Question

find p(0) p(1) p(-2) for each of the following polynomials .(i) ( p(y)=(y+2) (y-2).​

Answer 1

Step-by-step explanation:

Given:-

Polynomial P(y) = (y+2) (y-2).

To find:-

Find the values of P(0),P(1) and P(-2)?

Solution:-

Given Polynomial is P(y) = (y+2) (y-2)---------(1)

I)Value of P(0):-

Put y = 0 in (1) then

=> P(0) = (0+2)(0-2)

=>P(0)=(2)(-2)

P(0)=-4

ii) Value of P(1):-

Put y = 1 in (1) then

=> P(1) = (1+2)(1-2)

=>P(1)=(3)(-1)

P(1)=-3

iii) Value of P(-2):-

Put y = -2 in (1) then

=> P(-2) = (-2+2)(-2-2)

=>P(-2)=(0)(-4)

P(-2)=0

Answer:-

The value of P(0) is -4

The value of P(1) is -3

The value of P(-2) is 0

Answer 2

$\sf\huge\bold{Answer:}$

Given :

A polynomial

⟶ p(y)=(y+2) (y-2).

To find :

Values of

⇒p(0)

⇒p(1)

⇒p(-2)

Solution :

Polynomial p(y)=(y+2) (y-2).

$\fbox{ For p(0) [when y = 0 ]}$

⟶p(y)=(y+2) (y-2)

⟶p(0)=(0+2) (0-2)

⟶p(0)=(2) (-2)

⟶p(0)= -4

$\fbox{ For p( 1 ) [when y = 1 ]}$

⟶p(y)=(y+2) (y-2)

⟶p( 1 )=(1+2) (1-2)

⟶p( 1 )=(3)(-1)

⟶p( 1 )= -3

$\fbox{ For p( -2 ) [when y = -2 ]}$

⟶p(y)=(y+2) (y-2)

⟶p(-2)=(-2+2) (-2-2)

⟶p(-2)=(0)(-4)

⟶p(-2)=0

$\therefore \\ \implies \: p(0) = - 4 \\ \implies \: p(1) = - 3 \\ \implies \: p( - 2) = 0$