My Math Forum Using the Rational Zero Test

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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?
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 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: 2,442 Thanks: 1203 $f(x) = k(x+1)\left(x+\frac{1}{2}\right)\left(x-\frac{1}{2}\right)(x-1)(x-2)$ 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: 114 Thanks: 44 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.
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 September 15th, 2016, 02:10 PM #7 Math Team   Joined: Jul 2011 From: Texas Posts: 2,442 Thanks: 1203 $f(x)=(x+1)(2x+1)(2x−1)(x−1)(x−2)$ $f(x)=(x+1)(x-1)(2x+1)(2x-1)(x-2)$ $f(x)=(x^2-1)(4x^2-1)(x-2)$ $f(x)=(4x^4-5x^2+1)(x-2)$ $f(x) = x(4x^4-5x^2+1) - 2(4x^4-5x^2+1)$ $f(x) = 4x^5 - 5x^3 + x - 8x^4 + 10x^2 - 2$ $f(x) = 4x^5 - 8x^4 - 5x^3 + 10x^2 + x - 2$ 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!

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