The roots of the equation $$x^3 + px^2 + qx + r = 0$$ are $\alpha, 2\alpha, 4\alpha$, where $p, q, r$ and $\alpha$ are non-zero real constants - CIE - A-Level Further Maths - Question 2 - 2018 - Paper 1

To show this, we will use Vieta's formulas, which relate the coefficients of the polynomial to the sums and products of its roots.

1. Sum of the Roots: According to Vieta's formula, the sum of the roots for the cubic polynomial is given by:

   $\alpha + 2\alpha + 4\alpha = 7\alpha = -p.$

   From this, we have: $7\alpha = -p.$
   Therefore, we can express $\alpha$ as: $\alpha = -\frac{p}{7}.$

2. Product of the Roots: The product of the roots is also given by Vieta's formula as:

   $\alpha \cdot 2\alpha \cdot 4\alpha = 8\alpha^3 = -r.$

   Substituting $\alpha$, we get: $8\left(-\frac{p}{7}\right)^3 = -r.$

3. Sum of Products of Roots Taken Two at a Time: The sum of the products of the roots taken two at a time is:

   $\alpha(2\alpha) + \alpha(4\alpha) + (2\alpha)(4\alpha) = 2\alpha^2 + 4\alpha^2 + 8\alpha^2 = 14\alpha^2.$

   This must equal $q$. Thus:

   $14\alpha^2 = q.$

4. Combining the Results: From the above expressions, we can calculate the values of $p$ and $q$ in relation to $\alpha$:

   Substitute $\alpha$ into $q$:

   $q = 14\left(-\frac{p}{7}\right)^2 = 14\left(\frac{p^2}{49}\right) = \frac{2p^2}{7}.$

5. Final Step: Now, substituting $\alpha$ back into the relationship $2p\alpha + q$:

   $2p\left(-\frac{p}{7}\right) + \frac{2p^2}{7} = -\frac{2p^2}{7} + \frac{2p^2}{7} = 0.$

   Hence, we have shown that: $2p\alpha + q = 0.$