Question

Answer please guys help me​

Answer 1

• By reminder theorm!!!

$\sf\mapsto 3x-2=0\\ \\ \sf\mapsto 3x=2\\ \\ \sf\mapsto x=\dfrac{2}{3}$

Now put the value of x in p(x)

$\sf\implies 3x^3+x^2-x-9\\ \\ \sf\implies 3\bigg[\dfrac{2}{3}\bigg]^3+\bigg[\dfrac{2}{3}\bigg]^2-\dfrac{2}{3}-9\\ \\ \implies\sf \bigg[\dfrac{3\times 8}{27}\bigg]+\bigg[\dfrac{4}{9}\bigg]-\dfrac{2}{3}-9\\ \\ \sf\implies \dfrac{2}{27}+\dfrac{4}{9}-\dfrac{2}{3}-\dfrac{9}{1}\\ \\ \sf\implies \dfrac{24+8-18-27}{27}\\ \\ \sf\implies \dfrac{32-45}{27}\\ \\ \sf\implies \dfrac{-13}{27}$

Answer = $\boxed{\sf{\dfrac{-13}{27}}}$