from the numerator and denominator of either expression.  If the factors are exact opposites, like x-8 and 8-x, cancel and put a -1 in the numerator.\n\n\nIn your example only the first expression is factorable:\n(x-8)(x+3)/(2x)(3+x) times 5x^2/(8-x)\n\nThe (x-8) cancels the (8-x) to become -1, then (x+3) cancels the (3+x), and the x from the (2x) cancels one of the x's from (5x^2)\n \nAnswer: -5x over 2
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:
230941
Write the coefficients of the variables\n-4__ 1 -3 -7 -31\n_____-4 28 -84\n____1 -7 21 -115\nSynthetic division is hard to type.\nThe remainder is the last number in the bottomw row, so\nP(-4) = -115\nTo check, just evaluate the polynomial P(x) at -4.  That is, replace all the x's with -4\nP(-4) = (-4)^3 - 3(-4)^2 - 7(-4) -31\n= -64 - 48 + 28 - 31\n= -115\nThus, this verifies that the correct answer is -115.
526107
You are only allowed to factor if the equation is equal to zero; your's is equal to 4.  You need to solve "x^3 - 6x^2 +9x - 4 = 0"  using synthetic division.
670366
Yes.  There are algorithms for factoring polynomials up to 4th degree.  These algorithms are guaranteed to work, however in the case of the 5th degree polynomials and above, there is no algorithm that is guaranteed to find the solution.  In part this is because the roots can be irrational numbers.  I was proved by Abel in 1824 that no such algorithm can even exist.\n\nThere are iterative methods to factor polynomials of 5th degree and above, but they provide approximations, they are not guaranteed to find the correct answer.  The iterative methods are quite complicated and are equivalent to finding eigenvalues of matrices.\n\nIf you are in high school or early college, all the problems they will assign you should be solvable with either the linear, quadratic, or cubic equations.
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!
945505
{(x^2+5x+4)/(x+3)^2}/(x+1/(x^2 -9))\nFactor everything that you can factor.\nx^2 + 5x + 4 = (x+1)(x+4)\nx^2 - 9 = (x+3)(x-3)\nYou have\n{(x+1)(x+4)/(x+3)^2} * {(x+3)(x-3)/(x+1)}\nYou can cancel the (x+3)'s and the (x+1)'s leaving you with\n(x+4)(x-3)/(x+3)
154296
Megami of Nekro is mostly correct, but I will try to clarify the situation a bit. \n\n1. factorize, or factor\n\nIf you have an algebraic expression that can be described as the product of some simpler expressions, to rewrite the expression as this product is to factorize it. \n\nSuppose you want to factorize the polynomial x^2 -3*x. In the case of polynomials of some degree, we regard polynomials of lower degree as simpler. In this case, the polynomial x^2-3*x is of degree 2, because the biggest power of x is 2 (the x^2 term). The lower degree polynomials in this case are of degree 0, called numbers, and of degree 1, called monomials. We can write the polynomial x^2-3*x as the product of lower order polynomials in the following way:\n\nx^2-3*x=x(x-3)\n\nThe two monomials x and (x-3) are factors of the polynomial x^2-3*x, and by expressing the polynomial as a product of two lower order polynomials, we have "factorized it" or "factored it". \n\n\nHere is another one:\n\nx^4+2x^2+1 is factorized as \n(x^2+1)(x^2+1)\n\nWith grade 8 mathematics, you cannot go any farther than this. We replaced a 4th degree polynomial with the product of its two factors, two 2nd degree polynomials (remember, we have decided that lower degree polynomials are simpler)\n\n\nAside: It is also interesting to note that one can factor whole numbers sometimes into a product of smaller whole numbers, which we regard as simpler. Therefore, \n\n10=5*2\n\nand 24=2*2*2*3\n\nThis is also called factoring. Numbers that can be expressed this way are called composite numbers. Numbers that cannot be expressed this way are called prime numbers (the real definition is slightly more complicated: you can look it up).\n\n2. Expand\n\nTo expand an expression is usually the opposite of factorizing. You multiply all the parts of the expression, at least as much as you can, using the distributive law, etc. \n\nIn the case of the expression x(x-3), we multiply our two monomials and return to the polynomial x^2-3*x:\n\nx(x-3)=x^2-3*x\n\nHere is another example. To expand\n\nx*(y-2)*(z+5)*y\n\nwe just multiply these factors together, \n\n x*(y-2)*(z+5)*y=\n(xy-2x)*(zy+5y)=\nxzy^2+5xy^2-2xzy-10xy\n\n\n\n3. Simplify\n\nThere is some ambiguity about what it means to simplify an expression. Computer algebra programs that simplify expressions sometimes use one definition of the simplest expression, and sometimes another. It all really depends on what shows up the features you are more interested in at the time. \n\nUsually to simplify an expression you have to do things like:\n\na. add up any like terms\n\nTo simplify  5x + 2y+ 2x+3+4, we add the terms with just numbers in them, and add the terms with just xs in them:\n\n5x + 2y+ 2x+3+4=7x+2y+7\n\nb. Multiply any numbers that can be multiplied\n\n5*2*x+2+y simplified is usually 10x+y+2\n\nc. Usually, when one simplifies an expression, the big powers are presented first, and the small powers follow (or sometimes vice versa).\n\nx+x^3+x^2 can be simplified as either \n\nx^3+x^2+x or x+x^2+x^3, depending on taste, the needs of the moment, etc.\n\n\nd. Usually when simplifying, any numbers or polynomials in ratios that can be divided will be\n\nSo 10/6 is often simplified as 5/3\n\nAnd (x^2-3*x)/x is often simplified as x-3\n\nHopefully that should get you started.
607601
I think you set the equation to zero then you have to factor it.  Go to the link below if you wanna know how to do it. OR look your your math book.
1109299
I do a lot of graphics calculator programming so could give you tips if you want. In any case you could write a program on the ti-83 + or 84+ by pressing prgm right right enter typing in a name then writing the equation which can be viewed from edit. You can also store them in the GDB I think it's called by going second prgm and across to store GDB. or You can write a program\n:Clrhome\n:Output(1,1,"Your info"\n:Pause\n:Clrhome\n:Output(1,1,"More info"