\(a^4+2a^3+3a^2+2a+1\) and \(a^6-2a^3+1\)?
Understanding the Problem: Finding the HCF of Polynomials
The question asks us to find the Highest Common Factor (HCF), also known as the Greatest Common Divisor (GCD), of two given polynomial expressions:
- Polynomial 1: \( P(a) = a^4+2a^3+3a^2+2a+1 \)
- Polynomial 2: \( Q(a) = a^6-2a^3+1 \)
To find the HCF, we need to factorize both polynomials completely and identify the common factors.
Step 1: Factorizing the Polynomial \( Q(a) = a^6-2a^3+1 \)
The polynomial \( Q(a) \) can be seen as a quadratic expression in terms of \( a^3 \). Let \( x = a^3 \). Then the expression becomes:
\( x^2 - 2x + 1 \)This is a perfect square trinomial, which factors as:
\( (x-1)^2 \)Now, substitute \( x = a^3 \) back into the factored expression:
\( (a^3 - 1)^2 \)We know the difference of cubes formula: \( a^3 - 1 = (a-1)(a^2+a+1) \). Substituting this:
\( [(a-1)(a^2+a+1)]^2 \)Expanding the square, we get the complete factorization of \( Q(a) \):
\( Q(a) = (a-1)^2 (a^2+a+1)^2 \)Step 2: Factorizing the Polynomial \( P(a) = a^4+2a^3+3a^2+2a+1 \)
The polynomial \( P(a) \) is a reciprocal polynomial because its coefficients are symmetric (1, 2, 3, 2, 1). We can factor it by dividing by \( a^2 \) (assuming \( a \neq 0 \)) and rearranging:
\( \frac{P(a)}{a^2} = \frac{a^4}{a^2} + \frac{2a^3}{a^2} + \frac{3a^2}{a^2} + \frac{2a}{a^2} + \frac{1}{a^2} \) \( = a^2 + 2a + 3 + \frac{2}{a} + \frac{1}{a^2} \)Group the terms:
\( = \left( a^2 + \frac{1}{a^2} \right) + 2\left( a + \frac{1}{a} \right) + 3 \)Let \( y = a + \frac{1}{a} \). Then \( y^2 = \left( a + \frac{1}{a} \right)^2 = a^2 + 2(a)\left(\frac{1}{a}\right) + \frac{1}{a^2} = a^2 + 2 + \frac{1}{a^2} \). Therefore, \( a^2 + \frac{1}{a^2} = y^2 - 2 \).
Substitute \( y \) and \( y^2 - 2 \) into the expression:
\( = (y^2 - 2) + 2y + 3 \) \( = y^2 + 2y + 1 \)This is a perfect square:
\( = (y+1)^2 \)Now, substitute \( y = a + \frac{1}{a} \) back:
\( = \left( a + \frac{1}{a} + 1 \right)^2 \)To get \( P(a) \), multiply by \( a^2 \):
\( P(a) = a^2 \left( a + \frac{1}{a} + 1 \right)^2 \) \( P(a) = \left[ a \left( a + \frac{1}{a} + 1 \right) \right]^2 \) \( P(a) = (a^2 + 1 + a)^2 \) \( P(a) = (a^2+a+1)^2 \)Step 3: Identifying the Highest Common Factor (HCF)
We have the factorizations:
- \( P(a) = (a^2+a+1)^2 \)
- \( Q(a) = (a-1)^2 (a^2+a+1)^2 \)
To find the HCF, we look for the common factors raised to the lowest power they appear in either factorization. The common factor is \( (a^2+a+1) \). The lowest power it appears with is 2 in both factorizations.
Therefore, the HCF is:
\( \text{HCF}(P(a), Q(a)) = (a^2+a+1)^2 \)Conclusion
Comparing our result with the given options, the HCF of the two polynomials is \( (a^2+a+1)^2 \).
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