Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
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.
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
