Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
To determine which of the given numbers is not a root of the polynomial [tex]\( f(x) = x^3 - 7x^2 + 14x - 8 \)[/tex], we need to evaluate the polynomial at each given candidate. A root of the polynomial is a number for which the polynomial evaluates to zero.
Given candidates: 1, 4, 2, 8
Let's evaluate [tex]\( f(x) \)[/tex] at each candidate:
1. Evaluating at [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 1^3 - 7 \cdot 1^2 + 14 \cdot 1 - 8 \][/tex]
[tex]\[ f(1) = 1 - 7 + 14 - 8 = 0 \][/tex]
So, [tex]\( x = 1 \)[/tex] is a root.
2. Evaluating at [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = 4^3 - 7 \cdot 4^2 + 14 \cdot 4 - 8 \][/tex]
[tex]\[ f(4) = 64 - 112 + 56 - 8 = 0 \][/tex]
So, [tex]\( x = 4 \)[/tex] is a root.
3. Evaluating at [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 2^3 - 7 \cdot 2^2 + 14 \cdot 2 - 8 \][/tex]
[tex]\[ f(2) = 8 - 28 + 28 - 8 = 0 \][/tex]
So, [tex]\( x = 2 \)[/tex] is a root.
4. Evaluating at [tex]\( x = 8 \)[/tex]:
[tex]\[ f(8) = 8^3 - 7 \cdot 8^2 + 14 \cdot 8 - 8 \][/tex]
[tex]\[ f(8) = 512 - 448 + 112 - 8 = 168 \][/tex]
Since [tex]\( f(8) = 168 \neq 0 \)[/tex], [tex]\( x = 8 \)[/tex] is not a root.
After evaluating the polynomial at all given candidates, it is clear that [tex]\( x = 8 \)[/tex] is not a root of [tex]\( f(x) = x^3 - 7x^2 + 14x - 8 \)[/tex].
Therefore, the number that is not a root of the polynomial [tex]\( f(x) = x^3 - 7x^2 + 14x - 8 \)[/tex] is [tex]\( \boxed{8} \)[/tex].
Given candidates: 1, 4, 2, 8
Let's evaluate [tex]\( f(x) \)[/tex] at each candidate:
1. Evaluating at [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 1^3 - 7 \cdot 1^2 + 14 \cdot 1 - 8 \][/tex]
[tex]\[ f(1) = 1 - 7 + 14 - 8 = 0 \][/tex]
So, [tex]\( x = 1 \)[/tex] is a root.
2. Evaluating at [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = 4^3 - 7 \cdot 4^2 + 14 \cdot 4 - 8 \][/tex]
[tex]\[ f(4) = 64 - 112 + 56 - 8 = 0 \][/tex]
So, [tex]\( x = 4 \)[/tex] is a root.
3. Evaluating at [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 2^3 - 7 \cdot 2^2 + 14 \cdot 2 - 8 \][/tex]
[tex]\[ f(2) = 8 - 28 + 28 - 8 = 0 \][/tex]
So, [tex]\( x = 2 \)[/tex] is a root.
4. Evaluating at [tex]\( x = 8 \)[/tex]:
[tex]\[ f(8) = 8^3 - 7 \cdot 8^2 + 14 \cdot 8 - 8 \][/tex]
[tex]\[ f(8) = 512 - 448 + 112 - 8 = 168 \][/tex]
Since [tex]\( f(8) = 168 \neq 0 \)[/tex], [tex]\( x = 8 \)[/tex] is not a root.
After evaluating the polynomial at all given candidates, it is clear that [tex]\( x = 8 \)[/tex] is not a root of [tex]\( f(x) = x^3 - 7x^2 + 14x - 8 \)[/tex].
Therefore, the number that is not a root of the polynomial [tex]\( f(x) = x^3 - 7x^2 + 14x - 8 \)[/tex] is [tex]\( \boxed{8} \)[/tex].
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.