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Thread: Simplifying expressions

  1. #1
    Member Jonboy's Avatar
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    Simplifying expressions

    Hi everyone! I would like you all to check these problems and if there's an easier way please do give the suggestion. The directions are to simplify.

    #1 $\displaystyle 8 + \frac{25}{\sqrt{5}}$

    I though I should get one fraction: $\displaystyle \frac{8\sqrt{5}\,+ 25}{\sqrt{5}}$

    Rationalize: $\displaystyle \frac{40\,+\,25\sqrt{5}}{5} = \boxed{8 + 5\sqrt{5}}$

    # 2 $\displaystyle 9x^3 - 35x^2 - 4x$

    $\displaystyle \boxed{x(9x^2 - 35x - 4)}$

    I don't see how that can simplify any more... do you agree?

    # 3 $\displaystyle 27x^6 + 125y^6 = (3x^2)^3 + (5y^2)^3$

    I factored this as: $\displaystyle \boxed{(3x^2 + 5y^2)(9x^4 - 3x^2 5y^2 + 25y^4)}$

    # 4 $\displaystyle 4x^3 - 4x^2 - 9x + 9$

    Factor by grouping: $\displaystyle 4x^2(x - 1) -9(x - 1) \Rightarrow (4x^2 - 9)(x - 1)$

    Which goes to: $\displaystyle \boxed{(2x - 3)(2x + 3)(x - 1)}$

    Look good guys/gals?
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  2. #2
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    Quote Originally Posted by Jonboy View Post
    Hi everyone! I would like you all to check these problems and if there's an easier way please do give the suggestion. The directions are to simplify.

    #1 $\displaystyle 8 + \frac{25}{\sqrt{5}}$

    I though I should get one fraction: $\displaystyle \frac{8\sqrt{5}\,+ 25}{\sqrt{5}}$

    Rationalize: $\displaystyle \frac{40\,+\,25\sqrt{5}}{5} = \boxed{8 + 5\sqrt{5}}$

    # 2 $\displaystyle 9x^3 - 35x^2 - 4x$

    $\displaystyle \boxed{x(9x^2 - 35x - 4)}$

    I don't see how that can simplify any more... do you agree?

    # 3 $\displaystyle 27x^6 + 125y^6 = (3x^2)^3 + (5y^2)^3$

    I factored this as: $\displaystyle \boxed{(3x^2 + 5y^2)(9x^4 - 3x^2 5y^2 + 25y^4)}$

    # 4 $\displaystyle 4x^3 - 4x^2 - 9x + 9$

    Factor by grouping: $\displaystyle 4x^2(x - 1) -9(x - 1) \Rightarrow (4x^2 - 9)(x - 1)$

    Which goes to: $\displaystyle \boxed{(2x - 3)(2x + 3)(x - 1)}$

    Look good guys/gals?
    Good job on #1. For the second one, now factor $\displaystyle 9x^2-35x-4$. The third one looks good, although multiply out the 3 and the 5 to get $\displaystyle 15x^2y^2$. And, the fourth one looks good. Nice job.
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  3. #3
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    Why not just notice that $\displaystyle \frac{{25}}{{\sqrt 5 }} = 5\sqrt 5 $?
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  4. #4
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    Quote Originally Posted by topsquark View Post
    I'll add a bit here to Plato's answer: An expression is typically considered to be unsimplified if there is a radical in the denominator.

    -Dan
    There are many ways to do this problem. Although he did unnecessary steps, his final answer does not have a radical in the denominator and is in simplified form.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by AfterShock View Post
    There are many ways to do this problem. Although he did unnecessary steps, his final answer does not have a radical in the denominator and is in simplified form.
    Oops! I didn't look down the page far enough.

    -Dan
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