Question

Find the value of p(-2/3) for p(y)= 2y^3 - y^2 - by - 6

Answer 1

◀HERE'S ⏬ YOUR ANSWER▶

$p(y) = 2 {y}^{3} - {y}^{2} - by - 6$
So,
$p( - \frac{2}{3} ) = 2 {( - \frac{2}{3} })^{3} - {( - \frac{2}{3} })^{2} - b( - \frac{2}{3} ) - 6 \\ = - \frac{16}{27} - \frac{4}{9} + \frac{2b}{3} - 6 \\ = \frac{ - 16 - 12 + 18b - 162}{27} \\ = \frac{ - 190 + 18b}{27}$

⏏HOPE THIS ⬆ HELPS YOU⏏