Verify that $3, - 1$ and $- \frac{1}{3}$ are the zeroes of the cubic polynomial $p (x) = 3 x^{3} - 5 x^{2} - 11 x - 3,$ and verify the relationship between the zeroes and coefficients.

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Solution

Step 1: Verify the zeroes of the polynomial:
Here, the polynomial is:
$p (x) = 3 x^{3} - 5 x^{2} - 11 x - 3$
Now we see,
$p (3) = 3 {(3)}^{3} - 5 {(3)}^{2} - 11 (3) - 3 = 81 - 45 - 33 - 3 = 0 p (- 1) = 3 {(- 1)}^{3'} - 5 {(- 1)}^{2} - 11 (- 1) - 3 = - 3 - 5 + 11 - 3 = 0 p (- \frac{1}{3}) = 3 {(- \frac{1}{3})}^{3} - 5 {(- \frac{1}{3})}^{2} - 11 (- \frac{1}{3}) - 3 = - \frac{1}{9} - \frac{5}{9} + \frac{11}{3} - 3 = 0$
Therefore $3, - 1$ and $- \frac{1}{3}$ are the zeroes of the given cubic polynomial.
Step 2: Relationship between zeroes and coefficients:
Sum of the zeroes $= 3 - 1 - \frac{1}{3}$
$= \frac{5}{3} = - (\frac{- 5}{3}) = - \frac{coefficientof x^{2}}{coefficientof x^{3}}$
Sum of the zeroes taken two at a time $= 3 (- 1) + (- 1) (- \frac{1}{3}) + 3 (- \frac{1}{3})$
$= - 3 + \frac{1}{3} - 1 = \frac{- 11}{3} = \frac{coefficientof x}{coefficientof x^{3}}$
Product of the zeroes $= 3 (- 1) (- \frac{1}{3})$
$= - (\frac{- 3}{3}) = - \frac{constantterm}{coefficientof x^{3}}$
Hence, relationship between zeroes and coefficients are verified.

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