Question

When p(y) =3 then the value of the polynomial ay3-y2+2y-5 is 1 then find the
value of a​

Answer 1

Step-by-step explanation:

ay

3

+3y

2

−3

Div by y−4 then put y=4

So (4)

3

a+3(4)

2

−3=64a+48−3=64a+45..............p

1

2y

3

−5y+a

Div by y−4 then put y=4

2(4)

3

−5(4)+a=128−20+a=a+108..............p

2

Then 2p

1

−p

2

=0 given

So 2(64a+45)−(a+108)=0

Or 128a+90−a−108=0

Or 127a=18

Or a=

127

18

Answer 2

Answer:

ANSWER

ay

3

+3y

2

−3

Div by y−4 then put y=4

So (4)

3

a+3(4)

2

−3=64a+48−3=64a+45..............p

1

2y

3

−5y+a

Div by y−4 then put y=4

2(4)

3

−5(4)+a=128−20+a=a+108..............p

2

Then 2p

1

−p

2

=0 given

So 2(64a+45)−(a+108)=0

Or 128a+90−a−108=0

Or 127a=18

Or a=

127

18