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Math Help - factors of a polynomial

  1. #1
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    factors of a polynomial

    The polynomial 2x^4-ax^3+19x^2-20x+12=0 has a factor in the form of (x-k)^2 , where k are natural numbers . Find the values of k and a .
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by mathaddict View Post
    The polynomial 2x^4-ax^3+19x^2-20x+12=0 has a factor in the form of (x-k)^2 , where k are natural numbers . Find the values of k and a .
    f(x)=2x^4-ax^3+19x^2-20x+12
    If it has such a factor, then f(k)=0 and f'(k)=0

    This should give you a system of 2 equations, and you can solve for a and k !
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    Quote Originally Posted by Moo View Post
    Hello,

    f(x)=2x^4-ax^3+19x^2-20x+12
    If it has such a factor, then f(k)=0 and f'(k)=0

    This should give you a system of 2 equations, and you can solve for a and k !

    Thanks Moo , just wondering f(k') , does it mean differentiate and why is it equals 0 .
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    Moo
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    Quote Originally Posted by mathaddict View Post
    Thanks Moo , just wondering f(k') , does it mean differentiate and why is it equals 0 .
    It's f'(k), and yes, it means that you differentiate f, and then you take the value x=k.

    Okay, let's say (x-k) is a factor for f.
    There exists a polynomial Q such that :
    f(x)=(x-k)^2 Q(x)
    It is obvious that f(k)=0
    Now if you differentiate, use the product rule and you'll have :
    f'(k)=(x-k)^2 Q'(x)+2(x-k)Q(x)=(x-k)[(x-k)Q'(x)+2Q(x)]
    Is it clear that f'(k)=0 ?

    In fact, it is true for higher powers : if f has a factor (x-k)^n, then f(k)=f'(k)=f^{(2)}(k)=\dots=f^{(n-1)}(k)=0
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    Re :

    Thanks a lot Moo for telling me these extra things which are not in my book .
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