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September 27th, 2014, 04:02 PM   #1
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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?
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September 27th, 2014, 04:22 PM   #2
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you might want to recheck the problem statement ... x = 1/3 is a root of multiplicity two, not x = 1/2.
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September 28th, 2014, 02:08 AM   #3
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$\displaystyle 9x^4 - 6x^3 + 10x^2 - 6x + 1 = (9x^2 - 6x + 1)(x^2 + 1)$
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September 28th, 2014, 02:45 AM   #4
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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?
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September 28th, 2014, 03:44 AM   #5
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Quote:
Originally Posted by Hoempa View Post
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
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September 28th, 2014, 03:56 AM   #6
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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:

$\displaystyle 9x^2+9=9\left(x^2+1\right)=0$
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September 28th, 2014, 04:44 AM   #7
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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.
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September 28th, 2014, 04:59 AM   #8
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Quote:
Originally Posted by MarkFL View Post
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.
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September 28th, 2014, 07:36 AM   #9
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Quote:
Originally Posted by jiasyuen View Post
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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