Answer :
To determine which of the given polynomials is a 3rd degree polynomial with exactly one real root, we will need to analyze each option.
Let's recall that a 3rd degree polynomial of the form [tex]\( F(x) = x^3 + ax^2 + bx + c \)[/tex] will have exactly one real root and two complex roots if its discriminant is less than 0. The discriminant [tex]\(\Delta\)[/tex] for a cubic polynomial [tex]\(x^3 + ax^2 + bx + c\)[/tex] can be quite complex, but for our purposes, we can determine the nature of roots by examining the polynomials directly.
### Option A: [tex]\(F(x) = x^3 + 9x^2 + 27x + 27\)[/tex]
Let's factorize [tex]\(F(x)\)[/tex]:
[tex]\[ F(x) = (x + 3)^3 \][/tex]
Here, you can see that the polynomial factors cleanly into [tex]\((x + 3)(x + 3)(x + 3)\)[/tex], which means it has a triple root at [tex]\(x = -3\)[/tex]. This indicates it has only one real root, repeated thrice.
### Option B: [tex]\(F(x) = x^3 + 3x^2 + 9x + 27\)[/tex]
To find the roots, we'll use synthetic division or trial-and-error to find if there is a real root. Let's try some simple values:
1. For [tex]\(x = -3\)[/tex]:
[tex]\[ (-3)^3 + 3(-3)^2 + 9(-3) + 27 = -27 + 27 - 27 + 27 = 0 \][/tex]
Thus, [tex]\(x = -3\)[/tex] is a root.
We could try factorizing it as:
[tex]\[ (x + 3) (x^2 + 0x + 9) \][/tex]
This gives us:
[tex]\[ x = -3 \][/tex]
[tex]\[ x^2 + 9 = 0 \quad \implies \quad x = \pm 3i \][/tex]
This means the polynomial has one real root and two complex roots.
### Option C: [tex]\(F(x) = x^3 - 9x^2 + 27x - 27\)[/tex]
Let's find the roots by evaluating the polynomial at possible candidates like [tex]\(x = 3\)[/tex], since the coefficients suggest that [tex]\(3\)[/tex] might be a root.
1. For [tex]\(x = 3\)[/tex]:
[tex]\[ (3)^3 - 9(3)^2 + 27(3) - 27 = 27 - 81 + 81 - 27 = 0 \][/tex]
Thus, [tex]\(x = 3\)[/tex] is a root.
Factorizing it as:
[tex]\[ (x - 3)(x^2 + 0x + 9) \][/tex]
This gives us:
[tex]\[ x = 3 \][/tex]
[tex]\[ x^2 + 9 = 0 \quad \implies \quad x = \pm 3i \][/tex]
So, this polynomial has one real root and two complex roots.
### Option D: [tex]\(F(x) = x^3 + 3x^2 - 9x - 27\)[/tex]
To find roots, we evaluate potential real roots like [tex]\(x = 3\)[/tex]:
1. For [tex]\(x = 3\)[/tex]:
[tex]\[ (3)^3 + 3(3)^2 - 9(3) - 27 = 27 + 27 - 27 - 27 = 0 \][/tex]
Thus, [tex]\(x = 3\)[/tex] is a root.
Factorizing it as:
[tex]\[ (x - 3)(x^2 + 6x + 9) \][/tex]
This gives:
[tex]\[ x = 3 \][/tex]
[tex]\[ x^2 + 6x + 9 = (x + 3)^2 \quad \implies \quad x = -3 \][/tex]
This polynomial actually has three real roots: [tex]\(x = 3\)[/tex] and [tex]\(x = -3\)[/tex] (double root).
### Conclusion
From the above analysis:
- Option A has one real root, [tex]\(x = -3\)[/tex] (triple root).
- Option B has one real root, [tex]\(x = -3\)[/tex], and two imaginary roots.
- Option C has one real root, [tex]\(x = 3\)[/tex], and two imaginary roots.
- Option D has three real roots.
Thus, the correct answer is:
B. [tex]\(F(x) = x^3 + 3x^2 + 9x + 27\)[/tex]
Let's recall that a 3rd degree polynomial of the form [tex]\( F(x) = x^3 + ax^2 + bx + c \)[/tex] will have exactly one real root and two complex roots if its discriminant is less than 0. The discriminant [tex]\(\Delta\)[/tex] for a cubic polynomial [tex]\(x^3 + ax^2 + bx + c\)[/tex] can be quite complex, but for our purposes, we can determine the nature of roots by examining the polynomials directly.
### Option A: [tex]\(F(x) = x^3 + 9x^2 + 27x + 27\)[/tex]
Let's factorize [tex]\(F(x)\)[/tex]:
[tex]\[ F(x) = (x + 3)^3 \][/tex]
Here, you can see that the polynomial factors cleanly into [tex]\((x + 3)(x + 3)(x + 3)\)[/tex], which means it has a triple root at [tex]\(x = -3\)[/tex]. This indicates it has only one real root, repeated thrice.
### Option B: [tex]\(F(x) = x^3 + 3x^2 + 9x + 27\)[/tex]
To find the roots, we'll use synthetic division or trial-and-error to find if there is a real root. Let's try some simple values:
1. For [tex]\(x = -3\)[/tex]:
[tex]\[ (-3)^3 + 3(-3)^2 + 9(-3) + 27 = -27 + 27 - 27 + 27 = 0 \][/tex]
Thus, [tex]\(x = -3\)[/tex] is a root.
We could try factorizing it as:
[tex]\[ (x + 3) (x^2 + 0x + 9) \][/tex]
This gives us:
[tex]\[ x = -3 \][/tex]
[tex]\[ x^2 + 9 = 0 \quad \implies \quad x = \pm 3i \][/tex]
This means the polynomial has one real root and two complex roots.
### Option C: [tex]\(F(x) = x^3 - 9x^2 + 27x - 27\)[/tex]
Let's find the roots by evaluating the polynomial at possible candidates like [tex]\(x = 3\)[/tex], since the coefficients suggest that [tex]\(3\)[/tex] might be a root.
1. For [tex]\(x = 3\)[/tex]:
[tex]\[ (3)^3 - 9(3)^2 + 27(3) - 27 = 27 - 81 + 81 - 27 = 0 \][/tex]
Thus, [tex]\(x = 3\)[/tex] is a root.
Factorizing it as:
[tex]\[ (x - 3)(x^2 + 0x + 9) \][/tex]
This gives us:
[tex]\[ x = 3 \][/tex]
[tex]\[ x^2 + 9 = 0 \quad \implies \quad x = \pm 3i \][/tex]
So, this polynomial has one real root and two complex roots.
### Option D: [tex]\(F(x) = x^3 + 3x^2 - 9x - 27\)[/tex]
To find roots, we evaluate potential real roots like [tex]\(x = 3\)[/tex]:
1. For [tex]\(x = 3\)[/tex]:
[tex]\[ (3)^3 + 3(3)^2 - 9(3) - 27 = 27 + 27 - 27 - 27 = 0 \][/tex]
Thus, [tex]\(x = 3\)[/tex] is a root.
Factorizing it as:
[tex]\[ (x - 3)(x^2 + 6x + 9) \][/tex]
This gives:
[tex]\[ x = 3 \][/tex]
[tex]\[ x^2 + 6x + 9 = (x + 3)^2 \quad \implies \quad x = -3 \][/tex]
This polynomial actually has three real roots: [tex]\(x = 3\)[/tex] and [tex]\(x = -3\)[/tex] (double root).
### Conclusion
From the above analysis:
- Option A has one real root, [tex]\(x = -3\)[/tex] (triple root).
- Option B has one real root, [tex]\(x = -3\)[/tex], and two imaginary roots.
- Option C has one real root, [tex]\(x = 3\)[/tex], and two imaginary roots.
- Option D has three real roots.
Thus, the correct answer is:
B. [tex]\(F(x) = x^3 + 3x^2 + 9x + 27\)[/tex]
