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Thread: polynomial question

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    Junior Member Joshmosh2's Avatar
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    polynomial question

    P(x) is a monic polynomial of the fourth degree. When P(x) is divided by x+1 and x-2 the remainders are 5 and -4 respectively. Given that P(x) is an even function ie. one where P(x)=P(-x)
    a. Express it in the form p0+p1x^1+p2x^2+p3x^3+p4x^4

    Well what I instantly thought was equating both by subbing in the zeroes, but what equation would I substitute it in?
    Maybe P(x)=A(x)Q(x)+R(x)?
    well what I know is P(-1) = 5
    and P(2) = -4
    and then at one point, I would have to use the condition P(x)=P(-x)
    Any help please? thanks.

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    Re: polynomial question

    P(x) = x^4 - 8x^2 + 12 ???
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    Junior Member Joshmosh2's Avatar
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    Re: polynomial question

    Quote Originally Posted by Drongoski View Post
    P(x) = x^4 - 8x^2 + 12 ???
    YES! can you please explain to me how you got that?

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    Re: polynomial question

    Quote Originally Posted by Joshmosh2 View Post
    YES! can you please explain to me how you got that?
    Use the fact that since P(x) is even, it can't contain and odd powers, so you're left with p4x^4 + p2x^2 + p0.
    Since it's monic, p4=1, so you can use the remainders given to create 2 simultaneous equations with variables p2 and p0.
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    Junior Member Joshmosh2's Avatar
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    Re: polynomial question

    Quote Originally Posted by panda15 View Post
    Use the fact that since P(x) is even, it can't contain and odd powers, so you're left with p4x^4 + p2x^2 + p0.
    Since it's monic, p4=1, so you can use the remainders given to create 2 simultaneous equations with variables p2 and p0.
    Ahh, just simple intuition. Thanks!

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    Junior Member Joshmosh2's Avatar
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    Re: polynomial question

    One more thing,
    when the polynomial P(x) is divided by x^2-1 the remainder is 3x-1. What is the remainder when P(x) is divided by x-1?
    I know that R(1) = 2, which is the answer, but what could be a nice proof?

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    Re: polynomial question

    Quote Originally Posted by Joshmosh2 View Post
    One more thing,
    when the polynomial P(x) is divided by x^2-1 the remainder is 3x-1. What is the remainder when P(x) is divided by x-1?
    I know that R(1) = 2, which is the answer, but what could be a nice proof?
    Why would you need a nice proof when the answer is done in like 5 seconds
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    Re: polynomial question

    Quote Originally Posted by Joshmosh2 View Post
    One more thing,
    when the polynomial P(x) is divided by x^2-1 the remainder is 3x-1. What is the remainder when P(x) is divided by x-1?
    I know that R(1) = 2, which is the answer, but what could be a nice proof?
    P(x)=Q(x)(x^2-1)+(3x-1)
    P(1)=0+3-1
    =2
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    Re: polynomial question

    Quote Originally Posted by panda15 View Post
    Use the fact that since P(x) is even, it can't contain and odd powers, so you're left with p4x^4 + p2x^2 + p0.
    Since it's monic, p4=1, so you can use the remainders given to create 2 simultaneous equations with variables p2 and p0.
    How do you make the two simultaneous equations?

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    Junior Member BenHowe's Avatar
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    Re: polynomial question

    Once you've used the fact that the polynomial is monic, so the coefficient of the x^4 is 1, and that it is an even function, so there is no odd power of x, you can write p(x)=x^4+bx^2+d. Now the polynomial has only two unknowns, b and d, and you have two pieces of information p(-1)=5 and p(2)=-4. When you use this you end up getting the two simulataneous equations b+d=4 and 4b+d=-20. Find the value of b & d and bob's your uncle
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    Re: polynomial question

    thanks

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