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September 15th, 2016, 01:11 PM   #1
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Using the Rational Zero Test

So I have this problem, I found (or I at least think I found) all the zeros of the function by looking at the graph. Once I did so, I needed to verify by multiplying it together and ending up with the function they give you. Where did I go wrong? I know I need to combine like terms at the end of my second attachment. But it wouldn't matter because I clearly won't get 4x^5. I end up with x^5 and the rest don't add up correctly either. Any suggestions on where I went wrong?
Attached Images IMG_0046.jpg (89.2 KB, 7 views) IMG_0049.jpg (85.1 KB, 8 views) September 15th, 2016, 01:28 PM #2 Senior Member   Joined: Aug 2016 From: morocco Posts: 273 Thanks: 32 $$f(x)=4(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)$$ don't forget the coefficient of the highest degree. Thanks from Taco September 15th, 2016, 01:28 PM #3 Math Team   Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 $f(x) = k(x+1)\left(x+\frac{1}{2}\right)\left(x-\frac{1}{2}\right)(x-1)(x-2)$ where $k$ is a constant. note that $f(0) = -2$ ... $-2 = k(0+1)\left(0+\frac{1}{2}\right)\left(0-\frac{1}{2}\right)(0-1)(0-2)$ $-2 = k\left(-\frac{1}{2}\right) \implies k = 4$ therefore ... $f(x) = 4(x+1)\left(x+\frac{1}{2}\right)\left(x-\frac{1}{2}\right)(x-1)(x-2)$ $f(x) = (x+1) \cdot 2\left(x+\frac{1}{2}\right) \cdot 2\left(x-\frac{1}{2}\right)(x-1)(x-2)$ $f(x) = (x+1)(2x+1)(2x-1)(x-1)(x-2)$ Thanks from Taco September 15th, 2016, 01:30 PM #4 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics You probably did nothing wrong. The factorization of a an integer polynomial with root 1/2 would be 2x - 1. Thus, if you replace (x-1/2)(x+1/2) with the factors (2x - 1)(2x+1) you will notice you get the required leading coefficient. Note that these aren't the same polynomial but they do have identical roots. Thanks from Taco September 15th, 2016, 02:01 PM #5 Newbie   Joined: Jun 2016 From: United States Posts: 10 Thanks: 1 So when I multiply the function of, f(x)=(x+1)(2x+1)(2x−1)(x−1)(x−2) because now I understand where you got 2x ± 1. I get f(x) = 2x^5 - 4x^4 + 8x -4. I would show you my work but apparently, I cannot find where to upload my file. And I don't understand how people are so good with LaTeX :P September 15th, 2016, 02:04 PM   #6
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Quote:
 Originally Posted by skeeter $f(x) = k(x+1)\left(x+\frac{1}{2}\right)\left(x-\frac{1}{2}\right)(x-1)(x-2)$ where $k$ is a constant. note that $f(0) = -2$ ... $-2 = k(0+1)\left(0+\frac{1}{2}\right)\left(0-\frac{1}{2}\right)(0-1)(0-2)$ $-2 = k\left(-\frac{1}{2}\right) \implies k = 4$ therefore ... $f(x) = 4(x+1)\left(x+\frac{1}{2}\right)\left(x-\frac{1}{2}\right)(x-1)(x-2)$ $f(x) = (x+1) \cdot 2\left(x+\frac{1}{2}\right) \cdot 2\left(x-\frac{1}{2}\right)(x-1)(x-2)$ $f(x) = (x+1)(2x+1)(2x-1)(x-1)(x-2)$

Oh, if I want to add another file I need to quote somebody. Okay, here's what I got.
Attached Images IMG_0051.jpg (92.4 KB, 2 views) September 15th, 2016, 02:10 PM #7 Math Team   Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 $f(x)=(x+1)(2x+1)(2x−1)(x−1)(x−2)$ $f(x)=(x+1)(x-1)(2x+1)(2x-1)(x-2)$ $f(x)=(x^2-1)(4x^2-1)(x-2)$ $f(x)=(4x^4-5x^2+1)(x-2)$ $f(x) = x(4x^4-5x^2+1) - 2(4x^4-5x^2+1)$ $f(x) = 4x^5 - 5x^3 + x - 8x^4 + 10x^2 - 2$ $f(x) = 4x^5 - 8x^4 - 5x^3 + 10x^2 + x - 2$ Thanks from Taco September 15th, 2016, 02:49 PM #8 Newbie   Joined: Jun 2016 From: United States Posts: 10 Thanks: 1 Oh, I should have changed the order according to the numbers. Alright, thank you so much that makes sense. October 23rd, 2016, 03:23 AM   #9
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Quote:
 Originally Posted by Taco And I don't understand how people are so good with LaTeX :P
Practice, practice, practice! Tags rational, test Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post leo255 Calculus 2 November 4th, 2014 11:55 AM semifold Advanced Statistics 0 January 23rd, 2010 02:17 PM johndeere19 Advanced Statistics 1 July 1st, 2009 11:34 AM JRock7997 Calculus 1 February 19th, 2009 05:32 PM symmetry Advanced Statistics 0 April 7th, 2007 03:05 AM

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