Question
Based on the degree of the polynomial f(x)=(x−1)^3(x+7), what is the greatest number of zeros it could have?
Answers
GPT-4o mini
To determine the greatest number of zeros for the polynomial \( f(x) = (x - 1)^3 (x + 7) \), we first need to calculate the degree of the polynomial.
1. The term \( (x - 1)^3 \) is a polynomial of degree 3 because it is raised to the power of 3.
2. The term \( (x + 7) \) is a polynomial of degree 1 because it is a linear polynomial.
To find the overall degree of the polynomial \( f(x) \), we add the degrees of the individual components:
\[
\text{Degree of } f(x) = \text{Degree of } (x - 1)^3 + \text{Degree of } (x + 7) = 3 + 1 = 4.
\]
According to the Fundamental Theorem of Algebra, a polynomial of degree \( n \) can have up to \( n \) zeros (counting multiplicities). Therefore, since the degree of \( f(x) \) is 4, the greatest number of zeros that \( f(x) \) could have is:
\[
\boxed{4}.
\]
Now, we can also consider the specific zeros:
- The zero from \( (x - 1)^3 \) occurs at \( x = 1 \) with multiplicity 3.
- The zero from \( (x + 7) \) occurs at \( x = -7 \) with multiplicity 1.
This means the zeros of the polynomial are:
- \( x = 1 \) (with multiplicity 3)
- \( x = -7 \) (with multiplicity 1)
Thus, the unique zeros here are \( x = 1 \) and \( x = -7 \). However, as noted earlier, counting multiplicities, the total number of zeros is 4, confirming that the answer is indeed:
\[
\boxed{4}.
\]
1. The term \( (x - 1)^3 \) is a polynomial of degree 3 because it is raised to the power of 3.
2. The term \( (x + 7) \) is a polynomial of degree 1 because it is a linear polynomial.
To find the overall degree of the polynomial \( f(x) \), we add the degrees of the individual components:
\[
\text{Degree of } f(x) = \text{Degree of } (x - 1)^3 + \text{Degree of } (x + 7) = 3 + 1 = 4.
\]
According to the Fundamental Theorem of Algebra, a polynomial of degree \( n \) can have up to \( n \) zeros (counting multiplicities). Therefore, since the degree of \( f(x) \) is 4, the greatest number of zeros that \( f(x) \) could have is:
\[
\boxed{4}.
\]
Now, we can also consider the specific zeros:
- The zero from \( (x - 1)^3 \) occurs at \( x = 1 \) with multiplicity 3.
- The zero from \( (x + 7) \) occurs at \( x = -7 \) with multiplicity 1.
This means the zeros of the polynomial are:
- \( x = 1 \) (with multiplicity 3)
- \( x = -7 \) (with multiplicity 1)
Thus, the unique zeros here are \( x = 1 \) and \( x = -7 \). However, as noted earlier, counting multiplicities, the total number of zeros is 4, confirming that the answer is indeed:
\[
\boxed{4}.
\]