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    yeah assuming
    f(x) = (x+1)(x+2)(x+3) + 3
    how do you show that this is completely divisible by 3
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    (Original post by MalaysianDude)
    yeah assuming
    f(x) = (x+1)(x+2)(x+3) + 3
    how do you show that this is completely divisible by 3
    (x+1), (x+2) and (x+3) are consecutive numbers . If you divide
    (x+1)(x+2)(x+3) by three you get (x+2)

    MB
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    (Original post by MalaysianDude)
    yeah assuming
    f(x) = (x+1)(x+2)(x+3) + 3
    how do you show that this is completely divisible by 3
    not sure how to show it mathematically, but if you think about it, out of three consecutive numbers, one of the numbers will always be a multiple of three, therefore the product of three consecutive numbers will always be divisible by three.....and then you just add 3 on, which means the number is still divisible by three.
    this will work similarly with other numbers, i.e. four consecutive numbers divided by four, etc...
    oh yeah....the numbers have to be integers, obviously
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    sorry for the late notification guys..
    i thought this was an easy q till one of my friends called me up and asked me
    i didnt know how to show her how to do it...

    so can someone tell me how i can write it out in the answer booklet?

    musicboy : i dont really get what you mean man..
    can you explain further?
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    (Original post by MalaysianDude)
    sorry for the late notification guys..
    i thought this was an easy q till one of my friends called me up and asked me
    i didnt know how to show her how to do it...

    so can someone tell me how i can write it out in the answer booklet?

    musicboy : i dont really get what you mean man..
    can you explain further?
    is this the same question from some past paper? becasue i was doing a similar question recently, and i wasn't able to do it. so when i checked the answers, they just exapected you to write "the product of 3 consecutive numbers is always divisible by three etc..." but you weren't required to prove it, because it was only a 2 mark question.
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    (Original post by MalaysianDude)
    yeah assuming
    f(x) = (x+1)(x+2)(x+3) + 3
    how do you show that this is completely divisible by 3
    the way id do is expend the braket n u gt ax^3 + bx^2 + cx + d = f(x)

    try to find 1 factor then use long division to find all the other factor!

    tho it may b wrong coz im tired bt im sure about expending the braket n using factor theorem!
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    (Original post by tammypotato)
    the way id do is expend the braket n u gt ax^3 + bx^2 + cx + d = f(x)

    try to find 1 factor then use long division to find all the other factor!

    tho it may b wrong coz im tired bt im sure about expending the braket n using factor theorem!
    i was trying that for ages when i did this question, but couldn't get anywhere...
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    (Original post by mockel)
    i was trying that for ages when i did this question, but couldn't get anywhere...
    lol so lets all hope this question doesnt come up
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    (Original post by mockel)
    is this the same question from some past paper? becasue i was doing a similar question recently, and i wasn't able to do it. so when i checked the answers, they just exapected you to write "the product of 3 consecutive numbers is always divisible by three etc..." but you weren't required to prove it, because it was only a 2 mark question.
    yups it was a past year question
    and frankly speaking,i dont really understand what they meant by "the product of 3 consecutive numbers is always divisible by three"
    i mean...
    is that all you have to write or something?
    cause i used a counter example method and i think it worked perfectly..
    but im still not too sure about that

    tammypotato : tried that man...didnt work
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    (Original post by MalaysianDude)
    yups it was a past year question
    and frankly speaking,i dont really understand what they meant by "the product of 3 consecutive numbers is always divisible by three"
    i mean...
    is that all you have to write or something?
    cause i used a counter example method and i think it worked perfectly..
    but im still not too sure about that

    tammypotato : tried that man...didnt work
    sorry hope i could help u bt im lost!
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    (Original post by MalaysianDude)
    yups it was a past year question
    and frankly speaking,i dont really understand what they meant by "the product of 3 consecutive numbers is always divisible by three"
    i mean...
    is that all you have to write or something?
    cause i used a counter example method and i think it worked perfectly..
    but im still not too sure about that

    tammypotato : tried that man...didnt work
    yeah, but it was only worth 2 marks
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    (Original post by mockel)
    not sure how to show it mathematically, but if you think about it, out of three consecutive numbers, one of the numbers will always be a multiple of three, therefore the product of three consecutive numbers will always be divisible by three.....and then you just add 3 on, which means the number is still divisible by three.
    this will work similarly with other numbers, i.e. four consecutive numbers divided by four, etc...
    oh yeah....the numbers have to be integers, obviously
    you could also explain it using this, if you want
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    (Original post by MalaysianDude)
    yeah assuming
    f(x) = (x+1)(x+2)(x+3) + 3
    how do you show that this is completely divisible by 3

    Any three consecutive numbers multiplied together to give a multiple of 3.. and as you add +3 at the end.. it is also a multiple of 3..

    so f(x) = (multiple of 3) + (multiple of 3) = still equals a multiple of 3
 
 
 
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