Question

p(x)= root x^3 + 1, is not a polynomial. give reasons​

Answer 1

Answer:

It is not a polynimial bybthe definition of a polynomial.

But what if it is equal to some polynimial as a functions on [math]\R[/math] . Notice that polynomials are equal to each other iff all their coefficients are equal (this is not true for polynomials over finite fields by the way).

We get:

[math]x^\frac{3}{2} =P(x)+1=Q(x)\Rightarrow x^3=Q^2(x)[/math]. But as soon as you try to check the degree of Q you find out that 1 is too small and 2 is too big. (Either we do not get the term with [math] x^3[/math] or we get the term with [math]x^4[/math] we should not have).

Thus it cannot be a polynomial even as a function.

Answer 2

Answer:

Because  the value of variable is not fined it can not divided by 3 so it is not a polynomial.

Step-by-step explanation: