Question

If p(x) equal to x4+1, then find p(2)-p(-2)

Answer 1

$p(x) = {x}^{4} + 1 \\ = > p(2) = {(2)}^{4} + 1 = 16 + 1 = 17$

Similarly,

$p( - 2) = {( - 2)}^{4} + 1 = 16 + 1 = 17$

Therefore,

$p(2) - p( - 2) = 17 - 17 = 0$

Answer 2

Answer:

Value of p(2) and p(-2) is 0.  

Step-by-step explanation:

Given

$p(x)=x^4+1$

we have to find the value of p(2)-p(-2)

$p(2)=2^4+1=16+1=17$

$p(-2)=(-2)^4+1=16+1=17$

As the given function is even value of p(2) and p(-2) is same

gives

p(2)-p(-2)=17-17=0

Hence, value of p(2) and p(-2) is 0.