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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^4-6x^3+10x^2-6x+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: 2,315 Thanks: 1138 you might want to recheck the problem statement ... x = 1/3 is a root of multiplicity two, not x = 1/2. Thanks from topsquark
 September 28th, 2014, 02:08 AM #3 Global Moderator   Joined: Dec 2006 Posts: 16,605 Thanks: 1201 $\displaystyle 9x^4 - 6x^3 + 10x^2 - 6x + 1 = (9x^2 - 6x + 1)(x^2 + 1)$
 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^4-6x^3+10x^2-6x+1 = (3x - 1)^2 * (x^2 + 1), any remaining roots of 9x^4-6x^3+10x^2-6x+1 are roots of x^2 + 1. Why? Thanks from topsquark
September 28th, 2014, 03:44 AM   #5
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 Originally Posted by Hoempa 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^4-6x^3+10x^2-6x+1 = (3x - 1)^2 * (x^2 + 1), any remaining roots of 9x^4-6x^3+10x^2-6x+1 are roots of x^2 + 1. Why?

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: 12,155 Thanks: 461 Math Focus: Calculus/ODEs You could use synthetic division twice: $\displaystyle \begin{array}{c|rr}& 9 & -6 & 10 & -6 & 1 \\ \tfrac{1}{3} & & 3 & -1 & 3 & -1 \\ \hline & 9 & -3 & 9 & -3 & 0 \end{array}$ $\displaystyle \begin{array}{c|rr}& 9 & -3 & 9 & -3 \\ \tfrac{1}{3} & & 3 & 0 & 3 \\ \hline & 9 & 0 & 9 & 0 \end{array}$ Now you know the remaining roots come from: $\displaystyle 9x^2+9=9\left(x^2+1\right)=0$ Thanks from topsquark
 September 28th, 2014, 04:44 AM #7 Math Team   Joined: Apr 2010 Posts: 2,770 Thanks: 356 Alternatively, $\displaystyle (9x^2 - 6x + 1)(ax^2 + bx + c) = 9x^4-6x^3+10x^2-6x+1 = 9ax^4 + (9b - 6a)x^3 + (9c + a - 6b)x^2 + (b - 6c)x + c$, 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
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 Originally Posted by MarkFL You could use synthetic division twice: $\displaystyle \begin{array}{c|rr}& 9 & -6 & 10 & -6 & 1 \\ \tfrac{1}{3} & & 3 & -1 & 3 & -1 \\ \hline & 9 & -3 & 9 & -3 & 0 \end{array}$ $\displaystyle \begin{array}{c|rr}& 9 & -3 & 9 & -3 \\ \tfrac{1}{3} & & 3 & 0 & 3 \\ \hline & 9 & 0 & 9 & 0 \end{array}$ Now you know the remaining roots come from: $\displaystyle 9x^2+9=9\left(x^2+1\right)=0$
I understand the synthetic division which you wrote. Then what the answer for this question will be? Why all the answers from others is different from you? I feel I'm very stupid.

September 28th, 2014, 07:36 AM   #9
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 Originally Posted by jiasyuen I understand the synthetic division which you wrote. Then what the answer for this question will be? Why all the answers from others is different from you? I feel I'm very stupid.
Using the synthetic division I found:

$\displaystyle 9x^4-6x^3+10x^2-6x+1=\left(x-\frac{1}{3}\right)^2\cdot9\left(x^2+1\right)$

Note, this can then be written:

$\displaystyle 9x^4-6x^3+10x^2-6x+1=\left(3\left(x-\frac{1}{3}\right)\right)^2\left(x^2+1\right)$

$\displaystyle 9x^4-6x^3+10x^2-6x+1=\left(3x-1\right)^2\left(x^2+1\right)$

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