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To determine the number of roots of the polynomial [tex]\( f(x) = x^4 + x^3 - 7x^2 - x + 6 \)[/tex], we need to find the values of [tex]\( x \)[/tex] that make the polynomial equal to zero. These values are known as the roots of the polynomial.
A polynomial of degree [tex]\( n \)[/tex] has exactly [tex]\( n \)[/tex] roots in the complex number system, considering multiplicity (where some roots can repeat).
Here, the polynomial [tex]\( f(x) = x^4 + x^3 - 7x^2 - x + 6 \)[/tex] is a fourth-degree polynomial ([tex]\( n = 4 \)[/tex]). Therefore, we expect it to have exactly 4 roots in the complex number system.
Let’s confirm this by finding the roots of the polynomial.
Step-by-Step:
1. Initial Analysis:
- The polynomial [tex]\( f(x) = x^4 + x^3 - 7x^2 - x + 6 \)[/tex] is of degree 4.
2. Finding Rational Roots:
- Use the Rational Root Theorem, which states that possible rational roots are the factors of the constant term (6) divided by the factors of the leading coefficient (1).
- Possible rational roots: [tex]\( \pm 1, \pm 2, \pm 3, \pm 6 \)[/tex].
3. Test Possible Roots:
- Substitute the possible rational roots into the polynomial to check if they satisfy [tex]\( f(x) = 0 \)[/tex].
- Testing [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 1^4 + 1^3 - 7(1)^2 - 1 + 6 = 1 + 1 - 7 - 1 + 6 = 0 \][/tex]
- [tex]\( x = 1 \)[/tex] is a root.
- Testing [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = (-1)^4 + (-1)^3 - 7(-1)^2 - (-1) + 6 = 1 - 1 - 7 + 1 + 6 = 0 \][/tex]
- [tex]\( x = -1 \)[/tex] is a root.
4. Polynomial Factorization:
- Since [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex] are roots, we can factor the polynomial as:
[tex]\[ f(x) = (x - 1)(x + 1)(quadratic\ polynomial) \][/tex]
5. Finding the Remaining Quadratic Polynomial:
- Perform polynomial division to divide [tex]\( f(x) \)[/tex] by [tex]\( (x - 1)(x + 1) = x^2 - 1 \)[/tex]:
[tex]\[ \frac{x^4 + x^3 - 7x^2 - x + 6}{x^2 - 1} = x^2 + x - 6 \][/tex]
6. Roots of Quadratic Polynomial:
- Solve [tex]\( x^2 + x - 6 = 0 \)[/tex] using the quadratic formula:
[tex]\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], [tex]\( c = -6 \)[/tex]:
[tex]\[ x = \frac{{-1 \pm \sqrt{{1^2 - 4 \cdot 1 \cdot (-6)}}}}{2 \cdot 1} = \frac{{-1 \pm \sqrt{{1 + 24}}}}{2} = \frac{{-1 \pm \sqrt{25}}}{2} = \frac{{-1 \pm 5}}{2} \][/tex]
[tex]\[ x = 2 \quad \text{and} \quad x = -3 \][/tex]
- Roots of the quadratic polynomial: [tex]\( x = 2 \)[/tex], [tex]\( x = -3 \)[/tex].
7. All Roots of the Polynomial:
- Combining all the roots: [tex]\( x = 1, x = -1, x = 2, x = -3 \)[/tex].
Thus, the polynomial [tex]\( f(x) = x^4 + x^3 - 7x^2 - x + 6 \)[/tex] has 4 roots.
Answer: The number of roots of the polynomial is 4.
A polynomial of degree [tex]\( n \)[/tex] has exactly [tex]\( n \)[/tex] roots in the complex number system, considering multiplicity (where some roots can repeat).
Here, the polynomial [tex]\( f(x) = x^4 + x^3 - 7x^2 - x + 6 \)[/tex] is a fourth-degree polynomial ([tex]\( n = 4 \)[/tex]). Therefore, we expect it to have exactly 4 roots in the complex number system.
Let’s confirm this by finding the roots of the polynomial.
Step-by-Step:
1. Initial Analysis:
- The polynomial [tex]\( f(x) = x^4 + x^3 - 7x^2 - x + 6 \)[/tex] is of degree 4.
2. Finding Rational Roots:
- Use the Rational Root Theorem, which states that possible rational roots are the factors of the constant term (6) divided by the factors of the leading coefficient (1).
- Possible rational roots: [tex]\( \pm 1, \pm 2, \pm 3, \pm 6 \)[/tex].
3. Test Possible Roots:
- Substitute the possible rational roots into the polynomial to check if they satisfy [tex]\( f(x) = 0 \)[/tex].
- Testing [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 1^4 + 1^3 - 7(1)^2 - 1 + 6 = 1 + 1 - 7 - 1 + 6 = 0 \][/tex]
- [tex]\( x = 1 \)[/tex] is a root.
- Testing [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = (-1)^4 + (-1)^3 - 7(-1)^2 - (-1) + 6 = 1 - 1 - 7 + 1 + 6 = 0 \][/tex]
- [tex]\( x = -1 \)[/tex] is a root.
4. Polynomial Factorization:
- Since [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex] are roots, we can factor the polynomial as:
[tex]\[ f(x) = (x - 1)(x + 1)(quadratic\ polynomial) \][/tex]
5. Finding the Remaining Quadratic Polynomial:
- Perform polynomial division to divide [tex]\( f(x) \)[/tex] by [tex]\( (x - 1)(x + 1) = x^2 - 1 \)[/tex]:
[tex]\[ \frac{x^4 + x^3 - 7x^2 - x + 6}{x^2 - 1} = x^2 + x - 6 \][/tex]
6. Roots of Quadratic Polynomial:
- Solve [tex]\( x^2 + x - 6 = 0 \)[/tex] using the quadratic formula:
[tex]\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], [tex]\( c = -6 \)[/tex]:
[tex]\[ x = \frac{{-1 \pm \sqrt{{1^2 - 4 \cdot 1 \cdot (-6)}}}}{2 \cdot 1} = \frac{{-1 \pm \sqrt{{1 + 24}}}}{2} = \frac{{-1 \pm \sqrt{25}}}{2} = \frac{{-1 \pm 5}}{2} \][/tex]
[tex]\[ x = 2 \quad \text{and} \quad x = -3 \][/tex]
- Roots of the quadratic polynomial: [tex]\( x = 2 \)[/tex], [tex]\( x = -3 \)[/tex].
7. All Roots of the Polynomial:
- Combining all the roots: [tex]\( x = 1, x = -1, x = 2, x = -3 \)[/tex].
Thus, the polynomial [tex]\( f(x) = x^4 + x^3 - 7x^2 - x + 6 \)[/tex] has 4 roots.
Answer: The number of roots of the polynomial is 4.
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