Verity that 3, −2, 1 are the zeros of the cubic polynomial p(x) = x³ − 2x² − 5x + 6 and verify the relation between its zeros and coefficients.

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Solution

$\begin{array}{l} The given polynomial is p (x) = (x^{3} - 2 x^{2} - 5 x + 6) \\ ∴ p (3) = (3^{3} - 2 \times 3^{2} - 5 \times 3 + 6) = (27 - 18 - 15 + 6) = 0 \\ p (- 2) = [(- 2^{3}) - 2 \times (- 2^{2}) - 5 \times (- 2) + 6] = (- 8 - 8 + 10 + 6) = 0 \\ p (1) = (1^{3} - 2 \times 1^{2} - 5 \times 1 + 6) = (1 - 2 - 5 + 6) = 0 \\ ∴ 3, - 2 and 1 are the zeroes of p (x), \\ Let α =3, β = - 2 and γ =1. Then we have: \\ (α + β + γ) = (3 - 2 + 1) = 2 = \frac{- (coefficient of x^{2})}{(coefficient of x^{3})} \\ (α β + β γ + γ α) = (- 6 - 2 + 3) = \frac{- 5}{1} = \frac{coefficient of x}{coefficient of x^{3}} \\ α β γ = {3 \times (- 2) \times 1} = \frac{- 6}{1} = \frac{- (constant term)}{(coefficient of x^{3})} \end{array}$

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Verity that 3, −2, 1 are the zeros of the cubic polynomial p(x) = x3 − 2x2 − 5x + 6 and verify the relation between its zeros and coefficients.

Verity that 3, −2, 1 are the zeros of the cubic polynomial p(x) = x³ − 2x² − 5x + 6 and verify the relation between its zeros and coefficients.