Question

if p(y)= y2+5y+4, find the value of p(-1) and p(1) .​

Answer 1

Given-

P(y) =y²+5y+4

To find -

The value of p(-1) and p(1)

Solution -

p(y) =y²+5y+4

p(-1)=(-1)²+5(-1)+4

p(-1)=1-5+4

p(-1)=5-5

p(-1)=0

Again,

p(y)=y²+5y+4

p(1)=(1)²+5(1)+4

p(1)=1+5+4

p(1)=10

Therefore the value of p(-1) is 0 and value of p(1) is 10

Answer 2

Answer:

x²-7x+12

Step-by-step explanation:

=(½)²=7(½)+12

=¼-7/2+12/1

=1-14+48/4

=49+14/4

=35/4

=8¾