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Math Help - How do you factor 8h^2-24h-320=0

  1. #1
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    How do you factor 8h^2-24h-320=0

    I keep reasoning like this:

    -8(-h^2+3h+40), find some numbers that mult. to -40 and add to 3, 8*-5=-40 and 8+-5=3, so -8(-h+8)(h-5) but this makes 8h^2-24h+320=0 which is a different statement.

    there seems to be something wrong with the -8(-h+8)(h-5) step. Somehow the 8 and -5 are the zeros and the real answer is 8(h-8)(h+5)=0 but that contradicts how every other quadratic is factored -normally c=mn and b=m+n and the "m"s and "n"s are always here (x+m)(x+n) away from the zeros.
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  2. #2
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    Why not just take out \displaystyle 8 as a factor?

    This would leave \displaystyle 8(h^2 - 3h - 40) = 0

    \displaystyle 8(h-8)(h+5) = 0...
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  3. #3
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    Quote Originally Posted by bournouli View Post
    I keep reasoning like this:

    -8(-h^2+3h+40), find some numbers that mult. to -40 and add to 3, 8*-5=-40 and 8+-5=3, so -8(-h+8)(h-5) but this makes 8h^2-24h+320=0 which is a different statement.

    there seems to be something wrong with the -8(-h+8)(h-5) step. Somehow the 8 and -5 are the zeros and the real answer is 8(h-8)(h+5)=0 but that contradicts how every other quadratic is factored -normally c=mn and b=m+n and the "m"s and "n"s are always here (x+m)(x+n) away from the zeros.
    this method applies only when the coefficient of the quadratic item is positive, i.e., u need to simplify the original equation like this:
    8(h^2-3h+40)=0;
    now I think u can handle it well and get the right answers
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  4. #4
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    Quote Originally Posted by Prove It View Post
    Why not just take out \displaystyle 8 as a factor?

    This would leave \displaystyle 8(h^2 - 3h - 40) = 0

    \displaystyle 8(h-8)(h+5) = 0...
    And why not tell the OP how to factorise rather than just do it for them.

    If a polynomial with integer coefficients has rational roots they are a ratio of a factor of the constant term to a factor of the leading coefficient.

    So for x^2-3h-40 we know all the rational roots are factors of 40, these are \pm1, \pm 2, \pm 4, \pm 8, \pm 5, \pm 10, \pm 20. Now check through this list to find the rational roots, then you can turn these into the factors.



    CB
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    And why not tell the OP how to factorise rather than just do it for them.

    If a polynomial with integer coefficients has rational roots they are a ratio of a factor of the constant term to a factor of the leading coefficient.

    So for x^2-3h-40 we know all the rational roots are factors of 40, these are \pm1, \pm 2, \pm 4, \pm 8, \pm 5, \pm 10, \pm 20. Now check through this list to find the rational roots, then you can turn these into the factors.



    CB
    The OP had given a full solution, which was nearly correct except for the sign error I pointed out ....
    Last edited by mr fantastic; March 26th 2011 at 03:12 PM.
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    Why should the sign of the coefficient matter? The textbook agrees with you but still isn't a positive coefficient (like +a) just a negative coefficient (-(-a))? So if the text is right, we could never factor since positive numbers are implicitly negative. Obviously I either don't know what coefficients are or there's no reason why one should worry about the signs -and I don't know which to choose.

    (sorry this was a response to kylehk)
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  7. #7
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    Quote Originally Posted by bournouli View Post
    Why should the sign of the coefficient matter? The textbook agrees with you but still isn't a positive coefficient (like +a) just a negative coefficient (-(-a))? So if the text is right, we could never factor since positive numbers are implicitly negative. Obviously I either don't know what coefficients are or there's no reason why one should worry about the signs -and I don't know which to choose.

    (sorry this was a response to kylehk)
    Hi bournouli,

    sorry for any misunderstanding about the 'positive' conclusion; it just could simplify the factorization for us but not necessarily to enforce a '+' sign.
    Just reread ur post, now I guess I find where u went wrong:
    you transformed the original equation to -8(-h^2+3H+40)=0, it's ok for the '-h^2';
    now suppose the factorization result is (-h+a)(h+b)=0
    spread it we'll get -h^2++(a-b)h+ab=0; compare it with -h^2+3H+40=0
    then we get two equations: a-b=3, ab=40 (exactly where u went wrong, as u considered ab=-40 in ur post)
    a=8, b=5.

    back to (-h+a)(h+b)=0, two solutions should be h=a or h=-b, i.e., h=8, or h=-5.

    hope it's clear for u now
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  8. #8
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    Quote Originally Posted by kylehk View Post
    Hi bournouli,

    sorry for any misunderstanding about the 'positive' conclusion; it just could simplify the factorization for us but not necessarily to enforce a '+' sign.
    Just reread ur post, now I guess I find where u went wrong:
    you transformed the original equation to -8(-h^2+3H+40)=0, it's ok for the '-h^2';
    now suppose the factorization result is (-h+a)(h+b)=0
    spread it we'll get -h^2++(a-b)h+ab=0; compare it with -h^2+3H+40=0
    then we get two equations: a-b=3, ab=40 (exactly where u went wrong, as u considered ab=-40 in ur post)
    a=8, b=5.

    back to (-h+a)(h+b)=0, two solutions should be h=a or h=-b, i.e., h=8, or h=-5.

    hope it's clear for u now
    for this, a-b=3, ab=40 how did you get the positive 40? I want to take the -1 coefficient on h^2 and mult. that by b to get -40 but that always leads to antinomies so that last part isn't clear. Thanks for the help though.
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  9. #9
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    Quote Originally Posted by bournouli View Post
    for this, a-b=3, ab=40 how did you get the positive 40? I want to take the -1 coefficient on h^2 and mult. that by b to get -40 but that always leads to antinomies so that last part isn't clear. Thanks for the help though.
    suppose the factorization result is (-h+a)(h+b)=0; spread it we'll get -h^2+(a-b)h+ab=0;
    as -h^2+(a-b)h+ab=0 is just another expression of the equation -h^2+3h+40=0, the coefficient of corresponding items (i.e., h^2, h, and the constant item) should be equivalent, right?
    -h^2 + (a-b)h + ab=0
    -h^2 + 3 h + 40=0
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  10. #10
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    thanks kylehk,

    but why do you think my original method failed -it seemed so straightforward?

    Is it only applicable when a=1?
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  11. #11
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    Quote Originally Posted by bournouli View Post
    thanks kylehk,

    but why do you think my original method failed -it seemed so straightforward?

    Is it only applicable when a=1?
    sorry, that was kinda stupid, I mean is it only applicable when a=positive number? It probably isn't because you explicitly stated that the sign doesn't matter but there is truly no other way I can find that explains why I was wrong.
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  12. #12
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    Quote Originally Posted by bournouli View Post
    I keep reasoning like this:

    -8(-h^2+3h+40), find some numbers that mult. to -40 and add to 3, 8*-5=-40 and 8+-5=3, so -8(-h+8)(h-5) but this makes 8h^2-24h+320=0 which is a different statement.

    Where you have gone wrong here is....

    you should be finding the factors of 40 that differ by 3, since "after" multiplying out


    (h+?)(-h+?)

    one of the numbers will multiply by "-h"



    there seems to be something wrong with the -8(-h+8)(h-5) step. Somehow the 8 and -5 are the zeros and the real answer is 8(h-8)(h+5)=0 but that contradicts how every other quadratic is factored -normally c=mn and b=m+n and the "m"s and "n"s are always here (x+m)(x+n) away from the zeros.
    Factoring is just the following process...

    (5)3=15

    (7-2)(1+2)=15

    7(1+2)-2(1+2)=7(1)+7(2)-2(1)-2(2)=7+14-2-4=21-6=15

    5(6)=30

    (3+2)(10-4)=3(10-4)+2(10-4)=30-12+20-8=50-20=30

    (h-8)(h+5)=?

    We do not know "h", so we do not know "h+8" etc,
    hence we multiply out "term-by-term".

    (h-8)(h+5)=h(h+5)-8(h+5)=h^2+5h-8h-40=h^2-3h-40

    Your equation is

    8h^2-24h-320=0\Rightarrow\ 8\left(h^2-3h-40\right)=0

    8\ne\ 0\Rightarrow\ h^2-3h-40=0

    which is also saying

    h^2-(3h+40)=0

    h^2=3h+40

    Both are equal, so if you subtract them, the result is 0.
    You can subtract either side, so you can have

    h^2-3h-40=0

    or

    3h+40-h^2=0

    Hence, if you a "minus" on the squared term, you can always change all the signs
    when you have an equality.

    h^2-3h-40=0

    h(h)=h^2

    so you can factor using

    (h+?)(h+?)=h^2-3h-40

    Hence, you need the factors of -40 that combine to give -3.

    If you have "negative h squared", your factors look like this...

    (h+?)(-h+?)

    and you must arrive at

    -h^2+3h+40=0

    so you need the factors of 40,
    such that multiplying out the brackets leaves you with +3h

    (h+a)(-h+b)=h(-h+b)+a(-h+b)

    =-h^2+hb-ah+ab=-h^2+h(b-a)+ab

    so

    b-a=3

    ab=40
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  13. #13
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    Quote Originally Posted by Archie Meade View Post
    Factoring is just the following process...

    (5)3=15

    (7-2)(1+2)=15

    7(1+2)-2(1+2)=7(1)+7(2)-2(1)-2(2)=7+14-2-4=21-6=15

    5(6)=30

    (3+2)(10-4)=3(10-4)+2(10-4)=30-12+20-8=50-20=30

    (h-8)(h+5)=?

    We do not know "h", so we do not know "h+8" etc,
    hence we multiply out "term-by-term".

    (h-8)(h+5)=h(h+5)-8(h+5)=h^2+5h-8h-40=h^2-3h-40

    Your equation is

    8h^2-24h-320=0\Rightarrow\ 8\left(h^2-3h-40\right)=0

    8\ne\ 0\Rightarrow\ h^2-3h-40=0

    which is also saying

    h^2-(3h+40)=0

    h^2=3h+40

    Both are equal, so if you subtract them, the result is 0.
    You can subtract either side, so you can have

    h^2-3h-40=0

    or

    3h+40-h^2=0

    Hence, if you a "minus" on the squared term, you can always change all the signs
    when you have an equality.

    h^2-3h-40=0

    h(h)=h^2

    so you can factor using

    (h+?)(h+?)=h^2-3h-40

    Hence, you need the factors of -40 that combine to give -3.

    If you have "negative h squared", your factors look like this...

    (h+?)(-h+?)

    and you must arrive at

    -h^2+3h+40=0

    so you need the factors of 40,
    such that multiplying out the brackets leaves you with +3h

    (h+a)(-h+b)=h(-h+b)+a(-h+b)

    =-h^2+hb-ah+ab=-h^2+h(b-a)+ab

    so

    b-a=3

    ab=40
    Wow, that helped.

    I suppose my fallacy was in assuming that the "a" in ab=40 was the same as the "a" in ax^2+bx+c so that if it was -ax^2+bx+c one had to use -1 or something?
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  14. #14
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    Ah yes.

    The "a", "b" and "c" in the quadratic refer to those specific multipliers of x^2 and x and the constant at the end.

    This allows direct placement of those constants into certain formulae,
    such as the quadratic formula and the perpendicular distance from a point to a line formula.
    If any of those constants are negative, then a negative value goes into those formulae.

    For example 2x^2-3x+4 has a=2, b=-3, c=4

    while -3x^2+2x+5 has a=-3, b=2, c=5.

    You'll get used to this.
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