Question

3.
If p(x) = x²+2x+1 then find the values of P(0), P(1)​

Answer 1

Answer:

at P(0)= 1 and at P(1)=4

Step-by-step explanation:

p(0)=0 ^2+2*0+1

     = 1

p(1)=(1)^2 + 2*1 + 1

     = 1 +2 +1

     = 4

Answer 2

Answer:

The value of p(0) is 1 and p(1) is 4.

Step-by-step explanation:

Given a polynomial, $p(x) = x^{2} +2x+1$

When x is a variable and $p(x)$ is a polynomial, the value $p(x)$ takes at any $x=a$ is referred to as $p(a)$.

So, to find the value of P(0), replace x with 0 in the polynomial,

$p(x) = x^{2} +2x+1$

$\implies p(0) = 0^{2} +2\times 0+1$

Anything multiplied with 0 is 0,

therefore, the value of p(x) at x = 0 is 1.

To find the value of p(1), replace x with 1 in the polynomial,

$p(x) = x^{2} +2x+1$

$\implies p(1) = 1^{2} +2\times 1+1$

$= 1 +2+1=4$

therefore, the value of p(x) at x = 1 is 4.

Therefore, the value of p(0) is 1 and p(1) is 4.