Verification of Zeros of a Cubic Polynomial

Video Explanation

Question

Verify that the numbers

\[ 2,\; 1,\; 1 \]

are the zeroes of the cubic polynomial

\[ g(x) = x^3 – 4x^2 + 5x – 2, \]

and verify the relationship between the zeroes and coefficients.

Solution

Step 1: Verification of Zeros

A number \(x=\alpha\) is a zero of the polynomial if \(g(\alpha)=0\).

(i) For \(x = 2\)

\[ g(2) = (2)^3 – 4(2)^2 + 5(2) – 2 \]

\[ = 8 – 16 + 10 – 2 = 0 \]

Hence, \(2\) is a zero of the polynomial.

(ii) For \(x = 1\)

\[ g(1) = (1)^3 – 4(1)^2 + 5(1) – 2 \]

\[ = 1 – 4 + 5 – 2 = 0 \]

Hence, \(1\) is a zero of the polynomial.

(iii) For \(x = 1\) (again)

\[ g(1) = 0 \]

Thus, \(1\) is a repeated zero of the polynomial.

Step 2: Verification of Relationship Between Zeros and Coefficients

Let the zeroes of the polynomial be

\[ \alpha = 2,\; \beta = 1,\; \gamma = 1. \]

The given polynomial is

\[ x^3 – 4x^2 + 5x – 2. \]

Comparing with \(ax^3 + bx^2 + cx + d\), we get

\[ a = 1,\quad b = -4,\quad c = 5,\quad d = -2. \]

(i) Sum of the zeroes

\[ \alpha + \beta + \gamma = 2 + 1 + 1 = 4 \]

\[ -\frac{b}{a} = -\frac{-4}{1} = 4 \]

Hence, the relation \[ \alpha + \beta + \gamma = -\frac{b}{a} \] is verified.

(ii) Sum of the products of zeroes taken two at a time

\[ \alpha\beta + \beta\gamma + \gamma\alpha = (2)(1) + (1)(1) + (1)(2) \]

\[ = 2 + 1 + 2 = 5 \]

\[ \frac{c}{a} = \frac{5}{1} = 5 \]

Hence, the relation \[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \] is verified.

(iii) Product of the zeroes

\[ \alpha\beta\gamma = (2)(1)(1) = 2 \]

\[ -\frac{d}{a} = -\frac{-2}{1} = 2 \]

Hence, the relation \[ \alpha\beta\gamma = -\frac{d}{a} \] is verified.

Conclusion

The numbers \[ 2,\; 1,\; 1 \] are the zeroes of the cubic polynomial \[ x^3 – 4x^2 + 5x – 2. \]

All the relationships between zeroes and coefficients are verified.