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Sagot :
To determine the multiplicity of the root [tex]\( x = 2 \)[/tex] for the polynomial [tex]\( y = x^3 - 3x^2 + 4 \)[/tex], we follow these steps:
1. Define the Polynomial:
[tex]\[ P(x) = x^3 - 3x^2 + 4 \][/tex]
2. Evaluate the Polynomial at [tex]\( x = 2 \)[/tex]:
[tex]\[ P(2) = (2)^3 - 3(2)^2 + 4 = 8 - 12 + 4 = 0 \][/tex]
Since [tex]\( P(2) = 0 \)[/tex], [tex]\( x = 2 \)[/tex] is indeed a root of the polynomial.
3. Find the First Derivative:
[tex]\[ P'(x) = 3x^2 - 6x \][/tex]
Evaluate the first derivative at [tex]\( x = 2 \)[/tex]:
[tex]\[ P'(2) = 3(2)^2 - 6(2) = 12 - 12 = 0 \][/tex]
4. Find the Second Derivative:
[tex]\[ P''(x) = 6x - 6 \][/tex]
Evaluate the second derivative at [tex]\( x = 2 \)[/tex]:
[tex]\[ P''(2) = 6(2) - 6 = 12 - 6 = 6 \][/tex]
Since [tex]\( P''(2) \neq 0 \)[/tex], this means that the second derivative is non-zero at [tex]\( x = 2 \)[/tex].
To summarize:
- [tex]\( P(2) = 0 \)[/tex]
- [tex]\( P'(2) = 0 \)[/tex]
- [tex]\( P''(2) \neq 0 \)[/tex]
These conditions imply that [tex]\( x = 2 \)[/tex] is a root with multiplicity 2, because the polynomial and its first derivative are zero at [tex]\( x = 2 \)[/tex], but the second derivative is non-zero. Therefore, the root [tex]\( x = 2 \)[/tex] has a multiplicity of [tex]\( 2 \)[/tex].
So, the correct answer is:
2
1. Define the Polynomial:
[tex]\[ P(x) = x^3 - 3x^2 + 4 \][/tex]
2. Evaluate the Polynomial at [tex]\( x = 2 \)[/tex]:
[tex]\[ P(2) = (2)^3 - 3(2)^2 + 4 = 8 - 12 + 4 = 0 \][/tex]
Since [tex]\( P(2) = 0 \)[/tex], [tex]\( x = 2 \)[/tex] is indeed a root of the polynomial.
3. Find the First Derivative:
[tex]\[ P'(x) = 3x^2 - 6x \][/tex]
Evaluate the first derivative at [tex]\( x = 2 \)[/tex]:
[tex]\[ P'(2) = 3(2)^2 - 6(2) = 12 - 12 = 0 \][/tex]
4. Find the Second Derivative:
[tex]\[ P''(x) = 6x - 6 \][/tex]
Evaluate the second derivative at [tex]\( x = 2 \)[/tex]:
[tex]\[ P''(2) = 6(2) - 6 = 12 - 6 = 6 \][/tex]
Since [tex]\( P''(2) \neq 0 \)[/tex], this means that the second derivative is non-zero at [tex]\( x = 2 \)[/tex].
To summarize:
- [tex]\( P(2) = 0 \)[/tex]
- [tex]\( P'(2) = 0 \)[/tex]
- [tex]\( P''(2) \neq 0 \)[/tex]
These conditions imply that [tex]\( x = 2 \)[/tex] is a root with multiplicity 2, because the polynomial and its first derivative are zero at [tex]\( x = 2 \)[/tex], but the second derivative is non-zero. Therefore, the root [tex]\( x = 2 \)[/tex] has a multiplicity of [tex]\( 2 \)[/tex].
So, the correct answer is:
2
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