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    Find the value of the constant k so that the polynomial P(x), where:
    P(x)=x^2+kx+11
    has a remainder 3 when it is divided by (x-2).
    Show that, with thsi value of k, P(x) is positive for all real x.

    I did the first part, but can't do the second. Here's what I did for the first bit:

    x^2+kx+11=3
    2^2+k(2)+11=3
    2k=-12
    k=-6

    Also, (29/32)^1/5 = 1/2(29)^1/5

    How?

    How do you make 29/32 into just 29, with 1/2 outside the bracket?

    Thanks a lot if you can help.
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    (Original post by TheQueen1986)
    Find the value of the constant k so that the polynomial P(x), where:
    P(x)=x^2+kx+11
    has a remainder 3 when it is divided by (x-2).
    Show that, with thsi value of k, P(x) is positive for all real x.

    I did the first part, but can't do the second. Here's what I did for the first bit:

    x^2+kx+11=3
    2^2+k(2)+11=3
    2k=-12
    k=-6

    Also, (29/32)^1/5 = 1/2(29)^1/5

    How?

    How do you make 29/32 into just 29, with 1/2 outside the bracket?

    Thanks a lot if you can help.
    so do you mean you cant do the 'show the real roots' bit?
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    (Original post by TheQueen1986)
    Also, (29/32)^1/5 = 1/2(29)^1/5

    How?

    How do you make 29/32 into just 29, with 1/2 outside the bracket?

    Thanks a lot if you can help.
    (29/32)^1/5 = (29^1/5)/(32^1/5)

    32^1/5 = 2 => (29/32)^1/5 = (1/2)29^1/5
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    (Original post by Nylex)
    (29/32)^1/5 = (29^1/5)/(32^1/5)

    32^1/5 = 2 => (29/32)^1/5 = (1/2)29^1/5
    u no the real roots bit yea? but its a negative answer so there are no real roots? :confused:
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    (Original post by TheWolf)
    lol ur right, I made a stupid mistake....
    It was wrong I think, cos I didn't show P(x) was +ve for all real x.. I took off a 3 with the equation he had underneath.
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    Ummm....if it is always positive, then it has no real roots. The discriminant is negative.
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    (Original post by Nylex)
    It was wrong I think, cos I didn't show P(x) was +ve for all real x.. I took off a 3 with the equation he had underneath.
    ahh true then i was right in the end, stop confusin me
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    Alternatively, complete the square
    p(x) = (x-3)^2 + 2
    squared terms are always greater or equal to 0, so p(x) is always positive.
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    (Original post by JamesF)
    Alternatively, complete the square
    p(x) = (x-3)^2 + 2
    squared terms are always greater or equal to 0, so p(x) is always positive.
    how do u do it with the b^2 - 4ac method? i did it and its a negative number
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    (Original post by TheWolf)
    how do u do it with the b^2 - 4ac method? i did it and its a negative number
    Exactly, it should. If it is positive, then it means that the graph crosses the x-axis - so goes negative at some point. If it is negative then there are no real roots, and so it doesnt cross the x-axis, therefore it is positive for all real x.
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    (Original post by JamesF)
    Exactly, it should. If it is positive, then it means that the graph crosses the x-axis - so goes negative at some point. If it is negative then there are no real roots, and so it doesnt cross the x-axis.
    Show that, with thsi value of k, P(x) is positive for all real x.
    does a negative answer proof that? i thought that only a positive answer show that it is positive for all real x
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    (Original post by TheWolf)
    ahh true then i was right in the end, stop confusin me
    You said real roots, that was what confused me.
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    (Original post by TheWolf)
    does a negative answer proof that? i thought that only a positive answer show that it is positive for all real x
    Yes a negative answer proves that. A positive answer shows that it isnt always positive, because then there exists real roots.
 
 
 
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