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simple way to calculate binomials if you know opposite sign answer? Watch

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    so if you know the answer to (2+3x)^4 or the expansion of it how would you calculate the expansion of (2-3x)^4 from this
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    (Original post by jonjoshelvey21)
    so if you know the answer to (2+3x)^4 or the expansion of it how would you calculate the expansion of (2-3x)^4 from this
    Take the original expansion and replace x with -x, since (2-3x)^4 \equiv (2+3(-x))^4. You'll notice that even powers remain unaffected but the odd powers switch signs.
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    (Original post by RDKGames)
    Take the original expansion and replace x with -x, since (2-3x)^4 \equiv (2+3(-x))^4. You'll notice that even powers remain unaffected but the odd powers switch signs.
    thanks lad
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    (Original post by RDKGames)
    Take the original expansion and replace x with -x, since (2-3x)^4 \equiv (2+3(-x))^4. You'll notice that even powers remain unaffected but the odd powers switch signs.
    what if the question were the other way round like if you knew the answer to (2-3x)^4 but wanted to know (2+3x)^4??
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    (Original post by jonjoshelvey21)
    what if the question were the other way round like if you knew the answer to (2-3x)^4 but wanted to know (2+3x)^4??
    It’s obvious isnt it? Every odd power will be negative so just make them positive. Every occurance of x you see, factor into (-x) then replace this by (x)
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    (Original post by RDKGames)
    It’s obvious isnt it? Every odd power will be negative so just make them positive. Every occurance of x you see, factor into (-x) then replace this by (x)
    what so wherever an x is present you plug in -x?
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    (Original post by jonjoshelvey21)
    what so wherever an x is present you plug in -x?
    No. Whereever x is, say you have -3x^3, you want to factor it in such a way that you have -x, so 3(-x)^3 then replace every -x with x, to get 3x^3
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    (Original post by RDKGames)
    No. Whereever x is, say you have -3x^3, you want to factor it in such a way that you have -x, so 3(-x)^3 then replace every -x with x, to get 3x^3
    thanks sorry to be confusing but for my very first question would you factor it in such a way to get -x as well ????
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    (2+3x)^4 = 81x^4 + 216^3 + 216x^2 + 96x + 16
    1 x 3x^4 x 2^0 = 81x^4
    4 x 3x^3 x 2^1 = 216x^3
    6 x 3x^2 x 2^2 = 216x^2
    4 x 3x^1 x 2^3 = 96x
    1 x 3x^0 x 2^4 = 16

    It's basically going to be the same answer however it will alternate between positive and negative coefficients, for example (-2^1) = -1 but (-2^2) = 4 and (-2^3) = -8

    (2-3x)^4
    1 x (-3x^4) x 2^0 = 81x^4
    4 x (-3x^3) x 2^1 = -216x^3
    6 x (-3x^2) x 2^2 = 216x^2
    4 x (-3x^1) x 2^3 = -96x
    1 x (-3x^0) x 2^4 = 16
 
 
 
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