find the bound on the real zeros of the polynomial function

f(x)=x^4+x^3-4x-6
cna you please show me how you got the answer please

you can use the synthetic division to find the smallest k such that (x-k) causes all the values in the bottom line to be positive.

Or, to provide an easier (but rougher) bound, see

http://www.mathsisfun.com/algebra/polynomials-bounds-zeros.html

To find the bounds on the real zeros of a polynomial function, you can use the Rational Root Theorem and the Descartes' Rule of Signs.

1. Rational Root Theorem:
The Rational Root Theorem states that if a rational number p/q (where p and q are relatively prime integers) is a root of a polynomial equation, then p must be a factor of the constant term, and q must be a factor of the leading coefficient.

In the given polynomial function, f(x) = x^4 + x^3 - 4x - 6, the constant term is -6, and the leading coefficient is 1. Therefore, the possible rational roots are the divisors of 6, which include ±1, ±2, ±3, and ±6.

2. Descartes' Rule of Signs:
The Descartes' Rule of Signs helps determine the number of positive or negative real roots of a polynomial equation.

In the given polynomial function, f(x) = x^4 + x^3 - 4x - 6, if you substitute -x for x, you get:
f(-x) = (-x)^4 + (-x)^3 - 4(-x) - 6
= x^4 - x^3 + 4x - 6

Counting the sign changes in f(x) and f(-x), we observe that there are 2 sign changes. Therefore, there are either 2 or 0 positive real roots.

Now, if we substitute (-x) for x and simplify, we get:
f(-x) = (-x)^4 + (-x)^3 - 4(-x) - 6
= x^4 + x^3 + 4x - 6

Again, counting the sign changes in f(x) and f(-x), we observe that there are no sign changes. Therefore, there are 0 negative real roots.

By applying Descartes' Rule of Signs, we have determined that there are either 2 or 0 positive real roots and 0 negative real roots.

Hence, the bounds on the real zeros of the given polynomial function, f(x) = x^4 + x^3 - 4x - 6, are either 0 or 2.