Question

If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).

Hint:

Express any symmetric function of roots in terms of $(\alpha + \beta)$ and $(\alpha\beta)$ to solve polynomial relation problems easily.

Answer 1

Given:
A quadratic polynomial \( px^2 + qx + r \) has zeroes \( \alpha \) and \( \beta \).

We know the standard relations:
\[ \alpha + \beta = -\frac{q}{p} \] \[ \alpha\beta = \frac{r}{p} \]

We need to find:
\[ \alpha^3\beta + \beta^3\alpha \]
Factor the expression:
\[ \alpha^3\beta + \beta^3\alpha = \alpha\beta(\alpha^2 + \beta^2) \]
Now use the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substitute the known values:
\[ \alpha^2 + \beta^2 = \left(-\frac{q}{p}\right)^2 - 2\left(\frac{r}{p}\right) \] \[ = \frac{q^2}{p^2} - \frac{2r}{p} \]
Now multiply by \( \alpha\beta = \frac{r}{p} \):
\[ \alpha^3\beta + \beta^3\alpha = \frac{r}{p}\left(\frac{q^2}{p^2} - \frac{2r}{p}\right) \] \[ = \frac{r q^2}{p^3} - \frac{2r^2}{p^2} \] Put over a common denominator \( p^3 \):
\[ = \frac{r q^2 - 2 r^2 p}{p^3} \]

Final Answer:
\[ \boxed{\alpha^3\beta + \beta^3\alpha = \frac{r q^2 - 2 r^2 p}{p^3}} \]