Verify whether the indicated number is zero of the polynomial given
$p (x) = 3 x^{3} - 5 x^{2} - 11 x - 3, x = 3, x = - 1$ and $x = - \frac{1}{3}$

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Solution

In the given polynomial $p (x) = 3 x^{3} - 5 x^{2} - 11 x - 3$ , substitute $x = 3$ , $x = - 1$ and $x = - \frac{1}{3}$ as shown :

$p (3) = (3 \times 3^{3}) - (5 \times 3^{2}) - (11 \times 3) - 3 = (3 \times 27) - (5 \times 9) - 33 - 3 = 81 - 45 - 33 - 3 = 81 - 81 = 0 p (- 1) = (3 \times (- 1)^{3}) - (5 \times (- 1)^{2}) - (11 \times (- 1)) - 3 = (3 \times (- 1)) - (5 \times 1) + 11 - 3 = - 3 - 5 + 11 - 3$
$= - 11 + 11$

$= 0 p (- \frac{1}{3}) = (3 \times {(- \frac{1}{3})}^{3}) - (5 \times {(- \frac{1}{3})}^{2}) - (11 \times (- \frac{1}{3})) - 3$
$= (3 \times (- \frac{1}{27})) - (5 \times \frac{1}{9}) + \frac{11}{3} - 3 = - \frac{1}{9} - \frac{5}{9} + \frac{11}{3} - 3$
$= \frac{- 1 - 5 + 33 - 27}{9} = \frac{33 - 33}{9} = 0$

Hence the indicated numbers are not zeroes of the polynomial $p (x) = 3 x^{3} - 5 x^{2} - 11 x - 3$ .

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