Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
To solve the problem and find the value of \( A \), we need to use some properties of H.C.F (Highest Common Factor) and L.C.M (Least Common Multiple) of polynomials.
Given:
- \( \text{H.C.F} \) of \( x^3 - 1 \) and \( A \) is \( x - 1 \)
- \( \text{L.C.M} \) of \( x^3 - 1 \) and \( A \) is \( x^6 - 1 \)
We use the relationship between H.C.F and L.C.M of two expressions \( f(x) \) and \( g(x) \):
[tex]\[ \text{H.C.F}(f(x), g(x)) \cdot \text{L.C.M}(f(x), g(x)) = f(x) \cdot g(x) \][/tex]
Applying this to our problem:
[tex]\[ (x - 1) \cdot (x^6 - 1) = (x^3 - 1) \cdot A \][/tex]
First, we factorize the expressions where necessary.
[tex]\[ x^3 - 1 = (x - 1)(x^2 + x + 1) \][/tex]
[tex]\[ x^6 - 1 = (x^3 - 1)(x^3 + 1) = (x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1) \][/tex]
Given:
[tex]\[ (x - 1) \cdot (x^6 - 1) = (x^3 - 1) \cdot A \][/tex]
Substitute the factored forms:
[tex]\[ (x - 1) \left((x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1)\right) = ((x - 1)(x^2 + x + 1)) \cdot A \][/tex]
[tex]\[ (x - 1)^2(x^2 + x + 1)(x + 1)(x^2 - x + 1) = (x - 1)(x^2 + x + 1) \cdot A \][/tex]
We can cancel \((x - 1)(x^2 + x + 1)\) from both sides, as they are common factors:
[tex]\[ (x - 1)(x + 1)(x^2 - x + 1) = A \][/tex]
So we need to find \( A \):
\( A = (x - 1)(x + 1)(x^2 - x + 1) \)
[tex]\[ A = (x^2 - 1)(x^2 - x + 1) \][/tex]
The given choices are:
1. \( x^3 + 1 \)
2. \( x^4 - x^3 + x - 1 \)
3. \( (x - 1)(x^2 - x + 1) \)
4. \( (x - 1)(x^2 + x + 1) \)
Let's test each choice to match our solution for \( A \).
By comparing our final expression of A with the provided options, it should be clear:
Given \( A = (x - 1)(x^2 - x + 1) \).
Therefore, the answer is option 3:
[tex]\[ (x - 1)(x^2 - x + 1) \][/tex]
The correct value of \( A \) is:
[tex]\[ \boxed{(x-1)(x^2 - x + 1)} \][/tex]
Given:
- \( \text{H.C.F} \) of \( x^3 - 1 \) and \( A \) is \( x - 1 \)
- \( \text{L.C.M} \) of \( x^3 - 1 \) and \( A \) is \( x^6 - 1 \)
We use the relationship between H.C.F and L.C.M of two expressions \( f(x) \) and \( g(x) \):
[tex]\[ \text{H.C.F}(f(x), g(x)) \cdot \text{L.C.M}(f(x), g(x)) = f(x) \cdot g(x) \][/tex]
Applying this to our problem:
[tex]\[ (x - 1) \cdot (x^6 - 1) = (x^3 - 1) \cdot A \][/tex]
First, we factorize the expressions where necessary.
[tex]\[ x^3 - 1 = (x - 1)(x^2 + x + 1) \][/tex]
[tex]\[ x^6 - 1 = (x^3 - 1)(x^3 + 1) = (x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1) \][/tex]
Given:
[tex]\[ (x - 1) \cdot (x^6 - 1) = (x^3 - 1) \cdot A \][/tex]
Substitute the factored forms:
[tex]\[ (x - 1) \left((x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1)\right) = ((x - 1)(x^2 + x + 1)) \cdot A \][/tex]
[tex]\[ (x - 1)^2(x^2 + x + 1)(x + 1)(x^2 - x + 1) = (x - 1)(x^2 + x + 1) \cdot A \][/tex]
We can cancel \((x - 1)(x^2 + x + 1)\) from both sides, as they are common factors:
[tex]\[ (x - 1)(x + 1)(x^2 - x + 1) = A \][/tex]
So we need to find \( A \):
\( A = (x - 1)(x + 1)(x^2 - x + 1) \)
[tex]\[ A = (x^2 - 1)(x^2 - x + 1) \][/tex]
The given choices are:
1. \( x^3 + 1 \)
2. \( x^4 - x^3 + x - 1 \)
3. \( (x - 1)(x^2 - x + 1) \)
4. \( (x - 1)(x^2 + x + 1) \)
Let's test each choice to match our solution for \( A \).
By comparing our final expression of A with the provided options, it should be clear:
Given \( A = (x - 1)(x^2 - x + 1) \).
Therefore, the answer is option 3:
[tex]\[ (x - 1)(x^2 - x + 1) \][/tex]
The correct value of \( A \) is:
[tex]\[ \boxed{(x-1)(x^2 - x + 1)} \][/tex]
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.