Question

If p(y)=7y³+8y²-7y-8 then find p(1)?​

Answer 1

$p(y) = 7 {y}^{3} + 8 {y}^{2} - 7y - 8 \\ \\ p(1) =( 7 \times {1}^{3} )+( 8 \times {1}^{2} ) - (7 \times 1) - 8 \\ \\ = > 7 + 8 - 7 - 8 \\ \\ = > 0$

Answer 2

Answer:

0

Step-by-step explanation:

P(1)=7(1)™+8(1)~-7(1)-8

=7+8-7-8

=0

™ Means=power 3

~ means= power 2