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Math Help - 3 homework question

  1. #1
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    3 homework question

    1) Use long division to find the remainder when x^4-3x^2+5x-1 is divided by x^2-3

    2) Solve the rational equation

    x(2x+1) = 10 - 5
    x - 2____ x-2_ 2

    3) The line x = 3 is the axis of symmetry for the graph of a parabola. If the parabola contains the points (5, -3) and (-1,9), what is the equation for the parabola?
    Vertex is (3,-7)
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by greenpumpkins View Post
    1) Use long division to find the remainder when x^4-3x^2+5x-1 is divided by x^2-3

    2) Solve the rational equation

    x(2x+1) = 10 - 5
    x - 2____ x-2_ 2

    3) The line x = 3 is the axis of symmetry for the graph of a parabola. If the parabola contains the points (5, -3) and (-1,9), what is the equation for the parabola?
    Vertex is (3,-7)
    See here for help with polynomial long division. The quotient is x^2+\frac{5x-1}{x^2-3}. The remainder is 5x-1


    2) Solve the rational equation:

    \frac{x(2x+1)}{x-2}=\frac{10}{x-2}-\frac{5}{2}

    Multiply each term by the LCD of 2(x-2)

    2(x-2)\left(\frac{x(2x+1)}{x-2}\right)=2(x-2)\left(\frac{10}{x-2}\right)-2(x-2)\left(\frac{5}{2}\right)

    2x(2x+1)=2(10)-5(x-2)

    4x^2+2x=20-5x+10

    4x^2+7x-30=0

    (x-2)(4x+15)=0

    x=2 \ \ or \ \ x=-\frac{15}{4}

    But x=2 makes the denominator of the rational equation to be 0. So this solution is extraneous.


    3) The line x = 3 is the axis of symmetry for the graph of a parabola. If the parabola contains the points (5, -3) and (-1,9), what is the equation for the parabola?
    Vertex is (3,-7)

    y=a(x-h)^2+k Vertex form of the parabola with V(h,k)

    Substitute one of the points on the parabola (5, -3) and the vertex (3, -7)

    -3=a(5-3)^2-7

    -3=a(4)-7

    a=1

    Therefore the equation for the parabola is

    \boxed{y=(x-3)^2-7} or \boxed{y=x^2-6x+2}

    [tex]
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