Find the value of p(-2/3) for p(y) = 2y^3 - y^2 - 13y - 6 ?
Answers
Answered by
1
substitue y = -2/3
p = 2(-2/3)^3 - (-2/3)^2 - 13(-2/3) - 6
p = 2(-8/27) - (4/3) + (26/3) - 6
p = -16/27 - 4/9 - 26/3 - 6
p = -16/27 - 12/27 - 234/27 - 162/27
p = -424/27
p = 2(-2/3)^3 - (-2/3)^2 - 13(-2/3) - 6
p = 2(-8/27) - (4/3) + (26/3) - 6
p = -16/27 - 4/9 - 26/3 - 6
p = -16/27 - 12/27 - 234/27 - 162/27
p = -424/27
Similar questions