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 October 10th, 2012, 09:28 AM #1 Newbie   Joined: Sep 2012 Posts: 8 Thanks: 0 Find f(x) from the following information Let f(x) be a polynomial of degree less than or equal to 5 which leaves remainders -1 and 1 respectively upon division by $(x-1)^{3}$ and $(x+1)^{3}$. Find f(x).
 October 10th, 2012, 04:23 PM #2 Global Moderator   Joined: Dec 2006 Posts: 18,593 Thanks: 1491 f(x) = (-3x^5 + 10x^3 - 15x)/8
October 10th, 2012, 05:00 PM   #3
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Quote:
 Originally Posted by skipjack f(x) = (-3x^5 + 10x^3 - 15x)/8
Can you help me on this please I having a hard time (x^3)+(5x^2)+(5x)+(19)/(x)+(2)

 October 10th, 2012, 06:38 PM #4 Global Moderator   Joined: Dec 2006 Posts: 18,593 Thanks: 1491 If you meant the divisor to be (x + 2) rather than just x, see the division below.             x^2 + 3x    -  1            _________________ x + 2 | x^3 + 5x^2 + 5x + 19            x^3 + 2x^2            -------------                     3x^2 + 5x                     3x^2 + 6x                     -----------                                -x + 19                                -x -   2                                --------                                       21
October 11th, 2012, 05:51 AM   #5
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Re:

Quote:
 Originally Posted by skipjack f(x) = (-3x^5 + 10x^3 - 15x)/8
Hey How you are solving this? I need some explanation.

 October 11th, 2012, 05:58 AM #6 Global Moderator   Joined: Dec 2006 Posts: 18,593 Thanks: 1491 I couldn't spot any simple method, so I wrote f(x) = (ax² + bx + c)(x - 1)³ - 1 = (px² + qx + r)(x + 1)³ + 1. On expanding and equating coefficients, one has six equations for the six coefficients of the quadratics. Solving these equations allows f(x) to be found.
October 11th, 2012, 06:14 AM   #7
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Quote:
 Originally Posted by skipjack I couldn't spot any simple method, so I wrote f(x) = (ax² + bx + c)(x - 1)³ - 1 = (px² + qx + r)(x + 1)³ + 1. On expanding and equating coefficients, one has six equations for the six coefficients of the quadratics. Solving these equations allows f(x) to be found.
So this means if f(x) is divided by $(x-1)^{3}$ then remainder will be -1. I know remainder theorem says that when f(x) is divided by x-a then remainder is f(a), but what if it is divided by an expression that is raised to a power of something?

 October 11th, 2012, 06:27 AM #8 Global Moderator   Joined: Dec 2006 Posts: 18,593 Thanks: 1491 In general, the remainder on dividing a function f(x) by (x - a)^n would be a non-constant function of x. If that function is R(x), you would know that R(a) = f(a).

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### when a polynomial f(x) is divided by (x-1) and (x 5) , the remainder are -6 and 6

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