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Math Help - ACT Math Section

  1. #1
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    ACT Math Section

    I wasn't sure whether this belonged in this section or the algebra section. Anyway, I'm studying for the math portion of the ACT and I can't remember how to solve certain problems. Like this one:

    Which of the following is NOT a factor of z^5 - 16z ?

    A. z^2 - 1
    B. z^2 - 4
    C. z + 2
    D. z
    E. z - 2

    Now I'm aware the answer is A but I have no idea why. If someone could provide an explanation for how to solve that would be great.

    Another One: If (2x - y) / (x + y) = 2/3, then x/y = ?

    (2x - y is the numerator, x + y is the denominator)

    F. 1/2
    G. 2/3
    H. 5/4
    J. 5/3
    K. 5

    One more: For all nonzero x, y, and z such that x = yz, which of the following must be equivalent to xy?

    A. z/x
    B. yz^2
    C. yz
    D. x^2/z
    E. x/y

    I'm sorry if I posted in the wrong place. Thanks in advance.
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  2. #2
    Master Of Puppets
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    You should post your attempts to each question.

     z^5-16z

    Take out a common factor of z.

     z(z^4-16)

     z((z^2)^2-4^2)

    By difference of 2 squares

     z(z^2-4)(z^2+4)

     z(z-2)(z+2)(z^2+4)

     z(z-2)(z+2)(z^2-(\sqrt{-4})^2)

     z(z-2)(z+2)(z^2-(\sqrt{4}i)^2)

     z(z-2)(z+2)(z-\sqrt{4}i)(z+\sqrt{4}i)

     z(z-2)(z+2)(z-2i)(z+2i)
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  3. #3
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    Quote Originally Posted by 5441lk View Post
    I wasn't sure whether this belonged in this section or the algebra section. Anyway, I'm studying for the math portion of the ACT and I can't remember how to solve certain problems. Like this one:

    Which of the following is NOT a factor of z^5 - 16z ?

    A. z^2 - 1
    B. z^2 - 4
    C. z + 2
    D. z
    E. z - 2

    Now I'm aware the answer is A but I have no idea why. If someone could provide an explanation for how to solve that would be great.

    Another One: If (2x - y) / (x + y) = 2/3, then x/y = ?

    (2x - y is the numerator, x + y is the denominator)

    F. 1/2
    G. 2/3
    H. 5/4
    J. 5/3
    K. 5
    Multiplying on both sides by 3(x+y), 3(2x-y)= 2(x+y) so 6x- 3y= 2x+ 2y. Adding 3y and subtracting 2x from both sides, 4x= 5y. Now, what is x/y?

    One more: For all nonzero x, y, and z such that x = yz, which of the following must be equivalent to xy?

    A. z/x
    B. yz^2
    C. yz
    D. x^2/z
    E. x/y

    I'm sorry if I posted in the wrong place. Thanks in advance.
    The obvious first step is to multiply on both sides by y to get xy= y^2z. But that isn't any of the given answers so lets try eliminating some: if y= 2 and z= 1, then x= (2)(1)= 2 and xy= 4. z/x= 2/1= 2, not 4 so it can't be that. yz^2= 2(1^2)= 2 so that is not it. yz= 2(1)= 2 so that is not it. x^2/z= 2^2/1= 4 so that might be it! x/y= 2/2= 1 so that is not it.

    If this were a multiple choice test where I did not have much time, I think I could, with confidence, mark "D" and go on. But, of course, just showing it is correct for one chosen value of y and z doesn't mean it is always true. How can we get from x=yz to xy= x^2/z? As before, multiplying both sides of x= yz by y gives xy= y^2z. Now that I see the "correct" answer has only x and z in it, I can argue that x= yz is equivalent to y= x/z (divide both sides by z) and so y^2= x^2/z^2. Then xy= y^2z= (x^2/z^2)a= x^2/z as claimed.
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  4. #4
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    A more simple approach to the last one is to simply note that x=yz \Rightarrow y = \frac{x}{z} now substitute this with y in the expression xy to get: xy = x\frac{x}{z} = \frac{x^2}{z}.
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  5. #5
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    [quote=pickslides;357360]You should post your attempts to each question.

    Ok, next time I will.
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