Question

One zero of a quadratic polynomial is twice the other. If the sum of zeroes is (-6), find the polynomial.

Hint:

For polynomial \(x^2 + bx + c\), the sum of zeroes is \(-b\) and the product is \(c\). This shortcut only works when the coefficient of \(x^2\) is 1.

Answer 1

Step 1: Let the Zeroes:
Let one zero be \(x\).
Then the other zero will be \(2x\).

Step 2: Use the Given Sum of Zeroes:
Sum of zeroes = \(-6\)
\[ x + 2x = -6 \] \[ 3x = -6 \] \[ x = -2 \]
So, the zeroes are:
\[ -2 \text{ and } -4 \]
Step 3: Form the Polynomial:
Polynomial with zeroes \(\alpha\) and \(\beta\) is:
\[ x^2 - (\alpha + \beta)x + \alpha\beta \]
Here,
Sum = \(-6\)
Product = \((-2)(-4) = 8\)

\[ x^2 - (-6)x + 8 \] \[ x^2 + 6x + 8 \]
Step 4: Final Answer:
The required quadratic polynomial is:
\[ \boxed{x^2 + 6x + 8} \]