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Thread: Distributing

  1. #1
    Member cmf0106's Avatar
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    Distributing

    Using the distributive law of multiplication over subtraction i have a question

    $\displaystyle 3\sqrt[3]{x}(6y-2x)=18\sqrt[3]xy-6x\sqrt[3]{x}$

    Why is the answer not $\displaystyle 18y\sqrt[3]x-6x\sqrt[3]{x}$?
    Because in the first half of the answer $\displaystyle 18\sqrt[3]xy$ the x is under the radicand. While in the second half $\displaystyle 6x\sqrt[3]{x}$ the x is outside the radicand. I do not understand why one x would be included under the radicand while the other would not.

    Please Help Clarify & Many Thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cmf0106 View Post
    Using the distributive law of multiplication over subtraction i have a question

    $\displaystyle 3\sqrt[3]{x}(6y-2x)=18\sqrt[3]xy-6x\sqrt[3]{x}$

    Why is the answer not $\displaystyle 18y\sqrt[3]x-6x\sqrt[3]{x}$?
    Because in the first half of the answer $\displaystyle 18\sqrt[3]xy$ the x is under the radicand. While in the second half $\displaystyle 6x\sqrt[3]{x}$ the x is outside the radicand. I do not understand why one x would be included under the radicand while the other would not.

    Please Help Clarify & Many Thanks
    you have the same thing
    $\displaystyle \sqrt[3]{x}y = y \sqrt[3]{x}$ since multiplication is commutative

    you are not to simplify $\displaystyle x \sqrt[3]{x}$ ?
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by cmf0106 View Post
    Using the distributive law of multiplication over subtraction i have a question

    $\displaystyle 3\sqrt[3]{x}(6y-2x)=18\sqrt[3]xy-6x\sqrt[3]{x}$

    Why is the answer not $\displaystyle 18y\sqrt[3]x-6x\sqrt[3]{x}$?
    That is the answer. The order of the variables in each term shouldn't change the answer; however, when we have expressions where there are terms that contain x and y, we usually write x first, and then y.

    $\displaystyle 18y\sqrt[3]{x}$ is correct, but so is this $\displaystyle 18\sqrt[3]xy$


    Because in the first half of the answer $\displaystyle 18\sqrt[3]xy$ the x is under the radicand. While in the second half $\displaystyle 6x\sqrt[3]{x}$ the x is outside the radicand. I do not understand why one x would be included under the radicand while the other would not.

    Please Help Clarify & Many Thanks
    Well, in the second term, the x term is inside and outside the radical.

    $\displaystyle 3\sqrt[3]{x}(6y-2x)\implies \left[3\sqrt[3]{x}\cdot 6y\right]-\left[3\sqrt[3]{x}\cdot 2x\right]\implies 18\sqrt[3]xy-6x\sqrt[3]x$

    We could, however, combine $\displaystyle x$ and $\displaystyle \sqrt[3]x$, but I believe that is beyond what the question is asking.

    Does this help?

    --Chris
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  4. #4
    Member cmf0106's Avatar
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    $\displaystyle x\sqrt[3]{x}$ Not it does not ask me to, but im curious as to how to simply this. Im guessing its Using $\displaystyle \sqrt[n]{a^m}=a^{\frac{m}{n}}$. But im not sure how it works out with another x right beside it.

    EDIT: And yes it does help chris, but i would like to go beyond what the question is asking to enhance my knowledge.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by cmf0106 View Post
    $\displaystyle x\sqrt[3]{x}$ Not it does not ask me to, but im curious as to how to simply this. Im guessing its Using $\displaystyle \sqrt[n]{a^m}=a^{\frac{m}{n}}$. But im not sure how it works out with another x right beside it.
    Keep in mind $\displaystyle x^a\cdot x^b=x^{a+b}$ where we have $\displaystyle x=x^1$ and $\displaystyle \sqrt[3]x=x^{\frac{1}{3}}$

    So, $\displaystyle x\cdot x^{\frac{1}{3}}=\dots$

    --Chris
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  6. #6
    Member cmf0106's Avatar
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    so $\displaystyle x^{1\frac{1}{3}}$?
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by cmf0106 View Post
    so $\displaystyle x^{1\frac{1}{3}}$?
    yes...or as an improper fraction $\displaystyle x^{\frac{4}{3}}$. In radical notation it would be $\displaystyle \sqrt[3]{x^4}$

    Does this make more sense now?

    --Chris
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  8. #8
    Member cmf0106's Avatar
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    Yes it does thanks for answering my questions Chris and everyone else!
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