use the remainder theorem to determine the remainder when d^4+2d^2+5d-10 is divided by d+4?
828966
This is exactly what I am doing in math right now:\nso with the remainder theorem you plug d+4 into the equation so it becomes -4\nso\n(-4)^4+2(-4)^2+5(-4)-10\n256+32-20-10=258
674900
When you have (a+b+c)/d, you can separate the terms of the numerator like so a/d + b/d + c/d.  If I do that and simplify, I get:\nx - (6/5) - (3/x).
230715
We know that 2^10=1024.1024/11 leaves 1.Let x=2^10.Since 11 is a factor of (x-1) which itself is a factor of (x^n-1) (to prove this put x=1 in x^n-1).Now 2^(10*n)-1 divides 11.\n        Its double 2^((10*n)+1)-2 exactly divides 11.\n        Hence remainder is 2 in the given question.
783248
This is a partial fractions type question. \nRule of thumb:  Whenever the degree of the numerator is greater than or equal to the degree of the denominator, you want to use long division to change the question.  In the above case, you'll want to divide X + 10 into X.  Due to the limitations of text you'll have to bear with me here:\n                1\n--------------------\nX+10 | X+0\n           x+10\n           -------\n              -10\nSo X/(X+10)  =  -10/(x+10) + 1\n\nNow, you integrate that.\n\nIntegral (0,5) [X/(X+10)]dx = \nIntegral (0,5) [-10/(x+10) + 1] dx\n\nSplit it up into two integrals, to get\n\nIntegral (0,5) [-10/(x+10)] dx - Integral (0,5) [1]dx\n\nPull out any constants.  The second integral is trivial to solve\n\n-10 * Integral (0,5) [1/(x+10)]dx - Integral (0,5) [1] dx\n\nThrough simple substitution, you'll learn that the integral of 1/(x+10) is equal to ln|x+10|.  The second integral is trivial.  So we get\n                         5             5\n-10 * (ln|x+10)] |        -  x |\n                         0             0\n=\n\n-10 * (ln|5+10| - ln|0+10|) - (5 - 0)\n-10 * (ln|15| - ln|10|) - 5\n\nRemember the log rule when dealing with subtraction.\n\n-10 * ln(15/10) - 5\n-10 * ln (3/2) - 5
866735
I believe you use the remainder theorem for this, which states that the remainder when a polynomial f(x) is divided by (x-k) is\nf(k)\n\nThus the remainder would be f(-1) = 11
497719
10-d = -34-5d\n10-d+5d = -34-5d+5d\n10+4d = -34\n10-10+4d = -34-10\n4d = -44\nd = -11
762587
use remainder theorem\nhttp://en.wikipedia.org/wiki/Remainder_theorem\n"When p(x) is divided by (x-a), the remainder is p(a)"\nUR QUESTION:\np(x) = x^3+3x+k\n(x-a) = (x-1)\np(a) = 6\nhence, p(1)=6\nso, (1)^3 + 3(1) + k = 6\nhence, 1+3+k = 6\nso, k = 2\n\nhence, k = 2
712942
x^6 - 6x^5 + 15x^4 - 30x^3 + 15x^2 - 6x + 1 = 0\nx^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1 - 10x^3 = 0\nSo (x - 1)^6 - 10x^3 = 0\nSo ((x - 1)²)³ - (³√10x)³ = 0\n\nSo ((x - 1)² - ³√10x)((x - 1)^4 + ³√10x((x - 1)² + ³√100x²) = 0\n\nSo ((x - 1)^4 + ³√10x((x - 1)² + ³√100x²) = 0 N0 real roots (Min value near x = 0.2 just more than 0.87)\n\nOR (x - 1)² - ³√10x = 0 \nie x² - (2 + ³√10)x + 1 = 0\nie x = ½[(2 + ³√10) ± √(4 + 4³√10 + ³√100 - 4)\n= ½[(2 + ³√10) ± √(4³√10 + ³√100)]\n\nSOLUTION\nx = ½[(2 + ³√10) ± √(4³√10 + ³√100)]
716901
You can use polynomial long division to find the value of k. Be very careful! For the first expression, after dividing the numerator by the denominator a remainder of  8 - 4k  is obtained. Set this equal to zero to find the desired value of k.\n\nTo review polynomial long division, see:\n\nhttp://en.wikipedia.org/wiki/Polynomial_long_division
775644
Duch, there's some confusion over what the problem is.  It could be a/(3+4b) - a/(3+b)  OR  a/3 +4b -a/3 -b\n\nIf the former, then it boils down to -(11ab)/(9 + 15b + 4b²)\n\nIf the later, 3b
321475
*(2x/(x^2-4)) + (5/2-x) - (1/2+x)\n*(2x/(x^2-4)) + (5/2-x)((2+x)/(2+x)) - (1/2+x)((2-x)/(2-x))\n*(2x/(x^2-4)) + (10+5x)/(4-x^2) - (2-x)/(4-x^2)\n*(2x/(x^2-4)) + (8+6x)/(4-x^2)\n\nyou should be able to get it from there
275765
you need to know the relation that the greeks found thousands years ago Δ=δπ+υ. this is the identity of the incomplete division Δ is the number you want to divide,\nδ is the divider, π is the quotient and υ is the remainder.\nyou just can´t round up or down the remainder
165476
How about (a+c)/(b+d)  ? more interesting!!\n\nbtw this only work when a/b != c/d
913289
You end up with a remainder, so it's not pretty, but you should be able to do long division and get:\n\n3x^2 - 4x + (16/3) - 23/(3*(3x+2))
867635
(-(3^10)) / (4^5) = -57.6650391\n--> http://www.google.com/search?hl=en&lr=&q=-3%5E10%2F4%5E5&btnG=Search\n\n(-(1^3)) / (-(4^3)) = 0.015625
139386
(X-c) here represents the number you are going to devide with. c can take any value here.\nLet you have to devide with (x-5) , then c will be 5.\nAnd if you have to devide with (x+10) then c will be -10. As (x-(-10))=(x+10).\nThe remaider therum says, If you devide a polynomial, P(x) by (x-c) then the remainder will be P(c).\nIf you want a proof of this theorum,\nlet, the divisor of divison of P(x) by (x-c) is A and remainder is B.\nSo P(x)=A*(x-c)+B.\nSo P(c)=A*(c-c)+b\nor, B=P(c) which is the remainder, I hope this will clear your concept.
473264
all factorials greater than 6! divide evenly since they contain 7\nso add 1! to 6! and divide by 7, find remainder\n=1!+2!+3!+4!+5!+6!\n=1+2+6+24+120+720\n=873/7\n=126 R 5\n\nanswer is 5