One solution of the equation x^3 + 5x^2 + 5x - 2 = 0

is -2. Find the sum of the remaining solutions.

Printable View

- Jul 29th 2008, 05:30 AMmagentaritaFind 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.

- Jul 29th 2008, 05:31 AMMoo
- Jul 29th 2008, 09:04 PMmagentaritaAre you saying....
Are you saying to replace

**n**with the given exponents in the function and solve for x^(**n**- 1)? - Jul 30th 2008, 12:10 AMflyingsquirrel
Hi

No, that's not what Moo meant. The degree of $\displaystyle P(x)=x^3 + 5x^2 + 5x - 2$ is 3 which means that this polynomial has 3 roots. (they can be real or complex numbers) Let's denote these roots by $\displaystyle r_1,\,r_2$ and $\displaystyle r_3$ and factor the polynomial : $\displaystyle P(x)=(x-r_1)(x-r_2)(x-r_3)$. Expanding this expression, what do we get ? Let's see :

$\displaystyle \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 $\displaystyle x^2$ is the opposite of the sum of the three roots. As this coefficient equals 5, ($\displaystyle P(x)=x^3 + {\color{blue}5}x^2 + 5x - 2$) we have $\displaystyle 5=-(r_1+r_2+r_3)$. 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 $\displaystyle x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_0$ then the sum of the roots would have been equal to $\displaystyle -a_{n-1}$.

Is it clearer ? - Jul 30th 2008, 05:11 AMmagentaritaMuch Better
This is much clearer.

Thanks!