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Sagot :
Alright, let's go through the factorization of the given polynomial expressions step-by-step to understand their factors and see how they are decomposed.
### Expression 1: [tex]\((x+1)(2+i x)(2-i x)\)[/tex]
1. Identify the roots: The factors are [tex]\((x+1)\)[/tex], [tex]\((2+ix)\)[/tex], and [tex]\((2-ix)\)[/tex].
2. The term [tex]\((x+1)\)[/tex] suggests a root at [tex]\(x = -1\)[/tex].
3. The terms [tex]\((2 + ix)\)[/tex] and [tex]\((2 - ix)\)[/tex] suggest complex roots. We can write these roots as [tex]\(x = -\frac{2}{i}\)[/tex] and [tex]\(x = \frac{2}{i}\)[/tex]. Simplifying, these become [tex]\(x = -2i\)[/tex] and [tex]\(x = 2i\)[/tex].
### Expression 2: [tex]\((x-1)(2+i x)(2-i x)\)[/tex]
1. Identify the roots: The factors are [tex]\((x-1)\)[/tex], [tex]\((2+ix)\)[/tex], and [tex]\((2-ix)\)[/tex].
2. The term [tex]\((x-1)\)[/tex] suggests a root at [tex]\(x = 1\)[/tex].
3. The terms [tex]\((2 + ix)\)[/tex] and [tex]\((2 - ix)\)[/tex] suggest complex roots [tex]\(x = -\frac{2}{i} = -2i\)[/tex] and [tex]\(x = \frac{2}{i} = 2i\)[/tex].
### Expression 3: [tex]\((x-1)[x-(3+i)][x-(3-i)]\)[/tex]
1. Identify the roots: The factors are [tex]\((x-1)\)[/tex], [tex]\((x - (3+i))\)[/tex], and [tex]\((x - (3-i))\)[/tex].
2. The term [tex]\((x-1)\)[/tex] suggests a root at [tex]\(x = 1\)[/tex].
3. The term [tex]\((x - (3+i))\)[/tex] suggests a root at [tex]\(x = 3+i\)[/tex].
4. The term [tex]\((x - (3-i))\)[/tex] suggests a root at [tex]\(x = 3-i\)[/tex].
### Expression 4: [tex]\((x+1)[x+(2+i)][x-(2-i)]\)[/tex]
1. Identify the roots: The factors are [tex]\((x+1)\)[/tex], [tex]\((x+(2+i))\)[/tex], and [tex]\((x-(2-i))\)[/tex].
2. The term [tex]\((x+1)\)[/tex] suggests a root at [tex]\(x = -1\)[/tex].
3. The term [tex]\((x+(2+i))\)[/tex] can be rewritten as [tex]\((x - (-2-i))\)[/tex], suggesting a root at [tex]\(x = -2 - i\)[/tex].
4. The term [tex]\((x-(2-i))\)[/tex] suggests a root at [tex]\(x = 2-i\)[/tex].
### Expression 5: [tex]\((x-1)[x+(2+i)][x-(2-i)]\)[/tex]
1. Identify the roots: The factors are [tex]\((x-1)\)[/tex], [tex]\((x+(2+i))\)[/tex], and [tex]\((x-(2-i))\)[/tex].
2. The term [tex]\((x-1)\)[/tex] suggests a root at [tex]\(x = 1\)[/tex].
3. The term [tex]\((x+(2+i))\)[/tex], rewritten as [tex]\((x - (-2-i))\)[/tex], suggests a root at [tex]\(x = -2 - i\)[/tex].
4. The term [tex]\((x-(2-i))\)[/tex] suggests a root at [tex]\(x = 2-i\)[/tex].
### Conclusion:
The factors provided each reflect a polynomial factored into linear terms, some of which include complex numbers. The complex conjugate pairs [tex]\((2+ix)\)[/tex] and [tex]\((2-ix)\)[/tex] or [tex]\((3+i)\)[/tex] and [tex]\((3-i)\)[/tex] suggest that these polynomials originally had complex roots. Each of these factorizations demonstrates the decomposition of a polynomial into irreducible factors.
### Expression 1: [tex]\((x+1)(2+i x)(2-i x)\)[/tex]
1. Identify the roots: The factors are [tex]\((x+1)\)[/tex], [tex]\((2+ix)\)[/tex], and [tex]\((2-ix)\)[/tex].
2. The term [tex]\((x+1)\)[/tex] suggests a root at [tex]\(x = -1\)[/tex].
3. The terms [tex]\((2 + ix)\)[/tex] and [tex]\((2 - ix)\)[/tex] suggest complex roots. We can write these roots as [tex]\(x = -\frac{2}{i}\)[/tex] and [tex]\(x = \frac{2}{i}\)[/tex]. Simplifying, these become [tex]\(x = -2i\)[/tex] and [tex]\(x = 2i\)[/tex].
### Expression 2: [tex]\((x-1)(2+i x)(2-i x)\)[/tex]
1. Identify the roots: The factors are [tex]\((x-1)\)[/tex], [tex]\((2+ix)\)[/tex], and [tex]\((2-ix)\)[/tex].
2. The term [tex]\((x-1)\)[/tex] suggests a root at [tex]\(x = 1\)[/tex].
3. The terms [tex]\((2 + ix)\)[/tex] and [tex]\((2 - ix)\)[/tex] suggest complex roots [tex]\(x = -\frac{2}{i} = -2i\)[/tex] and [tex]\(x = \frac{2}{i} = 2i\)[/tex].
### Expression 3: [tex]\((x-1)[x-(3+i)][x-(3-i)]\)[/tex]
1. Identify the roots: The factors are [tex]\((x-1)\)[/tex], [tex]\((x - (3+i))\)[/tex], and [tex]\((x - (3-i))\)[/tex].
2. The term [tex]\((x-1)\)[/tex] suggests a root at [tex]\(x = 1\)[/tex].
3. The term [tex]\((x - (3+i))\)[/tex] suggests a root at [tex]\(x = 3+i\)[/tex].
4. The term [tex]\((x - (3-i))\)[/tex] suggests a root at [tex]\(x = 3-i\)[/tex].
### Expression 4: [tex]\((x+1)[x+(2+i)][x-(2-i)]\)[/tex]
1. Identify the roots: The factors are [tex]\((x+1)\)[/tex], [tex]\((x+(2+i))\)[/tex], and [tex]\((x-(2-i))\)[/tex].
2. The term [tex]\((x+1)\)[/tex] suggests a root at [tex]\(x = -1\)[/tex].
3. The term [tex]\((x+(2+i))\)[/tex] can be rewritten as [tex]\((x - (-2-i))\)[/tex], suggesting a root at [tex]\(x = -2 - i\)[/tex].
4. The term [tex]\((x-(2-i))\)[/tex] suggests a root at [tex]\(x = 2-i\)[/tex].
### Expression 5: [tex]\((x-1)[x+(2+i)][x-(2-i)]\)[/tex]
1. Identify the roots: The factors are [tex]\((x-1)\)[/tex], [tex]\((x+(2+i))\)[/tex], and [tex]\((x-(2-i))\)[/tex].
2. The term [tex]\((x-1)\)[/tex] suggests a root at [tex]\(x = 1\)[/tex].
3. The term [tex]\((x+(2+i))\)[/tex], rewritten as [tex]\((x - (-2-i))\)[/tex], suggests a root at [tex]\(x = -2 - i\)[/tex].
4. The term [tex]\((x-(2-i))\)[/tex] suggests a root at [tex]\(x = 2-i\)[/tex].
### Conclusion:
The factors provided each reflect a polynomial factored into linear terms, some of which include complex numbers. The complex conjugate pairs [tex]\((2+ix)\)[/tex] and [tex]\((2-ix)\)[/tex] or [tex]\((3+i)\)[/tex] and [tex]\((3-i)\)[/tex] suggest that these polynomials originally had complex roots. Each of these factorizations demonstrates the decomposition of a polynomial into irreducible factors.
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