Question

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

Answer 1

Answer:

$p(\frac{-2}{3} )\\=2*\frac{-8}{27} -2*\frac{4}{9} +13*\frac{2}{3} -6\\=\frac{-16-36+234-162}{27} \\=\frac{20}{27}$