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September 15th, 2016, 01:11 PM  #1 
Newbie Joined: Jun 2016 From: United States Posts: 10 Thanks: 1  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?

September 15th, 2016, 01:28 PM  #2 
Senior Member Joined: Aug 2016 From: morocco Posts: 273 Thanks: 32 
$$f(x)=4(xx_1)(xx_2)(xx_3)(xx_4)(xx_5)$$ don't forget the coefficient of the highest degree. 
September 15th, 2016, 01:28 PM  #3 
Math Team Joined: Jul 2011 From: Texas Posts: 2,205 Thanks: 1049 
$f(x) = k(x+1)\left(x+\frac{1}{2}\right)\left(x\frac{1}{2}\right)(x1)(x2)$ where $k$ is a constant. note that $f(0) = 2$ ... $2 = k(0+1)\left(0+\frac{1}{2}\right)\left(0\frac{1}{2}\right)(01)(02)$ $2 = k\left(\frac{1}{2}\right) \implies k = 4$ therefore ... $f(x) = 4(x+1)\left(x+\frac{1}{2}\right)\left(x\frac{1}{2}\right)(x1)(x2)$ $f(x) = (x+1) \cdot 2\left(x+\frac{1}{2}\right) \cdot 2\left(x\frac{1}{2}\right)(x1)(x2)$ $f(x) = (x+1)(2x+1)(2x1)(x1)(x2)$ 
September 15th, 2016, 01:30 PM  #4 
Member Joined: Sep 2016 From: USA Posts: 98 Thanks: 36 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 (x1/2)(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.

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  
Newbie Joined: Jun 2016 From: United States Posts: 10 Thanks: 1  Quote:
Oh, if I want to add another file I need to quote somebody. Okay, here's what I got.  
September 15th, 2016, 02:10 PM  #7 
Math Team Joined: Jul 2011 From: Texas Posts: 2,205 Thanks: 1049 
$f(x)=(x+1)(2x+1)(2x−1)(x−1)(x−2)$ $f(x)=(x+1)(x1)(2x+1)(2x1)(x2)$ $f(x)=(x^21)(4x^21)(x2)$ $f(x)=(4x^45x^2+1)(x2)$ $f(x) = x(4x^45x^2+1)  2(4x^45x^2+1)$ $f(x) = 4x^5  5x^3 + x  8x^4 + 10x^2  2$ $f(x) = 4x^5  8x^4  5x^3 + 10x^2 + x  2$ 
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 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,245 Thanks: 559  

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