Find Sum of Remaining Solutions

**magentarita** · July 29th 2008, 05:30 AM

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

**Moo** · July 29th 2008, 05:31 AM

Hello,

Originally Posted by magentarita

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 x² (if it was a polynomial beginning by x^n, it would be the coefficient of x^(n-1)).

**magentarita** · July 29th 2008, 09:04 PM

Are you saying to replace n with the given exponents in the function and solve for x^(n - 1)?

**flyingsquirrel** · July 30th 2008, 12:10 AM

Hi

Originally Posted by magentarita

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 :

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

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 ?

**magentarita** · July 30th 2008, 05:11 AM

This is much clearer.

Thanks!

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