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Choose whether the given statement is true or false.

3, -1, $\frac{- 1}{3}$ are the zeroes of the cubic polynomial $p (x) = 3 x^{3} - 5 x^{2} - 11 x - 3$ .

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True

False

answer is A.

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Detailed Solution

Here we are given the polynomial p(x) as

3 x^{3} - 5 x^{2} - 11 x - 3

.
Zeroes of the polynomial are given as 3, -1,

\frac{- 1}{3}

.
We need to verify them first. For this, let us put their values in place of x in p(x).
Putting x = 3 in p(x) we get,
p(3)=

3 {(3)}^{3} - 5 {(3)}^{2} - 11 (3) - 3

We know that

{(3)}^{3} = 27

and

{(3)}^{2} = 9

so we get,
p(3)=3(27)−5(9)−11(3)−3
⇒p(3)=81−45−33−3
⇒p(3)=0
Value of x as 3 gives p(x) as zero, therefore 3 is a zero of the polynomial.
Putting x = -1 in polynomial, we get,
p(−1)=

3 {(- 1)}^{3} - 5 {(- 1)}^{2} - 11 (- 1) - 3

⇒p(−1)=3(−1)−5(−1)+11−3
⇒p(−1)=−3−5+11−3
⇒p(−1)=0
Value of x as -1 gives p(x) as zero, therefore, -1 is a zero of the polynomial.
Putting x=

\frac{- 1}{3}

.in polynomial we get,
p(

\frac{- 1}{3}

.)=

3 {(\frac{- 1}{3})}^{3} - 5 {(\frac{- 1}{3})}^{2} - 11 (\frac{- 1}{3}) - 3

⇒p(

\frac{- 1}{3}

)=3(

\frac{- 1}{27}

)−5(

\frac{1}{9}

\frac{11}{3}

−3
⇒p(

\frac{- 1}{3}

\frac{- 1}{9} - \frac{5}{9} + \frac{11}{3}

−3
Taking LCM as 9 we get,
⇒p(

\frac{- 1}{3}

\frac{- 1 - 5 + 33 - 27}{9}

⇒p(

\frac{- 1}{3}

\frac{- 33 + 33}{9}

⇒p(

\frac{- 1}{3}

)=0
The given statement is true.

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