Question

Find the quadratic polynomial the sum of whose zeroes is 1 and their product is –12. Hence find the zeroes of the polynomial.

Hint:

Always verify your answers:
Sum of zeroes: \(4 + (-3) = 1\) (Matches given)
Product of zeroes: \(4 \times (-3) = -12\) (Matches given)

Answer 1

Step 1: Build the polynomial from the sum and product of zeroes.
For sum $S=1$ and product $P=-12$, the polynomial is $f(x) = x^2 - Sx + P = x^2 - x - 12$.
Step 2: Solve for the zeroes using the quadratic formula instead of factorising.
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-12)}}{2(1)} = \frac{1 \pm \sqrt{1+48}}{2} = \frac{1 \pm 7}{2} \]
Step 3: Compute both roots.
\[ x = \frac{8}{2} = 4 \quad \text{or} \quad x = \frac{-6}{2} = -3 \]
\[ \boxed{x^2 - x - 12, \text{ with zeroes } 4 \text{ and } -3} \]