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Please Help Mathematician - Pure 2 Simple Question. watch

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    Given That G(-5) = G(2) = 0 And That

    G(x) = (x - B)(x + C)(x^2 + 7x - 1)

    Where B And C Are Positive Integers, Find B And C.

    Help? Where Do I Start?
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    i'll have a bash at it
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    (Original post by nimzy)
    GO ON THEN, HAVE A GO.
    May I suggest you look up the factor theorem in your textbook? And if I'm not mistaken, that's a Pure 1 question, not pure 2...

    Here's a clue:
    If the function equals zero when x takes a certain value n, then (x-n) is a factor of the function.
    Replace n with your two values and voila.
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    (Original post by meepmeep)
    Last time someone said that around me, I got punched in the face.

    Anyway, may I suggest you look up the factor theorem in your textbook? And if I'm not mistaken, that's a Pure 1 question, not pure 2...

    Here's a clue:
    If the function equals zero when x takes a certain value n, then (x-n) is a factor of the function.
    Replace n with your two values and voila.
    its in my pure 2 textbook - and it isnt the factor theorum.
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    (Original post by nimzy)
    its in my pure 2 textbook - and it isnt the factor theorum.
    Well, if you take the second part of what I said:
    If the function equals zero when x takes a certain value n, then (x-n) is a factor of the function.
    we have G(-5)=0 and G(2)=0
    Comparing with the above, we get (x+5) and (x-2) as factors.
    Comparing to G(x) = (x - B)(x + C)(x^2 + 7x - 1), we find that (x+5) must be referring to C and (x-2) must be referring to (x-B). Therefore, C=5 and B=2.
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    (Original post by meepmeep)
    Well, if you take the second part of what I said:


    we have G(-5)=0 and G(2)=0
    Comparing with the above, we get (x+5) and (x-2) as factors.
    Comparing to G(x) = (x - B)(x + C)(x^2 + 7x - 1), we find that (x+5) must be referring to C and (x-2) must be referring to (x-B). Therefore, C=5 and B=2.
    I HAVE NEVER DONE THAT BEFORE, WE HAVENT COVERED THAT.

    PLEASE COULD YOU EXPLAIN, COZ I DONT GET ANY OF THAT AT ALL.
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    (Original post by nimzy)
    I HAVE NEVER DONE THAT BEFORE, WE HAVENT COVERED THAT.

    PLEASE COULD YOU EXPLAIN, COZ I DONT GET ANY OF THAT AT ALL.
    Sorry, no can do, as to explain it I would need paper to show how it works. Best go e-mail your maths teacher if he hasn't taught you it. The maths department at your school's good anyway!
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    (Original post by joicey)
    its bloody christmas holls why the hell you doing that?????
    i need to catch up on maths, a little behind, and i also have a physics exam on jan 12th
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    (Original post by nimzy)
    Given That G(-5) = G(2) = 0 And That

    G(x) = (x - B)(x + C)(x^2 + 7x - 1)

    Where B And C Are Positive Integers, Find B And C.

    Help? Where Do I Start?
    This is definitely in P1/P2 sumwhere.... if G (a) = 0, then
    (x - a) is a factor of G (x).

    knowing that the third term (x^2 + 7x - 1) wont factorize to any integer (b^2 - 4ac =/= 0), u know that -5 and 2 must be one of -B and +C. since u know that B and C are both positive integers, it must be in the form x - B, and x - C, so u choose sub in -5 and 2 for B and C such that both end up positive.

    well clearly B = 2 and C = 5.
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    answer is B= -5, C= -2 :confused: right?
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    (Original post by nimzy)
    I HAVE NEVER DONE THAT BEFORE, WE HAVENT COVERED THAT.

    PLEASE COULD YOU EXPLAIN, COZ I DONT GET ANY OF THAT AT ALL.
    I suggest you ask your teacher then, or at least do a little work yourself.

    Also, people are more inclined to help when you do not type in capitals - it comes across as shouting and is arrogant.
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    (Original post by Absolution)
    answer is B= -5, C= -2 :confused: right?
    Erm, B and C are both positive, so you have to take C=5 and B=2 for it to work.
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    (Original post by meepmeep)
    Erm, B and C are both positive, so you have to take C=5 and B=2 for it to work.
    oh right, ah well i'm rubbish at maths anyway at least i got the numbers
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    this isnt that hard........ u just need to know why B and C are one each of 2 and 5, and which one is which..........
 
 
 

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