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Math Help - Find Sum of Remaining Solutions

  1. #1
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    Find Sum of Remaining Solutions

    One solution of the equation x^3 + 5x^2 + 5x - 2 = 0
    is -2. Find the sum of the remaining solutions.

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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by magentarita View Post
    One solution of the equation x^3 + 5x^2 + 5x - 2 = 0
    is -2. Find the sum of the remaining solutions.
    The sum of all the solutions is the coefficient of x (if it was a polynomial beginning by x^n, it would be the coefficient of x^(n-1)).
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  3. #3
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    Are you saying....

    Are you saying to replace n with the given exponents in the function and solve for x^(n - 1)?
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by magentarita View Post
    Are you saying to replace n with the given exponents in the function and solve for x^(n - 1)?
    No, that's not what Moo meant. The degree of  P(x)=x^3 + 5x^2 + 5x - 2 is 3 which means that this polynomial has 3 roots. (they can be real or complex numbers) Let's denote these roots by r_1,\,r_2 and r_3 and factor the polynomial : P(x)=(x-r_1)(x-r_2)(x-r_3). Expanding this expression, what do we get ? Let's see :

    \begin{aligned}<br />
P(x)&=(x-r_1)(x-r_2)(x-r_3) \\<br />
&=(x-r_1)(x^2-(r_2+r_3)x+r_2r_3)\\<br />
&=x^3-(r_2+r_3)x^2+r_2r_3x-r_1x^2+r_1(r_2+r_3)x-r_1r_2r_3\\<br />
&=x^3{\color{blue}-(r_1+r_2+r_3)}x^2+(r_1r_2+r_1r_3+r_2r_3)x-r_1r_2r_3\\<br />
\end{aligned}

    What we learn from this is that the coefficient of x^2 is the opposite of the sum of the three roots. As this coefficient equals 5, (  P(x)=x^3 + {\color{blue}5}x^2 + 5x - 2) we have 5=-(r_1+r_2+r_3). Knowing that one of the roots is -2, you can solve this for the sum of the two remaining roots.

    What Moo meant is that if the polynomial were x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_0 then the sum of the roots would have been equal to -a_{n-1}.

    Is it clearer ?
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  5. #5
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    Much Better

    This is much clearer.

    Thanks!
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