Question

according to the fundamental theorem of algebra, how many zeros does the function f(x)= (x-15)(x+1)(x-10)?

Answer 1

Answer: 3 zeros.

Step-by-step explanation:

According the Fundamental theorem of algebra, a polynomial of degree $n$ has $n$ roots.

When you apply the distributive property in the function given in the problem, you obtain the following polynomial:

$f(x)=(x-15)(x+1)(x-10)=x^{3}-24x^{2}+125x+150$

The highest exponent is 3, therefore the its degree is 3.

If the degree is 3 it has 3 zeros.

Answer 2

Answer:

The correct answer is number of zeros o f polynomial is 3

Step-by-step explanation:

According to the fundamental theorem of algebra,the number of zeros is same as the degree of polynomial.

If the degree is n then number of zeros is n.

To find the degree of f(x)

f(x) = (x-15)(x+1)(x-10)

f(x) = x³ - 24x² + 125x + 15

From this we get degree of this polynomial is 3.

Therefore the number of zeros is 3