Verify that $1, - 1$ and $- 3$ are the zeroes of the cubic polynomial $x^{3}$ $+ 3 x^{2} - x - 3$ and check the relationship between zeroes and the coefficient.

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Solution

To verify roots, put root inside equations $x^{3} + 3 x^{2} - x - 3$
Root $1 \to 1^{3} + 3 (1)^{2} - 1 - 3 = 1 + 3 - 1 - 3 = 0$
$Root (- 1) = - 1^{3} + 3 (- 1)^{2} + 1 - 3 = - 1 + 3 + 1 - 3 = 0$
Root $(- 3) = (- 3)^{3} + 3 (- 3)^{2} - (- 3) - 3 = - 27 + 27 + 3 - 3 = 0$
Sum of root $= - b / a = - 3 / 1 = - 3$
From roots $= (1 - 1 - 3) = - 3$
Product of roots $= - d / a = 3 / 1 = 3$
From roots $= 1^{} - 1 - 3 = 3$
Product of roots $= p q + q r + r p = c / a = - 1 / 1 = - 1$
From roots $= 1 * - 1 + - 1 * - 3 + - 3 * 1 = - 1 + 3 - 3 = - 1$ Hence verified.

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