How can I do a mathematical proof of Synthetic division?
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.
55565
Your computer has a calculator built in.  Failing that, try long division.  You're gonna have to learn how to do it anyway.\n\nSheesh.
943473
the name i learned for this is fraction splitting\n\nyou can split into A/x + (Bx+c)/(x^2 + 1) \n\n( because x^2+1 is irreducible\nfor x = 1 and -1 the fractions should be zero that speeds up the calculation od A,B C, \nfinally write out and eqalt the corresponding fractions
66583
Polynomials are expressions involving x raised to a whole number power (exponent). Some examples are: \n\n\n\nIn this lesson we consider division of polynomials such as: \n and . \nThere are two ways to calculate a division of polynomials. One is long division and a second method is called synthetic division. Synthetic division can be used when the polynomial divisor such as x-2 has the highest power of x as 1 and the coefficient of x is also 1. Otherwise, such as , a long division process must be used. Both methods will be demonstrated. \n\nWe will now look at some examples showing how these two procedures are done
1091850
This is just something the person who said PEMDAS left out. \n\nYou need to solve it like this: Parenthesis, Exponents, Multiplication-Division, Addition-Subtraction. \n\nSay you have this problem(an example):\n\n3x4/5\n\nYou have both multiplication and division in this problem, which are tied for the third step. Once this happens, you have to solve the problem left-to-right. In this case, you would multiply 3 and 4 first to get 12, then divide 12 by 5.  Addition and Subtraction works the same way.
887334
You do this kind of division the same way you do normal long division, with the powers of x analogous to the powers of ten.  But, like normal long division, you have to be careful with the signs!\n\nStart at the left - x-3 goes 2x^3 times into 2x^4, so multiply x-3 by 2x^3, which gives us 2x^4 - 6x^3.  So the first term in our quotient is 2x^3.  We subtract 2x^4 -6x^3 from the dividend, leaving 5x^3+15x^2+6x-18. \n\nWe carry on in the same vein, so we multiply x-3 by 5x^2, the second term in our quotient, getting 5x^3-15x^2, which again we subtract from the dividend.  This leaves us with 6x-18.  Now x-3 goes 6 times into 6x-18, so the third term in our quotient is 6.  6 times x-3 is 6x-18, which we subtract from the dividend, which leaves us with zero, so there is no remainder.  The result of the division is the sum of the three terms we generated, which is 2x^3+5x^2+6.\n\nAs for the factorisation, you will remember that x^2-1 has factors x+1 and x-1, so x^4-1 will have factors x^2-1 aand x^2+1.  Then x^2-1 has factors x-1 and x+1.  So the factorisation of x^4-1 is (x-1)(x+1)(x^2+1).  If you are OK with complex numbers the x^2+1 term may be further factorised to x+i and x-i, so then the factors are (x+1)(x-1)(x+i)(x-i).\n\nYou could have got straight there though by recognising that the fourth powers of +1, -1, +i and -i are all 1!
179350
ok well u gotta do long division for this. but first you need to find a factor. theres some theorem or rule or somethin that sayz, basically check all the numbers that are factors of the last number. so the last number is -12. factors are +-1, +-2, +-3, +-4, +-6. so put each one into the equation for x  and eventually, 1 of them will make the equation =0. so i think 1 works. so that means the factor is x-1. now u do long division.\n\nso first you ignore the -1 and u just see what do u multiply x by to get x^4. and its x^3. and then you multiply the whole   (x-1) by x^3   and subtract it from your equation. and then bring down the next term. and repeat it and you should not get a remainder.\n\n....................x^3 - 2x^2 - 11x + 12\n(x-1)...%...(x^4 - 3x^3 - 9x^2 + 23x - 12)\n.............x^4 - x^3\n\n...............0 -2x^3 - 9x^2\n.................-2x^3 + 2x^2\n\n...................0   -11x^2 + 23x\n.......................-11x^2 + 11x\n\n........................0     + 12x - 12\n.................................0 + 0\n\nand so there. your first factor is (x-1) and now u gotta repeat the same thing for your quotient. which is  \nx^3 - 2x^2 - 11x + 12.  so 1 works here again. which basically means (x-1) factors out twice.    so divide again\n\n....................x^2 - x - 12\n(x-1)...%...(x^3 - 2x^2 - 11x + 12)\n.............x^3 - x^2\n\n..................-x^2 - 11x\n..................-x^2 + x\n\n........................-12x + 12\n........................-12x + 12\n\nand now u have a quadratic equation x^2 - x - 12. which isnt that hard to solve. 2 numbers that add to -1 and multiply to -12  is just -4 and 3. so (x-4)(x+3)\n\nso the whole equation = (x-1)^2 (x-4)(x+3)    so the   zeros   are  1, 4, -3\n\nhope u get it. let meknow if u need more help. laterz
1140302
Here is an interactive page that discusses one way to do it:\n\nhttp://dilc.upd.edu.ph/loDeploy/math/math10a.html\n\nFor large numbers, bequalmings technique using logs is an easier way to go.
1079776
you can simply the first one to (x-a)(x-a) then you can divide that by (x-a) and it equals (x-a)\n\nDo you know synthetic division; cuz you can divide the second problem by (x+2)^2 if you mulitply (x+2)^2 out.
538734
It's just like long division... but it is hard to show in text form here.\n\nLet's start with the second one, since it is simpler:\n\n2x^2 + 6x - 5 / 2x - 4\n\nWrite it as you would for long division:\n\n```````` ____________\n2x - 4 ) 2x^2 + 6x - 5\n\nStart with the first couple terms.  How many times does 2x-4 go into 2x^2 + 6x.  It goes in once (times x), with a remainder of 2.\n\n```````` _______1x____\n2x - 4 ) 2x^2 + 6x - 5\n`````````-` 2x^2 - 4x\n`````````----------------------\n````````````````````10x - 5\n\nAnd how many times does 2x-4 go into 10x-5... 5 times.\n\n```````` _______1x_+5_\n2x - 4 ) 2x^2 + 6x - 5\n`````````-` 2x^2 - 4x\n`````````----------------------\n````````````````````10x - 5\n``````````````````-`10x -20\n`````````----------------------\n```````````````````````````15\n\nSo the answer is x + 5 with a remainder of 15/(2x-4)\n\nTry the first on your own... it doesn't even have a remainder...\n\nHere's some hints:\n\nx + 11 ) 3 x^3 + 35 x^2 + 19 x - 33\n\nIgnoring x for a moment\n\n1 11 ) 3 35 19 -33\n\nFirst one goes in 3 times\n\n1 11 ) 3 35 19 -33\n```````- 3 33\n```````-----------------\n`````````````2 19\n\nNext goes in 2 times\n\n1 11 ) 3 35 19 -33\n```````- 3 33\n```````-----------------\n`````````````2 19\n```````````- 2 22\n```````-----------------\n```````````````-3 -33\n\nThis one goes in -3 times\n\n1 11 ) 3 35 19 -33\n```````- 3 33\n```````-----------------\n`````````````2 19\n```````````- 2 22\n```````-----------------\n```````````````-3 -33\n`````````````- -3 -33\n```````-----------------\n````````````````````0\n\nPut x back in you have 3x^2 + 2x - 3
520168
Try this:\n\nhttp://www.algebra.com/
1094108
Use a calculator and raise it to the 1/3 power.\nDon't try to do it by hand...
1148035
Division, if you do not know how to divide can be done easily.\n\nTake the problem 32 divided by 8.  You would divide this out the long way.  The easiest way might be to subtract 8 until you get to zero.   You notice it take you 4 times of subtracting 8 to get to zero.  So your answer would be 4.
352026
I would like to be able to give you the answer but I don't think I would be doing you a favour.  If you are to get practice in division, it is better for you to work it out on your own.  Does your teacher know that you are getting homework answers here on Yahoo?
269746
Make 2 7/15 an improper fraction, flip it over to its reciprical and mulitply the two.\n\nBoy...I hope I got that.  *bits fingernails*
235381
in dividing fractions, you just multiply them by its reciprocal.\n\nso \n\n[(x+5) / (16x)]  * [ (4x ^2) / [(3x -2)] \n\nperform the operations: \n\n[ (x+5) / 4 ] * [ x / (3x-2) ]\n\nthe answer is:\n\n[x(x+5)] / [ 4(3x-2) ]
386814
They are correct.  Observe the order of operations or PEMDAS\n\nParenthesis\nExponents\nMultiplication-Division\nAddition-Subtraction\n\nso first multiply   -89 * 658 =   -58562\nThen add 2   =                    -58560\nThen add 6   =                    -58554\nThen subtract 89 (or add -89) =   -58643