Question

Find the zeroes of the polynomial \(p(x) = 3x^2 + x - 10\) and verify the relationship between zeroes and its coefficients.

Answer 1

Finding Zeroes of the Polynomial

Given polynomial:

\[ p(x) = 3x^2 + x - 10 \]

Step 1: Set \( p(x) = 0 \)

\[ 3x^2 + x - 10 = 0 \]

Step 2: Factor the Polynomial

Find two numbers whose product is \( 3 \times (-10) = -30 \) and whose sum is 1. These numbers are 6 and -5: \[ 3x^2 + 6x - 5x - 10 = 0 \Rightarrow 3x(x + 2) - 5(x + 2) = 0 \Rightarrow (3x - 5)(x + 2) = 0 \]

Therefore, the zeroes are:

\[ 3x - 5 = 0 \Rightarrow x = \frac{5}{3}, \quad x + 2 = 0 \Rightarrow x = -2 \] \[ \text{Zeroes: } \alpha = \frac{5}{3}, \quad \beta = -2 \]

Step 3: Verify Relationship Between Zeroes and Coefficients

General form: \( ax^2 + bx + c \), where: \[ a = 3,\quad b = 1,\quad c = -10 \]

Sum of Zeroes:

\[ \alpha + \beta = \frac{5}{3} + (-2) = \frac{5}{3} - \frac{6}{3} = -\frac{1}{3} \] From coefficients: \[ -\frac{b}{a} = -\frac{1}{3} \] ✅ Verified

Product of Zeroes:

\[ \alpha \cdot \beta = \frac{5}{3} \cdot (-2) = -\frac{10}{3} \] From coefficients: \[ \frac{c}{a} = \frac{-10}{3} \] ✅ Verified

✅ Final Answer:

• Zeroes: \( \boxed{\frac{5}{3} \text{ and } -2} \)
• Both relationships \( \alpha + \beta = -\frac{b}{a} \) and \( \alpha \beta = \frac{c}{a} \) are verified.