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Answer :
To find which 3rd degree polynomial has exactly one real root, we need to consider the nature of the roots of a cubic polynomial. A cubic polynomial generally can have up to three real roots. However, it can also have one real root with the other two being complex conjugates.
Let's look at the options provided:
A. [tex]\( F(x)=x^3+3x^2-9x-27 \)[/tex]
B. [tex]\( F(x)=x^3+9x^2+27x+27 \)[/tex]
C. [tex]\( F(x)=x^3+3x^2+9x+27 \)[/tex]
D. [tex]\( F(x)=x^3-9x^2+27x-27 \)[/tex]
To determine which polynomial has exactly one real root, we consider the nature and number of roots. We are informed that:
- Option B, [tex]\( F(x)=x^3+9x^2+27x+27 \)[/tex], has exactly 1 real root.
The presence of exactly one real root in this polynomial implies that the other two roots are complex conjugates (since complex roots in polynomials with real coefficients come in conjugate pairs).
Therefore, the polynomial [tex]\( F(x)=x^3+9x^2+27x+27 \)[/tex] is the required polynomial with exactly one real root.
Let's look at the options provided:
A. [tex]\( F(x)=x^3+3x^2-9x-27 \)[/tex]
B. [tex]\( F(x)=x^3+9x^2+27x+27 \)[/tex]
C. [tex]\( F(x)=x^3+3x^2+9x+27 \)[/tex]
D. [tex]\( F(x)=x^3-9x^2+27x-27 \)[/tex]
To determine which polynomial has exactly one real root, we consider the nature and number of roots. We are informed that:
- Option B, [tex]\( F(x)=x^3+9x^2+27x+27 \)[/tex], has exactly 1 real root.
The presence of exactly one real root in this polynomial implies that the other two roots are complex conjugates (since complex roots in polynomials with real coefficients come in conjugate pairs).
Therefore, the polynomial [tex]\( F(x)=x^3+9x^2+27x+27 \)[/tex] is the required polynomial with exactly one real root.
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