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Math Help - Using Descartes' Rule of Signs

  1. #1
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    Using Descartes' Rule of Signs

    Use Descartes' rule of signs to determine the possible number of:
    a) positive real roots of 4x^3 + 2x^2 + 7x + 9 = 0
    b) negative real roots of 4x^3 + 2x^2 + 7x + 9 = 0


    My answer to a) (zero positive real roots), was correct.

    Zero positive real roots because there are zero sign changes..

    and I had thought there would be zero negative roots as well. But it came back wrong.

    Can someone explain to me why? Thanks.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by Savior_Self View Post
    Use Descartes' rule of signs to determine the possible number of:
    a) positive real roots of 4x^3 + 2x^2 + 7x + 9 = 0
    b) negative real roots of 4x^3 + 2x^2 + 7x + 9 = 0


    My answer to a) (zero positive real roots), was correct.

    Zero positive real roots because there are zero sign changes..

    and I had thought there would be zero negative roots as well. But it came back wrong.

    Can someone explain to me why? Thanks.
    Hi Savior_Self,

    f(-x)=4(-x)^3+2(-x)^2+7(-x)+9

    f(-x)=-4x^3+2x^2-7x+9

    Now, how many sign changes do you see?
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  3. #3
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    To find the maximum number of positive real roots count the number of sign changes. To find the number of negative real roots replace all x values with a (-x) and perform the same operation of counting sign changes.

    For example,

    x^2+x+2 becomes...
    (-x)^2+(-x)+2
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  4. #4
    MHF Contributor ebaines's Avatar
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    Using Descarte's method, to determine the number of negative roots you replace the x in the polynomial with -x, and see how many sign changes there are in the resulting equation:

    <br />
4(-x)^3 + 2(-x)^2 + 7(-x) + 9 = <br />
    <br />
-4x^3 + 2x^2 - 7x + 9<br />

    Note that there are 3 changes of sign as you read left to right. Thus there may be 3 negative roots, or possibly just 1 (since it's possible that 2 of the roots may actually be complex).

    If you plot this function you'll see that there is one real negtive root. The two other roots are therefore complex.
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  5. #5
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    Awesome. Thanks guys. I didn't pick up on the x to (-x) step for some reason. I won't forget it now.
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  6. #6
    MHF Contributor ebaines's Avatar
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    There's another "trick" that you could have used. Remember that a polynomial of order N must have N roots. So since you already knew that there were zero positive roots, you know that there must be either (a) 3 negative roots, or (b) 1 negative root and 2 complex roots.
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