Answer :
To find the Highest Common Factor (HCF) of the polynomials [tex]\(12(x^3 + x^2 + x + 1)\)[/tex] and [tex]\(18(x-1)\)[/tex], let's go through the process step-by-step.
### Step 1: Factorize Each Polynomial
1. Factorize [tex]\(12(x^3 + x^2 + x + 1)\)[/tex]:
First, let's simplify the polynomial inside the parenthesis:
[tex]\[ x^3 + x^2 + x + 1 \][/tex]
We can factor this polynomial by recognizing it as a sum of terms that can be grouped and simplified. Notice that:
[tex]\[ x^3 + x^2 + x + 1 = (x^3 + 1) + (x^2 + x) \][/tex]
However, this is not straightforward. A better approach is to use the fact that [tex]\(x^3 + x^2 + x + 1\)[/tex] can be factored directly:
[tex]\[ x^3 + x^2 + x + 1 = (x + 1)(x^2 + 1) \][/tex]
But actually, another form that is more practical here involves the polynomial expressions and synthetic division. Recognizing special forms, we'll simplify step-by-step. Here:
[tex]\[ x^3 + x^2 + x + 1 = (x + 1)(x^3 + 1) \][/tex]
Then:
[tex]\[ 12(x^3 + x^2 + x + 1) = 12(x + 1)(x^2 + 1) \][/tex]
2. Factorize [tex]\(18(x-1)\)[/tex]:
This polynomial is simpler:
[tex]\[ 18(x - 1) \][/tex]
### Step 2: Find the Common Factors
Now, combine the coefficients and factor terms:
- The coefficient term from [tex]\(12(x + 1)(x^2 + 1)\)[/tex] is [tex]\(12\)[/tex].
- The coefficient term from [tex]\(18(x - 1)\)[/tex] is [tex]\(18\)[/tex].
Finding the HCF of the coefficients [tex]\(12\)[/tex] and [tex]\(18\)[/tex]:
- [tex]\(12 = 2^2 \times 3\)[/tex]
- [tex]\(18 = 2 \times 3^2\)[/tex]
- The HCF of [tex]\(12\)[/tex] and [tex]\(18\)[/tex] is [tex]\(6\)[/tex].
### Step 3: Determine the Common Polynomial Factor Terms
- For the polynomial terms, we observe:
- [tex]\(12(x + 1)(x^2 + 1)\)[/tex]
- [tex]\(18(x - 1)\)[/tex]
Notice that [tex]\(x+1\)[/tex] and [tex]\(x-1\)[/tex] are different; there appear no common polynomial factors on deeper comparison, especially reducing higher-order polynomial interactions meaningfully in expanded form. Hence, highest polynomial term essentially shared is valor multiplicatively through scalar contribution but uniquely divergent polynomial arrangements.
### Step 4: Combine Common Factors
The HCF thus formed by combining numerical coefficient [tex]\(6\)[/tex] and simplistics without deeper evaluated common polynomial indicates primarily coefficient contributes. Realize higher intricate intro polynomial terms less numerically evident alignment contextually meaning higher dimensional polynomial factors may seem non-trivially impactful.
### Conclusion
Thus, the HCF of these polynomials is primarily reductive scalar [tex]\(6\)[/tex], symbolizing shared basis numerically rather polynomial quantities.
### Final Answer:
The HCF of [tex]\(12(x^3 + x^2 + x + 1)\)[/tex] and [tex]\(18(x-1)\)[/tex] is primarily coherent simplistically as effective scalar 6 shared, signifying the integer specific leading coefficient primarily represented shared numerical domain evaluative.
### Step 1: Factorize Each Polynomial
1. Factorize [tex]\(12(x^3 + x^2 + x + 1)\)[/tex]:
First, let's simplify the polynomial inside the parenthesis:
[tex]\[ x^3 + x^2 + x + 1 \][/tex]
We can factor this polynomial by recognizing it as a sum of terms that can be grouped and simplified. Notice that:
[tex]\[ x^3 + x^2 + x + 1 = (x^3 + 1) + (x^2 + x) \][/tex]
However, this is not straightforward. A better approach is to use the fact that [tex]\(x^3 + x^2 + x + 1\)[/tex] can be factored directly:
[tex]\[ x^3 + x^2 + x + 1 = (x + 1)(x^2 + 1) \][/tex]
But actually, another form that is more practical here involves the polynomial expressions and synthetic division. Recognizing special forms, we'll simplify step-by-step. Here:
[tex]\[ x^3 + x^2 + x + 1 = (x + 1)(x^3 + 1) \][/tex]
Then:
[tex]\[ 12(x^3 + x^2 + x + 1) = 12(x + 1)(x^2 + 1) \][/tex]
2. Factorize [tex]\(18(x-1)\)[/tex]:
This polynomial is simpler:
[tex]\[ 18(x - 1) \][/tex]
### Step 2: Find the Common Factors
Now, combine the coefficients and factor terms:
- The coefficient term from [tex]\(12(x + 1)(x^2 + 1)\)[/tex] is [tex]\(12\)[/tex].
- The coefficient term from [tex]\(18(x - 1)\)[/tex] is [tex]\(18\)[/tex].
Finding the HCF of the coefficients [tex]\(12\)[/tex] and [tex]\(18\)[/tex]:
- [tex]\(12 = 2^2 \times 3\)[/tex]
- [tex]\(18 = 2 \times 3^2\)[/tex]
- The HCF of [tex]\(12\)[/tex] and [tex]\(18\)[/tex] is [tex]\(6\)[/tex].
### Step 3: Determine the Common Polynomial Factor Terms
- For the polynomial terms, we observe:
- [tex]\(12(x + 1)(x^2 + 1)\)[/tex]
- [tex]\(18(x - 1)\)[/tex]
Notice that [tex]\(x+1\)[/tex] and [tex]\(x-1\)[/tex] are different; there appear no common polynomial factors on deeper comparison, especially reducing higher-order polynomial interactions meaningfully in expanded form. Hence, highest polynomial term essentially shared is valor multiplicatively through scalar contribution but uniquely divergent polynomial arrangements.
### Step 4: Combine Common Factors
The HCF thus formed by combining numerical coefficient [tex]\(6\)[/tex] and simplistics without deeper evaluated common polynomial indicates primarily coefficient contributes. Realize higher intricate intro polynomial terms less numerically evident alignment contextually meaning higher dimensional polynomial factors may seem non-trivially impactful.
### Conclusion
Thus, the HCF of these polynomials is primarily reductive scalar [tex]\(6\)[/tex], symbolizing shared basis numerically rather polynomial quantities.
### Final Answer:
The HCF of [tex]\(12(x^3 + x^2 + x + 1)\)[/tex] and [tex]\(18(x-1)\)[/tex] is primarily coherent simplistically as effective scalar 6 shared, signifying the integer specific leading coefficient primarily represented shared numerical domain evaluative.
