Question

Verify that the numbers given alongside of the cubic polynomial below are its zeroes. Also, verify the relationship between the zeroes and coefficients.

$f (x) = 2 x^{3} + x^{2} - 5 x + 2; \frac{1}{2}, 1, - 2$

Accepted Answer

We have,

$f (x) = 2 x^{3} + x^{2} - 5 x + 2$ ,

When $x = \frac{1}{2}$ ,

$f (\frac{1}{2}) = 2 {(\frac{1}{2})}^{3} + {(\frac{1}{2})}^{2} - 5 (\frac{1}{2}) + 2$

$= \frac{1}{4} + \frac{1}{4} - \frac{5}{2} + 2$

$= 0$

Hence, $\frac{1}{2}$ is a zero of the polynomial.

When $x = 1$ ,

$f (1) = 2 (1)^{3} + (1)^{2} - 5 (1) + 2$

$= 2 + 1 - 5 + 2$

$= 0$

Hence, $1$ is a zero of the polynomial.

When, $x = - 2$ ,

$f (- 2) = 2 (- 2)^{3} + (- 2)^{2} - 5 (- 2) + 2$

$= - 16 + 4 + 10 + 2$

$= 0$

Hence, $- 2$ is a zero of the polynomial.

Let, $α = \frac{1}{2}, β = 1$ and $γ = - 2$ . Then

$α + β + γ = \frac{1}{2} + 1 - 2$

$= \frac{1 + 2 - 4}{2}$

$= \frac{- 1}{2}$

For $f (x) = 2 x^{3} + x^{2} - 5 x + 2$ ,

$\Rightarrow α + β + γ = \frac{- Coefficient of x^{2}}{Coefficient of x^{3}}$

$\Rightarrow α + β + γ = - (\frac{1}{2})$

And, $α β + β γ + γ α = \frac{1}{2} (1) + 1 (- 2) - 2 (\frac{1}{2})$

$= \frac{1 - 4 - 2}{2}$

$= \frac{- 5}{2}$

For $f (x) = 2 x^{3} + x^{2} - 5 x + 2$ ,

$α β + β γ + γ α = \frac{Coefficient of x}{Coefficient of x^{3}}$

$\Rightarrow α β + β γ + γ α = \frac{- 5}{2}$

And, $α β γ = \frac{1}{2} \times 1 \times (- 2)$

$= - 1$

For $f (x) = 2 x^{3} + x^{2} - 5 x + 2$ ,

$α β γ = \frac{- Constant term}{Coefficient of x^{3}}$

$\Rightarrow α β γ = - (\frac{2}{2})$

$\Rightarrow α β γ = - 1$

Hence, verified.

R. D. Sharma Solutions for Chapter: Polynomials, Exercise 2: EXERCISE 2.2

R. D. Sharma Mathematics Solutions for Exercise - R. D. Sharma Solutions for Chapter: Polynomials, Exercise 2: EXERCISE 2.2

Questions from R. D. Sharma Solutions for Chapter: Polynomials, Exercise 2: EXERCISE 2.2 with Hints & Solutions

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