Verify that 5, −2 and $\frac{1}{3}$ are the zeros of the cubic polynomial p(x) = 3x³ − 10x² − 27x + 10 and verify the relation between its zeros and coefficients.

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Solution

$\begin{array}{l} p (x) = (3 x^{3} - 10 x^{2} - 27 x + 10) \\ p (5) = (3 \times 5^{3} - 10 \times 5^{2} - 27 \times 5 + 10) = (375 - 250 - 135 + 10) = 0 \\ p (- 2) = [3 \times (- 2^{3}) - 10 \times (- 2^{2}) - 27 \times (- 2) + 10] = (- 24 - 40 + 54 + 10) = 0 \\ p (\frac{1}{3}) = {3 \times {(\frac{1}{3})}^{3})^{3} - 10 \times {(\frac{1}{3})}^{2} - 27 \times \frac{1}{3} + 10} = (3 \times \frac{1}{27} - 10 \times \frac{1}{9} - 9 + 10) \\ = (\frac{1}{9} - \frac{10}{9} + 1) = (\frac{1 - 10 + 9}{9}) = (\frac{0}{9}) = 0 \\ ∴ 5, - 2 and \frac{1}{3} are the zeroes of p (x) . \\ Let α =5, β = - 2 and γ = \frac{1}{3} . Then we have : \\ (α + β + γ) = (5 - 2 + \frac{1}{3}) = (\frac{10}{3}) = \frac{- (coefficient of x^{2})}{(coefficient of x^{3})} \\ (α β + β γ + γ α) = (- 10 - \frac{2}{3} + \frac{5}{3}) = \frac{- 27}{3} = \frac{coefficient of x}{coefficient of x^{3}} \\ α β γ = {5 \times (- 2) \times \frac{1}{3}} = \frac{- 10}{3} = \frac{- (constant term)}{(coefficient of x^{3})} \end{array}$

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