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Math Help - Need help translating standard form quadratic equations into factored form?

  1. #1
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    Need help translating standard form quadratic equations into factored form?

    Could someone please translate these equations and explain the steps for doing it?

    y = x - 8x + 15
    y = 3x + 25x + 42

    Thanks.
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  2. #2
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    It's not magic. You just have to look.

    x - 8x + 15 =

    Obvious part:

    (x __ ___)(x __ ___)

    Factors of 15

    1,15
    3,5

    That's it.

    Can one get 8 from these factors?

    1+15 = 16 -- Nope.
    1-14 = -14 -- Nope
    3-5 = -2 -- Nope
    3+5 = 8 -- Yes!!

    (x __ 3)(x __ 5)


    15 > 0, so 3 and 5 need the same sign.
    -8 < 0, so both must be negative.

    x - 8x + 15 = (x - 3)(x - 5)

    You do the other one.
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  3. #3
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    Quote Originally Posted by mathdonkey View Post
    Could someone please translate these equations and explain the steps for doing it?

    y = x - 8x + 15
    y = 3x + 25x + 42

    Thanks.
    I'm guessing you mean in vertex form, that is, y=a(x-h)^2+k

    To do the we need to complete the square, to do this we take half of the linear coeffeint and square it.

    y=x^2-8x+15

    So the linear coeffeint is -8 half of it is \frac{-8}{2}=-4 and now we square it to get (-4)^2=16 This is what we need to complete the square.


    We will now add zero to the euqation in the form of 0=(16-16)

    so we get

    y=x^2-8x+15+0=x^2-8x+15+(16-16)


    Will will now rearrange the terms to get
    y=x^2-8x+16 +15-16

    Grouping the first three terms together and factoring gives

    y=[x^2-8x+16]-1=(x-4)^2-1

    So we get

    y=(x-4)^2-1 We can now identify a, h, and k

    a=1 h=4 and k=-1

    So we know that the vertext is located at the point (h,k) \to (4,-1) and the the parabola opens up because a is postitive.

    I hope this helps and is what you wanted.

    If so here is a hint for #2 factor like this

    y=ax^2+bx+c=a(x^2+\frac{b}{a}x)+c

    and try to adapt the method used above.

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  4. #4
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    Quote Originally Posted by TKHunny View Post
    x - 8x + 15 = (x - 3)(x - 5)
    How do I know it's (x - 3)(x - 5) and not (x - 5)(x - 3)?

    Quote Originally Posted by TKHunny
    You do the other one.
    In the other one the coefficient of x is 3. This will change how it's done, so it would be more helpful to see someone else do it and then I can learn from their method.
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  5. #5
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    Quote Originally Posted by mathdonkey View Post
    How do I know it's (x - 3)(x - 5) and not (x - 5)(x - 3)?
    Nice joke
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  6. #6
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    Quote Originally Posted by mathdonkey View Post
    How do I know it's (x - 3)(x - 5) and not (x - 5)(x - 3)?



    In the other one the coefficient of x is 3. This will change how it's done, so it would be more helpful to see someone else do it and then I can learn from their method.

    Multiplication is commutative, so (x - 3)(x - 5) = (x - 5)(x - 3)

    3x^2+25x+42

    To factor this trinomial, first multiply leading coef. times constant: 3\times42

    Result: 126

    Find two factors of 126 that add up to the middle term's coef. of 25.

    That'd be 18 and 7.

    Replace the middle term with these two factors:

    3x^2+18x+7x+42

    Factor by grouping: (3x^2+18x)+(7x+42)

    3x(x+6)+7(x+6) = (3x+7)(x+6)
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  7. #7
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    Quote Originally Posted by masters View Post
    Multiplication is commutative, so (x - 3)(x - 5) = (x - 5)(x - 3)

    3x^2+25x+42

    To factor this trinomial, first multiply leading coef. times constant: 3\times42

    Result: 126

    Find two factors of 126 that add up to the middle term's coef. of 25.

    That'd be 18 and 7.

    Replace the middle term with these two factors:

    3x^2+18x+7x+42

    Factor by grouping: (3x^2+18x)+(7x+42)

    3x(x+6)+7(x+6) = (3x+7)(x+6)
    Thanks, that helped a lot!

    Quote Originally Posted by masters
    Multiplication is commutative, so (x - 3)(x - 5) = (x - 5)(x - 3)
    Oops, forgot to ask this:

    Wont swapping the binomials change the way the parabola looks?
    Last edited by mathdonkey; May 29th 2008 at 10:46 AM.
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  8. #8
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    Swapping the binomials will not have an effect. However, if you were to multiply ax^2+bx+c=0 by -1, then the parabola would open in the opposite direction. The vertex would change, but the x-intercepts would be the same.

    Did you need to see y=3x^2+25x+42 in vertex form? You can determine the zeros from the factored version in my previous post. They are \{-6,\frac{-7}{3}\}
    Last edited by masters; May 29th 2008 at 12:10 PM.
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    Quote Originally Posted by TKHunny View Post
    15 > 0, so 3 and 5 need the same sign.
    -8 < 0, so both must be negative.
    These are good rules to know, so:

    What if 15 was a negative number, how would it affect the signs?

    What would happen if 8 was a positive number?
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  10. #10
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    I'm having some trouble with this question too.

    y = 20x - 3x - 9

    I get this:

    = 20x + 12x - 15x - 9
    = (20x + 12x) (-15x - 9)
    = 2(10x + 12x) 3(-5x - 3)
    (I don't know what to do next)

    But the answer is supposed to be:

    y = (4x - 3) (5x + 3)

    What am I doing right/wrong?
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  11. #11
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    Quote Originally Posted by mathdonkey View Post
    These are good rules to know, so:

    What if 15 was a negative number, how would it affect the signs?

    What would happen if 8 was a positive number?
    -15 < 0, so factors must have opposite signs.

    If they've opposite signs, we're barking up the wrong tree since 5-3 = 2, not 8.
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  12. #12
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    Quote Originally Posted by mathdonkey View Post
    y = 20x - 3x - 9
    Just make a catalogue.

    20

    1*20
    2*10
    4*5

    9

    1*9
    3*3

    -9 < 0, so opposite signs
    We're looking for a -3

    1*1 - 9*20 -- too big
    1*20 - 9*1 = 11
    1*3 - 9*3 = -26
    2*1 - 10*20 -- too big
    2*20 - 10*1 = 30
    2*3 - 10*3 = -27
    4*1 - 5*20 -- too big
    4*20 - 5*1 = 75
    4*3 - 5*3 = -3 -- Aha!!!!

    (4x - 3)(5x + 3)

    Be systematic.

    Note: I started witht the most extreme values, just for the practice. I don't recommend that if you REALLY want the quickest path to the answer.
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  13. #13
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    Sorry TKHunny, I'm a little confused by your method.

    Quote Originally Posted by TKHunny View Post
    Just make a catalogue.

    20

    1*20
    2*10
    4*5

    9

    1*9
    3*3

    -9 < 0, so opposite signs
    Using those numbers I can write (4x - 3) (5x + 3). Is that all I have to do?

    We're looking for a -3

    1*1 - 9*20 -- too big
    1*20 - 9*1 = 11
    1*3 - 9*3 = -26
    2*1 - 10*20 -- too big
    2*20 - 10*1 = 30
    2*3 - 10*3 = -27
    4*1 - 5*20 -- too big
    4*20 - 5*1 = 75
    4*3 - 5*3 = -3 -- Aha!!!!
    And then when I'm finished do I do this to confirm my answer?
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  14. #14
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    Quote Originally Posted by mathdonkey View Post
    I'm having some trouble with this question too.

    y = 20x - 3x - 9

    I get this:

    = 20x + 12x - 15x - 9
    = (20x + 12x) + (-15x - 9)
    = 2(10x + 12x) - 3(5x + 3)
    (I don't know what to do next)

    But the answer is supposed to be:

    y = (4x - 3) (5x + 3)

    What am I doing right/wrong?
    In your third step you should've factored a 4x instead of 2.

    y=4x(5x+3)-3(5x+3)

    y=(4x-3)(5x+3)

    As far as those other questions, it is best if you get a graphing calculator and try those different transformations you're talking about.
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  15. #15
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    Quote Originally Posted by masters View Post
    In your third step you should've factored a 4x instead of 2.

    y=4x(5x+3)-3(5x+3)

    y=(4x-3)(5x+3)
    Once again, thank you! This helped so much.
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