Let's determine the total number of roots for each polynomial given in their factored forms by counting their distinct factors, taking into account the multiplicity of each factor:
1. First Polynomial:
[tex]\[
f(x) = (x+1)(x-3)(x-4)
\][/tex]
- The factors are [tex]\(x + 1\)[/tex], [tex]\(x - 3\)[/tex], and [tex]\(x - 4\)[/tex].
- Each factor corresponds to one root.
- Therefore, the total number of roots is [tex]\(3\)[/tex].
2. Second Polynomial:
[tex]\[
f(x) = (x-6)^2(x+2)^2
\][/tex]
- The factors are [tex]\((x - 6)^2\)[/tex] and [tex]\((x + 2)^2\)[/tex].
- Each factor [tex]\((x - 6)^2\)[/tex] and [tex]\((x + 2)^2\)[/tex] has multiplicity 2.
- Therefore, the total number of roots is [tex]\(2 + 2 = 4\)[/tex].
3. Third Polynomial:
[tex]\[
f(x) = (x+5)^3(x-9)(x+1)
\][/tex]
- The factors are [tex]\((x + 5)^3\)[/tex], [tex]\((x - 9)\)[/tex], and [tex]\((x + 1)\)[/tex].
- The factor [tex]\((x + 5)^3\)[/tex] has multiplicity 3, while [tex]\((x - 9)\)[/tex] and [tex]\((x + 1)\)[/tex] each have multiplicity 1.
- Therefore, the total number of roots is [tex]\(3 + 1 + 1 = 5\)[/tex].
4. Fourth Polynomial:
[tex]\[
f(x) = (x + 2)(x - 1)[x - (4 + 3i)][x - (4 - 3i)]
\][/tex]
- The factors are [tex]\(x + 2\)[/tex], [tex]\(x - 1\)[/tex], [tex]\(x - (4 + 3i)\)[/tex], and [tex]\(x - (4 - 3i)\)[/tex].
- Each factor corresponds to one root.
- Therefore, the total number of roots is [tex]\(1 + 1 + 1 + 1 = 4\)[/tex].
Summarizing:
- For the polynomial [tex]\(f(x) = (x + 1)(x - 3)(x - 4)\)[/tex], the total number of roots is 3.
- For the polynomial [tex]\(f(x) = (x - 6)^2(x + 2)^2\)[/tex], the total number of roots is 4.
- For the polynomial [tex]\(f(x) = (x + 5)^3(x - 9)(x + 1)\)[/tex], the total number of roots is 5.
- For the polynomial [tex]\(f(x) = (x + 2)(x - 1)[x - (4 + 3i)][x - (4 - 3i)]\)[/tex], the total number of roots is 4.
