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Math Help - binomial expansion question - don't know what is being asked?

  1. #1
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    binomial expansion question - don't know what is being asked?

    Hi

    I have this question:
    (i) show that (2 + x)^4 = 16 + 32x + 24x^2 + 8x^3 + x^4 for all x.

    for all x? What does that mean?

    (ii) Find the values of x for which (2 + x)^4 = 16 + 16x + x^4

    ??

    Can someone please explain?

    Angus
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  2. #2
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    i) For all \displaystyle x means it's true no matter what the value of \displaystyle x is Just perform the binomial expansion.

    ii) Expand \displaystyle (2+x)^4 which you should have done for part i) and then set it equal to \displaystyle 16+16x+x^4 and solve.
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  3. #3
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    Quote Originally Posted by angypangy View Post
    (i) show that (2 + x)^4 = 16 + 32x + 24x^2 + 8x^3 + x^4 for all x.
    for all x? What does that mean?

    (ii) Find the values of x for which (2 + x)^4 = 16 + 16x + x^4
    We know \left( {a + b} \right)^4  = \sum\limits_{k = 0}^4 {\binom{4}{k}a^{4 - k} b^k }.
    So for the first part let a=2~\&~b=x.

    For part two solve 16x+24x^2+8x^2=0.
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  4. #4
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    i think this is the equation for second one

    x(x^2 + 3x +2) = 0

    the values of x are 0, -1 ,-2
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  5. #5
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    Hi

    I am still a little confused with this question. I can see how to get factors 0, -1 and -2. But if I use them in (2 + x)^4 - ie 2+ -2 = 0 - 0^4 = 0 - not 16 + 16x + x^4

    Can someone please explain?

    Angus
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    (2+x)^4 = x^4+8 x^3+24 x^2+32 x+16

    x^4+8 x^3+24 x^2+32 x+16 = x^4+16x+16[/tex]

    x^4 and 16 cancel and when you take 16x from both sides you get

    8x^3+24x^2+16x = 0

    You can factor out 8x from this equation to give 8x(x^2+3x+2) = 8x(x+2)(x+1) = 0


    edit: You have to sub in 0, -1 and 2 on the RHS too

    For example using 0

    (2+0)^4 = 16+16(0)+(0)^4 \implies 16=16 and so x=0 is a solution
    Last edited by e^(i*pi); January 7th 2011 at 07:20 AM. Reason: see posr
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  7. #7
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    Quote Originally Posted by angypangy View Post
    Hi

    I am still a little confused with this question. I can see how to get factors 0, -1 and -2. But if I use them in (2 + x)^4 - ie 2+ -2 = 0 - 0^4 = 0 - not 16 + 16x + x^4

    Can someone please explain?

    Angus
    (2+x)^4=x^4+8x^3+24x^2+32x+16

    (2+x)^4=x^4+16x+16\Rightarrow\ \left(x^4+8x^3+24x^2+32x+16\right)-\left(x^4+16x+16\right)=0

    \Rightarrow\ 8x^3+24x^2+16x=0\Rightarrow\ x^3+3x^2+2x=0

    You can only verify that x=0,\;-1,\;-2 cause the equation x^3+3x^2+2x to be zero.
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  8. #8
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    Ah you sub in val of x on right too. Yes understand now. Thanks.
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