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Math Help - Determine the value of a

  1. #1
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    Question Determine the value of a

    Given that (3x-a)(x-2)=3x^3-32x^2+81x-70, determine the value of a. My answer was 3 but not if it is right. thanks for any help with this one.
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  2. #2
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    Hello, kwtolley!

    Obviously, there is a typo in the problem . . .

    Given that (3x-a)(x-2) \:= \:3x^3-32x^2+81x-70, determine the value of a.
    The left side is a quadratic
    . . and cannot possibly equal the cubic on the right.
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  3. #3
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    Quote Originally Posted by kwtolley
    Given that (3x-a)(x-2)=3x^3-32x^2+81x-70, determine the value of a. My answer was 3 but not if it is right. thanks for any help with this one.
    It is possible to solve in terms of x but this number would not be a constant, is that what you're looking for or is there a typo in the way you wrote the problem?
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    Question redo of question sorry about that

    Given that (3x-a)(x-2)(x-7)=3x^3-32x^2+81x-70, determine the value of a. My answer is 3 is that right. Thanks again for checking with me.
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  5. #5
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    Hello, kwtolley!

    You blew it again: that's 87x . . .


    Given that (3x-a)(x-2)(x-7) \:=\:3x^3-32x^2+87x-70, determine the value of a.

    Multiply out the left side:
    . . 3x^3 - (a + 27)x^2 + (9a + 42)x - 14a \;= \;3x^3 - 32x^2 + 87x - 70

    Two polynomials are equal if their corresponding coefficients are equal.

    So we have:
    \begin{array}{cccc}3\:=\:3 \\ -(a + 27) \:= \:-32 \\ 9a + 42 \:= \:87 \\ -14a \:= \:-70\end{array}

    And the solution is: . a = 5

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