Which values of n degree polynomials must there definitely exist an x-intercept. Please explain all the reasoning

There must be definitely at most n value of x - intercept

Step-by-step explanation: If a polynomial has a value of n degree, then, there must be definitely at most n value of x - intercept.

For instance, The polynomial has a degree of 8, so there are at most 8 x-intercepts and at most 8–1 = 7 turning points.

There will definitely exist an x-intercept for all n degree polynomials (a·xⁿ + c) that have real roots where n>0 or c = -a

Step-by-step explanation:

The x=intercept of a polynomial is the coordinates of the point on the polynomial where the equation of the polynomial f(x) = 0

Therefore, the coordinates of the x-intercept = (k, 0)

Where x = k is the solution of the f(x) = 0

Hence, there will definitely exist an x-intercept for all n degree polynomials (a·xⁿ + c) that have real roots where n>0 or c = -a.

d 192

step-by-step explanation:

we find our area to find area we multiply thoses numbers to get our answer so our area is 192