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Sagot :
To solve for the remaining root(s) of the polynomial [tex]\( f(x) = x^3 - 7x - 6 \)[/tex], we start by taking into consideration the given roots and the properties of polynomials.
### Step-by-Step Solution:
1. Fundamental Theorem of Algebra:
- This theorem states that a polynomial of degree [tex]\( n \)[/tex] will have exactly [tex]\( n \)[/tex] roots (including repeated roots) in the complex number system. Since [tex]\( f(x) = x^3 - 7x - 6 \)[/tex] is a cubic polynomial (degree 3), it will have exactly 3 roots.
2. Given Roots:
- We are given that two of those roots are [tex]\( -2 \)[/tex] and [tex]\( 3 \)[/tex].
3. Forming a Polynomial Factor:
- Knowing the roots, we can express part of the polynomial as products of linear factors corresponding to the given roots:
[tex]\[ (x + 2)(x - 3) \][/tex]
4. Determining the Polynomial from Given Factors:
- First, multiply the factors of the given roots:
[tex]\[ (x + 2)(x - 3) = x^2 - x - 6 \][/tex]
- This result represents a quadratic polynomial that is a factor of [tex]\( f(x) \)[/tex].
5. Polynomial Division:
- Since [tex]\( f(x) \)[/tex] is known to have [tex]\( (x + 2) \)[/tex] and [tex]\( (x - 3) \)[/tex] as factors, we need to perform polynomial division to find the other factor.
- Divide the original polynomial [tex]\( f(x) = x^3 - 7x - 6 \)[/tex] by [tex]\( (x^2 - x - 6) \)[/tex]:
[tex]\[ x^3 - 7x - 6 \div (x^2 - x - 6) \][/tex]
- This divides exactly with no remainder, leading to:
[tex]\[ f(x) / (x^2 - x - 6) = x - 1 \][/tex]
6. Finding the Remaining Root:
- The result of the division [tex]\( x - 1 \)[/tex] reveals the remaining root of the polynomial.
Therefore, the third root of the polynomial [tex]\( f(x) = x^3 - 7x - 6 \)[/tex] is:
[tex]\[ 1 \][/tex]
### Summary:
Given that the polynomial [tex]\( f(x) = x^3 - 7x - 6 \)[/tex] has the roots [tex]\( -2 \)[/tex], [tex]\( 3 \)[/tex], and [tex]\( 1 \)[/tex], we conclude that these roots account for all the roots of this cubic polynomial. All the roots are real numbers.
### Step-by-Step Solution:
1. Fundamental Theorem of Algebra:
- This theorem states that a polynomial of degree [tex]\( n \)[/tex] will have exactly [tex]\( n \)[/tex] roots (including repeated roots) in the complex number system. Since [tex]\( f(x) = x^3 - 7x - 6 \)[/tex] is a cubic polynomial (degree 3), it will have exactly 3 roots.
2. Given Roots:
- We are given that two of those roots are [tex]\( -2 \)[/tex] and [tex]\( 3 \)[/tex].
3. Forming a Polynomial Factor:
- Knowing the roots, we can express part of the polynomial as products of linear factors corresponding to the given roots:
[tex]\[ (x + 2)(x - 3) \][/tex]
4. Determining the Polynomial from Given Factors:
- First, multiply the factors of the given roots:
[tex]\[ (x + 2)(x - 3) = x^2 - x - 6 \][/tex]
- This result represents a quadratic polynomial that is a factor of [tex]\( f(x) \)[/tex].
5. Polynomial Division:
- Since [tex]\( f(x) \)[/tex] is known to have [tex]\( (x + 2) \)[/tex] and [tex]\( (x - 3) \)[/tex] as factors, we need to perform polynomial division to find the other factor.
- Divide the original polynomial [tex]\( f(x) = x^3 - 7x - 6 \)[/tex] by [tex]\( (x^2 - x - 6) \)[/tex]:
[tex]\[ x^3 - 7x - 6 \div (x^2 - x - 6) \][/tex]
- This divides exactly with no remainder, leading to:
[tex]\[ f(x) / (x^2 - x - 6) = x - 1 \][/tex]
6. Finding the Remaining Root:
- The result of the division [tex]\( x - 1 \)[/tex] reveals the remaining root of the polynomial.
Therefore, the third root of the polynomial [tex]\( f(x) = x^3 - 7x - 6 \)[/tex] is:
[tex]\[ 1 \][/tex]
### Summary:
Given that the polynomial [tex]\( f(x) = x^3 - 7x - 6 \)[/tex] has the roots [tex]\( -2 \)[/tex], [tex]\( 3 \)[/tex], and [tex]\( 1 \)[/tex], we conclude that these roots account for all the roots of this cubic polynomial. All the roots are real numbers.
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