Question

Find the zeroes of the quadratic polynomial $x^2 + 7x + 10$, and verify the relationship between the zeroes and its coefficients.

Hint:

Always double check the signs when splitting terms and calculating zeroes.
A common mistake is forgetting that a factor of $(x+a)$ gives a zero of $x = -a$.

Answer 1

Step 1: Find the zeroes using the quadratic formula instead of factoring.
For $p(x) = x^2 + 7x + 10$, we have $a = 1$, $b = 7$, $c = 10$. The quadratic formula is
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Step 2: Compute the discriminant.
\[ b^2 - 4ac = 7^2 - 4(1)(10) = 49 - 40 = 9 \]
Step 3: Substitute and simplify.
\[ x = \frac{-7 \pm \sqrt{9}}{2} = \frac{-7 \pm 3}{2} \]
Step 4: Find both zeroes.
\[ x = \frac{-7 + 3}{2} = \frac{-4}{2} = -2 \qquad \text{or} \qquad x = \frac{-7 - 3}{2} = \frac{-10}{2} = -5 \] So the zeroes are $\alpha = -2$ and $\beta = -5$, the same values found by factoring, confirming the method works either way.

Step 5: Verify the sum of zeroes.
\[ \alpha + \beta = -2 + (-5) = -7, \qquad -\frac{b}{a} = -\frac{7}{1} = -7 \] Both sides match.

Step 6: Verify the product of zeroes.
\[ \alpha \times \beta = (-2)(-5) = 10, \qquad \frac{c}{a} = \frac{10}{1} = 10 \] Both sides match.

Final Answer:
The zeroes of the polynomial are $-2$ and $-5$, and both the sum ($-7$) and product ($10$) agree with the coefficient-based formulas. \[ \boxed{x = -2, -5} \]