I am using Vieta's formula on a cubic polynomial to solve for the sum of the roots $p_1+p_2+p_3$ and the product $p_1\times p_2\times p_3$. I've solved for both $p_1+p_2+p_3$ and $p_1\times p_2\times p_3$.
Now I'm asked to solve for $p_1p_2 + p_2p_3 + p_1p_3.$ And I'm sure how to properly do this. I cheated and did it backwards and found that that value is equal to the c value of the equation, but I'm looking for a more proper way of solving for that.
Any ideas?
The value $p_{1}p_{2} + p_{2}p_{3} + p_{3}p_{1}$ is independent of the sum and product of the roots. You can easily see this by Vieta's Theorem. Let $S_{2} = p_{1}p_{2} + p_{2}p_{3} + p_{3}p_{1}.$ Then by Vieta's on $y = ax^{3} + bx^{2} + cx + d,$ $S_{2} = \frac{c}{a}.$ Notice that changing $c$ will change $S_{2}.$