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  1. MHB Craftsman

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    #1
    Please assist me in this problem.


    $\frac{2}{3}b^5-\frac{1}{6}b^3+\frac{4}{9}b^2-1$

    I tried grouping but still could not find anything factorable form of the expression.

    Regards.

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    #2
    Quote Originally Posted by bergausstein View Post
    Please assist me in this problem.


    $\frac{2}{3}b^5-\frac{1}{6}b^3+\frac{4}{9}b^2-1$

    I tried grouping but still could not find anything factorable form of the expression.

    Regards.
    Hi bergausstein,

    After thinking it over and cheating a bit (using Wolfram), I'm not sure if a great factorization exists for this polynomial. You can definitely rewrite it a bit, but simpler is relative here. Is this problem from a textbook? What methods are you covering? I'm curious how simplified the answer is intended to be.

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    #3
    As Jameson says, a great factorization is impossible for this polynomial. We can write $$p(x)=(2/3)x^5-(1/6)x^3+(4/9)x^2-1=\frac{1}{18}(\underbrace{12x^5-3x^3+8x^2-18}_{q(x)}).$$ We have $q(1)=\ldots <0$ and $q(2)=\ldots>0,$ so by Bolzano's theorem there is a root $\xi\in (1,2).$
    It is not trivial to prove that $q(x)$ has only one real root, but even in this case and using Ruffini algorthim:

    $$\begin{array}{r|rrrrrr}
    & 12 & 0 & -3 & 8 & 0 & -18 \\
    \xi & & 12\xi & 12\xi^2 & 12\xi^3-3\xi & 12\xi^4-3\xi^2+8\xi & 12\xi^5-3\xi^3+8\xi^2\\
    \hline & 12 & 12\xi & 12\xi^2-3 & 12\xi^3-3\xi+8 & 12\xi^4-3\xi^2+8\xi & \underbrace{12\xi^5-3\xi^3+8\xi^2-18}_{=0} \end{array}$$ So, the factorization of $p(x)$ in $\mathbb{R}[x]$ would be $$p(x)=\dfrac{1}{18}(x-\xi)\left(12x^4+12\xi x^3+(12\xi^2-3)x^2+(12\xi^3-3\xi+8)x+12\xi^4-3\xi^3+8\xi\right).$$ That is, a complete disaster.
    Last edited by Fernando Revilla; January 3rd, 2017 at 13:38.

  4. MHB Craftsman
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    #4
    As Fernando Revilla pointed out, we can write this polynomial as $ \displaystyle \frac{1}{18}\left(12x^5- 3x^3+ 8x^2- 18\right)$ so this becomes a question of factoring $ \displaystyle 12x^5- 3x^3+ 8x^2- 18$.

    By the "rational root theorem", if there is a rational root (so that there is a factor with integer coefficients) then it must be of the form $ \displaystyle \frac{m}{n}$ (and the factor of the form (nx- m))where n is an integer factor of the leading coefficient, 12, and m is an integer factor of the constant term, 18. The integer factors of 12 are 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 12, and -12 and the integer coefficients of 18 are 1, -1, 2, -2, 3, -3, 6, -6, 9, -9, 18, and -18.

    Form all the different fractions with those numerators and denominators (there not as many as there might seem- a lot of cancelling occurs) and put them into the polynomial to see If they work.

  5. MHB Apprentice

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    #5
    Quote Originally Posted by bergausstein View Post
    Please assist me in this problem.


    $\frac{2}{3}b^5-\frac{1}{6}b^3+\frac{4}{9}b^2-1$

    I tried grouping but still could not find anything factorable form of the expression.

    Regards.
    Short answer, it is not factorizble with real coefficients. Of course every polynomial with complex roots is factorizable with complex coeffcients. I dont know how to get those complex roots but looks like thats not what you are looking for. And what you seem to be looking for is not possible

  6. MHB Apprentice

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    #6
    $\dfrac{2}{3}b^5 - \dfrac{1}{6}b^3+\dfrac{4}{9}b^2 - 1$

    $=\dfrac{12}{18}b^5 - \dfrac{3}{18}b^3 + \dfrac{8}{18}b^2 - \dfrac{18}{18}$

    $=\dfrac{1}{18}\left(12b^5 - 3b^3 + 8b^2 - 18\right)$

    $=\dfrac{1}{18}\left(3b^3(4b^2-1)+2(4b^2-9)\right)$

    $=\dfrac{1}{18}\left(3b^3(4b^2-1)+2\left(4b^2-9 \cdot\ \dfrac{4b^2-1}{4b^2-1}\right)\right)$

    $=\dfrac{1}{18}\left(3b^3(4b^2-1)+2\left(\dfrac{4b^2-9}{4b^2-1}\right)(4b^2-1)\right)$
    $=\dfrac{1}{18}\left((4b^2 - 1)\left(3b^3 + \dfrac{2(4b^2 - 9)}{4b^2 - 1}\right)\right)$
    $=\dfrac{1}{18}\left((2b+1)(2b-1)\left(3b^3 + \dfrac{2(2b + 3)(2b-3)}{(2b+1)(2b-1)}\right)\right)$
    Last edited by bwpbruce; January 4th, 2017 at 14:57.

  7. MHB Journeyman
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    Fernando Revilla's Avatar
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    #7
    Quote Originally Posted by bwpbruce View Post
    $\dfrac{2}{3}b^5 - \dfrac{1}{6}b^3+\dfrac{4}{9}b^2 - 1$
    $=\dfrac{1}{18}\left((2b+1)(2b-1)\left(3b^3 + \dfrac{2(2b + 3)(2b-3)}{(2b+1)(2b-1)}\right)\right)$
    According to your solution, $b=\pm\dfrac{1}{2}$ are roots. Have you checked them?

  8. MHB Apprentice

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    #8
    Quote Originally Posted by Fernando Revilla View Post
    According to your solution, $b=\pm\dfrac{1}{2}$ are roots. Have you checked them?
    Is there not a distinction between factoring a polynomial EXPRESSION and finding the roots of a polynomial EQUATION? Furthermore, even if we were tasked with finding roots, in accordance with the zero product property, $b=\pm\dfrac{1}{2}$ would be $\textit{possible}$ roots. But it is clear that neither value of b would work as a root. It is clear that we have to guess what the root is. If we assume that b = 1, then for the final binomial we have

    $3b^3 + \dfrac{2(2b + 3)(2b-3)}{(2b+1)(2b-1)} = 0$
    $3(1)^3 + \dfrac{2(2(1) + 3)(2(1)-3)}{(2(1)+1)(2(1)-1)} = 0$

    $3(1)^3 + \dfrac{2(5)(-1)}{(3)(1)} = 0$

    $3- \dfrac{10}{3} = 0$

    $3- 3.33 = 0$

    $-(3.33 - 3) = 0$

    $-0.33 \approx 0$

    So the value of b is close to 1.
    Last edited by bwpbruce; January 4th, 2017 at 16:46.

  9. MHB Journeyman
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    #9
    In these problems, we are supposed to provide a closed expression. For example if $p(x)=x^2-2$, then $1.5^2-2=0.25\approx 0.$ A closed expression is $p(x)=(x-\sqrt{2})(x+\sqrt{2})$, not $p(x)=(x-1.5)(x+1.5).$

  10. MHB Apprentice

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    #10
    Quote Originally Posted by Fernando Revilla View Post
    In these problems, we are supposed to provide a closed expression. For example if $p(x)=x^2-2$, then $1.5^2-2=0.25\approx 0.$ A closed expression is $p(x)=(x-\sqrt{2})(x+\sqrt{2})$, not $p(x)=(x-1.5)(x+1.5).$
    Which of my expressions are you implying is something other than closed form?

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