Answer:

For p(x)=1

p(1)=3(1)*3+5(1)*2-11(1)+3

=3(1)+5(1)-11+3

=3+5-11+3

=3+5+3-11

=11-11

p(1)=0

Now for

p(x)=-3

p(-3)=3(-3)*3+5(-3)*2-11(-3)+3 (x)*power odd (-) ve

=3(-27)+5(9)-11(-3)+3

=-81+45+33+3

=-81+45+36

=-81+81

=0

Answer:

[tex]\boxed{ \bf{ \: {3x}^{3} + {5x}^{2} - 11x + 3 = (x - 1 )(x + 3 )(3x - 1) \: }}\\ \\ [/tex]

Step-by-step explanation:

Let assume that

[tex]\sf \: f(x) = {3x}^{3} + {5x}^{2} - 11x + 3 \\ \\ [/tex]

Now, Consider

[tex]\sf \: f(1) = {3(1)}^{3} + {5(1)}^{2} - 11(1) + 3 \\ \\ [/tex]

[tex]\qquad\sf \: = \: 3 + 5 - 11 + 3 \\ \\ [/tex]

[tex]\qquad\sf \: = \: 11 - 11\\ \\ [/tex]

[tex]\qquad\sf \: = \: 0\\ \\ [/tex]

[tex]\bf\implies \: 1 \: is \: the \: zero \: of \: f(x) \\ \\ [/tex]

Now, Consider

[tex]\sf \: f( - 3) = {3( - 3)}^{3} + {5( - 3)}^{2} - 11( - 3) + 3 \\ \\ [/tex]

[tex]\qquad\sf \: = \: - 3(27) + 5(9) + 33 + 3 \\ \\ [/tex]

[tex]\qquad\sf \: = \: - 81 + 45 + 36\\ \\ [/tex]

[tex]\qquad\sf \: = \: - 81 + 81\\ \\ [/tex]

[tex]\qquad\sf \: = \:0\\ \\ [/tex]

[tex]\bf\implies \: - 3\: is \: the \: zero \: of \: f(x) \\ \\ [/tex]

Let assume that

[tex]\sf \: \alpha,\beta,\gamma \: be \: zeroes \: of \: f(x) = {3x}^{3} + {5x}^{2} - 11x + 3 \\ \\ [/tex]

such that

[tex]\sf \: \alpha = 1 \\ \\ [/tex]

[tex]\sf \: \beta = - 3 \\ \\ [/tex]

Now, we know

[tex]\boxed{{\sf Product\ of\ the\ zeroes= - \frac{Constant}{coefficient\ of\ x^{3}}}} \\ \\ [/tex]

So,

[tex]\sf \: \alpha \beta \gamma = - \dfrac{3}{3} \\ \\ [/tex]

[tex]\sf \: (1)( - 3) \gamma = - 1 \\ \\ [/tex]

[tex]\sf\implies \: \gamma = \dfrac{1}{3} \\ \\ [/tex]

Now, given polynomial

[tex]\sf \: f(x) = {3x}^{3} + {5x}^{2} - 11x + 3 \\ \\ [/tex]

can also be written as

[tex]\sf \: {3x}^{3} + {5x}^{2} - 11x + 3 = 3(x - \alpha )(x - \beta )(x - \gamma ) \\ \\ [/tex]

[tex]\sf \: {3x}^{3} + {5x}^{2} - 11x + 3 = 3(x - 1 )(x + 3 )\left(x - \dfrac{1}{3}\right)\\ \\ [/tex]

[tex]\sf \: {3x}^{3} + {5x}^{2} - 11x + 3 = 3(x - 1 )(x + 3 )\left(\dfrac{3x - 1}{3}\right)\\ \\ [/tex]

[tex]\sf \: {3x}^{3} + {5x}^{2} - 11x + 3 = (x - 1 )(x + 3 )(3x - 1)\\ \\ [/tex]

Hence,

[tex]\boxed{ \bf{ \: {3x}^{3} + {5x}^{2} - 11x + 3 = (x - 1 )(x + 3 )(3x - 1) \: }}\\ \\ [/tex]