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September 27th, 2014, 04:02 PM  #1 
Senior Member Joined: Sep 2013 From: Earth Posts: 821 Thanks: 35  Find the solution.
Find the solution set if the numbers in parentheses are roots of the given equation. 9x^46x^3+10x^26x+1=0 ; (1/2,1/2) I don't understand the question. How to solve? 
September 27th, 2014, 04:22 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,104 Thanks: 980 
you might want to recheck the problem statement ... x = 1/3 is a root of multiplicity two, not x = 1/2.

September 28th, 2014, 02:08 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 16,194 Thanks: 1147 
$\displaystyle 9x^4  6x^3 + 10x^2  6x + 1 = (9x^2  6x + 1)(x^2 + 1)$

September 28th, 2014, 02:45 AM  #4 
Math Team Joined: Apr 2010 Posts: 2,770 Thanks: 356 
Suppose, (1/3, 1/3) was given initially. You are supposed to factor out polynomials that have these roots and find the remaining roots using the factorisation. As 9x^46x^3+10x^26x+1 = (3x  1)^2 * (x^2 + 1), any remaining roots of 9x^46x^3+10x^26x+1 are roots of x^2 + 1. Why?

September 28th, 2014, 03:44 AM  #5  
Senior Member Joined: Sep 2013 From: Earth Posts: 821 Thanks: 35  Quote:
Don't understand  
September 28th, 2014, 03:56 AM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,131 Thanks: 438 Math Focus: Calculus/ODEs 
You could use synthetic division twice: $\displaystyle \begin{array}{crr}& 9 & 6 & 10 & 6 & 1 \\ \tfrac{1}{3} & & 3 & 1 & 3 & 1 \\ \hline & 9 & 3 & 9 & 3 & 0 \end{array}$ $\displaystyle \begin{array}{crr}& 9 & 3 & 9 & 3 \\ \tfrac{1}{3} & & 3 & 0 & 3 \\ \hline & 9 & 0 & 9 & 0 \end{array}$ Now you know the remaining roots come from: $\displaystyle 9x^2+9=9\left(x^2+1\right)=0$ 
September 28th, 2014, 04:44 AM  #7 
Math Team Joined: Apr 2010 Posts: 2,770 Thanks: 356 
Alternatively, $\displaystyle (9x^2  6x + 1)(ax^2 + bx + c) = 9x^46x^3+10x^26x+1 = 9ax^4 + (9b  6a)x^3 + (9c + a  6b)x^2 + (b  6c)x + c$, from which we have 5 equations and 3 variables. After evaluating from left to right a = 1, b = 0, c = 1. This works out for the last two equations hence no remainder.

September 28th, 2014, 04:59 AM  #8  
Senior Member Joined: Sep 2013 From: Earth Posts: 821 Thanks: 35  Quote:
 
September 28th, 2014, 07:36 AM  #9  
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,131 Thanks: 438 Math Focus: Calculus/ODEs  Quote:
$\displaystyle 9x^46x^3+10x^26x+1=\left(x\frac{1}{3}\right)^2\cdot9\left(x^2+1\right)$ Note, this can then be written: $\displaystyle 9x^46x^3+10x^26x+1=\left(3\left(x\frac{1}{3}\right)\right)^2\left(x^2+1\right)$ $\displaystyle 9x^46x^3+10x^26x+1=\left(3x1\right)^2\left(x^2+1\right)$  

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