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Math Help - Can you Check This please?

  1. #1
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    Question Can you Check This please?

    Need someone to check this and go over it with me please.

    Equation in factored form for a polynomial?


    1. Determine an equation in factored form for a polynomial function with zeros -1 (order 2) and 3 (order 3) that passes through the point (4,5).


    y = a [ x + 1 ] [ x - 3 ]^3 with 5 = a [ 5][1^3]

    I don't quite understand the bold part. How Do I incorporate it into my final function/answer.

    Am I solving for a?
    ....5=a(25)(1)
    ....5=a(25)
    ....1/5=a
    ?

    2. Determine an equation in factored form for the polynomial function with zeros 2 (order 2), 2/3 and 3 that passes through the point (1,6).


    y = a [ x - 2 ] [ 3x - 2 ][ x - 3 ] ...with 6 = a [ 1-2] [ 3 - 2 ] [ 1 - 3 ]

    How do we go from (the green) 2/3 to [3x-2]? And again, i'm confused with the bold part and how to put it into a?
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  2. #2
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    Re: Can you Check This please?

    Quote Originally Posted by tdotodot View Post
    Need someone to check this and go over it with me please.

    Equation in factored form for a polynomial?


    1. Determine an equation in factored form for a polynomial function with zeros -1 (order 2) and 3 (order 3) that passes through the point (4,5).


    y = a [ x + 1 ] [ x - 3 ]^3 with 5 = a [ 5][1^3]

    I don't quite understand the bold part. How Do I incorporate it into my final function/answer.

    Am I solving for a?
    ....5=a(25)(1)
    ....5=a(25)
    ....1/5=a
    ?
    Yes, that's correct. Now just replace "a" in the formula "y = a [ x + 1 ] [ x - 3 ]^3" with that number.

    Do you see how "y = a [ x + 1 ] [ x - 3 ]^3" and "passes through the point (4, 5)" results in "5= a[5^2][1]^3"?


    2. Determine an equation in factored form for the polynomial function with zeros 2 (order 2), 2/3 and 3 that passes through the point (1,6).


    y = a [ x - 2 ] [ 3x - 2 ][ x - 3 ] ...with 6 = a [ 1-2] [ 3 - 2 ] [ 1 - 3 ]

    How do we go from (the green) 2/3 to [3x-2]? And again, i'm confused with the bold part and how to put it into a?
    Saying that x= 2/3 is a zero means that it has a factor of the form "x- 2/3". Multiply through by 3.

    What do you get if you do the arithmetic indicated in "6= a[1- 2]^3 [3- 2][1- 3]"?
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  3. #3
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    Re: Can you Check This please?

    Quote Originally Posted by HallsofIvy View Post
    Yes, that's correct. Now just replace "a" in the formula "y = a [ x + 1 ] [ x - 3 ]^3" with that number.

    Do you see how "y = a [ x + 1 ] [ x - 3 ]^3" and "passes through the point (4, 5)" results in "5= a[5^2][1]^3"?



    Saying that x= 2/3 is a zero means that it has a factor of the form "x- 2/3". Multiply through by 3.

    What do you get if you do the arithmetic indicated in "6= a[1- 2]^3 [3- 2][1- 3]"?
    #1.
    ...............So my final answer would then be; y = 1/5 [ x + 1 ] [ x - 3 ]^3 ?
    ...............5 = a [5^2][1]^3.... Why are you squaring 5? Why are you putting "[1]^3"? How come the "4" is not incorporated?

    #2.
    6= a[1- 2]^3 [3- 2][1- 3]
    6= a(-1)^3 (1) (-2)
    6= a(1)(1)(-2)
    6= a(-2)
    -3= a

    So then... y = a [ x - 2 ] [ 3x - 2 ][ x - 3 ]
    .............. y= -3[ x - 2 ] [ 3x - 2 ][ x - 3 ] ...would be my final answer?
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  4. #4
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    Re: Can you Check This please?

    Quote Originally Posted by tdotodot View Post
    #1.
    ...............So my final answer would then be; y = 1/5 [ x + 1 ] [ x - 3 ]^3 ?
    ...............5 = a [5^2][1]^3.... Why are you squaring 5? Why are you putting "[1]^3"? How come the "4" is not incorporated?
    The "4" is incorporated- that is where the "5" and "3" came from. You had
    y= a[x+ 1]^2[x- 3]^3 and you want y= 5 when x= 4 . Replacing y by 5 and x by 4 in the equation
    5= a[4+ 1]^2[4- 3]^3= a[5]^2[1]^3.

    #2.
    6= a[1- 2]^3 [3- 2][1- 3]
    This should be [1- 2]^2, not to third power. Other than that, yes setting x= 1, y= 6 is correct.

    6= a(-1)^3 (1) (-2)
    6= a(1)(1)(-2)
    (-1)^3= -1, not 1 but since that was supposed to be squared, this is correct!

    6= a(-2)
    -3= a

    So then... y = a [ x - 2 ] [ 3x - 2 ][ x - 3 ]
    .............. y= -3[ x - 2 ] [ 3x - 2 ][ x - 3 ] ...would be my final answer?
    Yes. You could, rather than have represented the "zero at 2/3" by 3x- 2 have used "x- 2/3" instead.
    That would give you y= a[x- 2]^2[x- 2/3][x- 3] and setting x= 1, y= 6
    6= a(1- 2)^2(1- 2/3)(1- 3)= a(1)(1/3)(-2)= -(2/3)a
    to -(2/3)a= 6 and now a= -6(3/2)= -9.
    That gives y= -9(x- 2)^2(x- 2/3)(x- 3) but factoring 9 into 3(3) and multiplying one of those "3"s into the (x- 2/3) term,
    y= -3(x- 2)^2(3x- 2)(x- 3) exactly the same as before!
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