Question

if p(x)=x³-x²+x+1,then find the value of ½{p(-1)+p(1)}

Answer 1

Answer:

0

Step-by-step explanation:

Hi,

Given p(x) = x³-x²+x+1,

We need to find the value of 1/2[p(-1) + p(1)]

Consider p(-1) = (-1)³ - (-1)² -1 + 1

= -1 -1 -1 + 1

= -2

Consider p(1) = 1³ - 1² + 1 + 1

= 1 - 1 + 1 + 1

= 2

So, p(-1) + p(1)

= -2 + 2

= 0

Hence, 1/2[p(-1) + p(1)] = 1/2*0

= 0.

Hope, it helps !

Answer 2

Answer:

$\text{Hence, the value is }\frac{1}{2}(p(-1)+p(1))\text{ is 0.}$

Step-by-step explanation:

$\text{Given the polynomial }p(x)=x^3-x^2+x+1$

$\text{we have to find the value of }\frac{1}{2}(p(-1)+p(1))$

$p(x)=x^3-x^2+x+1$

Put x=-1 and x=1

$p(-1)=(-1)^3-(-1)^2+(-1)+1=-1-1-1+1=-2$

$p(1)=(1)^3-(1)^2+(1)+1=1-1+1+1=2$

$\frac{1}{2}(p(-1)+p(1))$

$=\frac{1}{2}(-2+2)=0$

$\text{Hence, the value is }\frac{1}{2}(p(-1)+p(1))\text{ is 0.}$