Question

find the value of p(-2/3) for p(y)=2y^3-y^2-13y-6

Answer 1

Given :-

• p(y) = 2y³ - y² - 13y - 16

To Find :-

• $\sf{p\frac{-2}{3}}$

Answer :-

p(y) = 2y³ - y² - 13y - 16

y = -2/3 put,

p( -2/3) = 2(-2/3)³ - ( -2/3 )² - 13 ( -2/3 ) - 6

= -16 / 27 - 4 / 9 + 26 / 3 - 6

= - 0.59 - 0.44 + 8.66 - 6

= 1.63

Conclusion :-

∴ The value of $\sf{p\frac{-2}{3}}$ = 1.63.