Show that 1/3 and 4/3 are zeroes of polynomial 9x^​3-6x²-11x+4.Also find third zero of polynomial>

Answer :

To show : $\frac{1}{3}$ and $\frac{4}{3}$ are zeroes of polynomial 9 x ³ - 6 x ² - 11 x  + 4   , We substitute 

x  = $\frac{1}{3}$ and get

$$\begin{gathered} 9{\left(\frac{1}{3}\right)}^3 - 6{\left(\frac{1}{3}\right)}^2 - 11\left(\frac{1}{3}\right) + 4 \\ = \frac{9}{27} - \frac{6}{9} - \frac{11}{3} + 4 \\ = \frac{1}{3} - \frac{2}{3} - \frac{11}{3} + 4 \\ = \frac{1 - 2 - 11 + 12}{3} \\ \Rightarrow \frac{13 -13}{3} \\ = \frac{0}{3} \\ \mathbf{=}\mathbf{ }\mathbf{ }\mathbf{0} \end{gathered}$$

So,
x  = $\frac{1}{3}$ is a zero of given polynomial .

And

At x  = $\frac{4}{3}$ and get

$$\begin{gathered} 9{\left(\frac{4}{3}\right)}^3 - 6{\left(\frac{4}{3}\right)}^2 - 11\left(\frac{4}{3}\right) + 4 \\ = 9 \times \frac{64}{27} - 6 \times \frac{16}{9} - \frac{44}{3} + 4 \\ = \frac{64}{3} - \frac{32}{3} - \frac{44}{3} + 4 \\ = \frac{64 - 32 - 44 + 12}{3} \\ = \frac{76 -76}{3} \\ = \frac{0}{3} \\ \mathbf{=}\mathbf{ }\mathbf{ }\mathbf{0} \end{gathered}$$

So,
x  = $\frac{4}{3}$ is a zero of given polynomial .

Therefore , ( x  - $\frac{1}{3}$ ) =  ( 3 x  -  1 )  and (  x - $\frac{4}{3}$ ) = ( 3 x  - 4 )  are factors of given polynomial

Now we divide given polynomial by ( 3 x  - 1 ) ( 3 x - 4 ) = 9 x ² - 15  x + 4  and get


So ,

$\frac{9 x{ }^3 - 6 x{ }^{2 }- 11 x + 4 }{ 9 x{ }^2 - 15 x + 4}$ =  x  + 1 

Then ,

Third zero  =  - 1                                  ( Ans )