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Sum of n Terms

Verify that ...

Question

Verify that $3, - 1, \frac{- 1}{3}$ are the zeros of the cubic polynomial $3 x^{3} - 5 x^{2} - 11 x - 3$ .

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Solution

$f (x) = 3 x^{3} - 5 x^{2} - 11 x - 3 f (3) = 3 {(3)}^{3} - 5 {(3)}^{2} - 11 (3) - 3 = 81 - 45 - 33 - 3 = 81 - 81 = 0 f (- 1) = 3 {(- 1)}^{3} - 5 {(- 1)}^{2} - 11 (- 1) - 3 = - 3 - 5 + 11 - 3 = - 11 + 11 = 0 f (\frac{- 1}{3}) = 3 {(\frac{- 1}{3})}^{3} - 5 {(\frac{- 1}{3})}^{2} - 11 (\frac{- 1}{3}) - 3 = 3 \times \frac{- 1}{27} - \frac{5}{9} + \frac{11}{3} - 3 = \frac{- 1}{9} - \frac{5}{9} + \frac{11 - 9}{3} = \frac{- 6 + 33 - 27}{9} = \frac{- 33 + 33}{9} = 0$
$∴ 3, - 1, \frac{- 1}{3}$ are the zeroes of $f (x)$ .

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3, - 1, - \frac{1}{3}

are the zeroes of the cubic polynomial

p (x) = 3 x^{3} - 5 x^{2} - 11 x - 3

and then verify the relationship between the zeroes and the coefficients

- \frac{1}{3}

is the zero of the cubic polynomial

f (x) = 3 x^{3} - 5 x^{2} - 11 x - 3

the other zeros are

Q. lf one zero of the cubic polynomial

P (x) = 3 x^{3} - 5 x^{2} - 11 x - 3

\frac{- 1}{3}

, then other two zeroes are ________.

One of the zeroes of cubic polynomial $(3 x^{3} - 5 x^{2} - 11 x - 3)$ is $\frac{- 1}{3}$ . Find the other zeroes.

Q. 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

x = - \frac{1}{3}

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