Answer :
To find all the roots of the polynomial [tex]\( f(x) = -3x^3 + 19x^2 - 9x + 1 \)[/tex], we follow these steps:
1. Rational Root Theorem: This theorem suggests that any rational root, which is a fraction [tex]\( \frac{p}{q} \)[/tex], of the polynomial is such that [tex]\( p \)[/tex] is a factor of the constant term and [tex]\( q \)[/tex] is a factor of the leading coefficient. For our polynomial, the constant term is 1, and the leading coefficient is -3.
- Factors of 1: [tex]\( \pm 1 \)[/tex]
- Factors of -3: [tex]\( \pm 1, \pm 3 \)[/tex]
Thus, the possible rational roots are [tex]\( \pm 1, \pm \frac{1}{3} \)[/tex].
2. Testing Possible Rational Roots: We substitute these values into the polynomial to check if they satisfy [tex]\( f(x) = 0 \)[/tex].
- [tex]\( f(1) = -3(1)^3 + 19(1)^2 - 9(1) + 1 = -3 + 19 - 9 + 1 = 8 \)[/tex] (not zero)
- [tex]\( f(-1) = -3(-1)^3 + 19(-1)^2 - 9(-1) + 1 = 3 + 19 + 9 + 1 = 32 \)[/tex] (not zero)
- [tex]\( f(\frac{1}{3}) = -3(\frac{1}{3})^3 + 19(\frac{1}{3})^2 - 9(\frac{1}{3}) + 1 = -\frac{1}{9} + \frac{19}{9} - 3 + 1 = 0 \)[/tex]
Therefore, [tex]\( x = \frac{1}{3} \)[/tex] is a root.
3. Polynomial Division: Now that we have found one root, [tex]\( x = \frac{1}{3} \)[/tex], we can perform synthetic or polynomial division to simplify the cubic polynomial.
Divide [tex]\( -3x^3 + 19x^2 - 9x + 1 \)[/tex] by [tex]\( x - \frac{1}{3} \)[/tex]. After division, we get a quadratic polynomial.
4. Solve the Quadratic: The division gives us a quadratic, let's assume it is of the form [tex]\( ax^2 + bx + c \)[/tex]. Solve this quadratic equation using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
5. Final Roots: Combining the solution [tex]\( x = \frac{1}{3} \)[/tex] with the solutions from the quadratic equation, you will get all the roots of the polynomial.
By performing these steps, you will have found all the roots of [tex]\( f(x) \)[/tex]. Each root will be either rational or irrational, depending on the discriminant of the quadratic obtained through division.
1. Rational Root Theorem: This theorem suggests that any rational root, which is a fraction [tex]\( \frac{p}{q} \)[/tex], of the polynomial is such that [tex]\( p \)[/tex] is a factor of the constant term and [tex]\( q \)[/tex] is a factor of the leading coefficient. For our polynomial, the constant term is 1, and the leading coefficient is -3.
- Factors of 1: [tex]\( \pm 1 \)[/tex]
- Factors of -3: [tex]\( \pm 1, \pm 3 \)[/tex]
Thus, the possible rational roots are [tex]\( \pm 1, \pm \frac{1}{3} \)[/tex].
2. Testing Possible Rational Roots: We substitute these values into the polynomial to check if they satisfy [tex]\( f(x) = 0 \)[/tex].
- [tex]\( f(1) = -3(1)^3 + 19(1)^2 - 9(1) + 1 = -3 + 19 - 9 + 1 = 8 \)[/tex] (not zero)
- [tex]\( f(-1) = -3(-1)^3 + 19(-1)^2 - 9(-1) + 1 = 3 + 19 + 9 + 1 = 32 \)[/tex] (not zero)
- [tex]\( f(\frac{1}{3}) = -3(\frac{1}{3})^3 + 19(\frac{1}{3})^2 - 9(\frac{1}{3}) + 1 = -\frac{1}{9} + \frac{19}{9} - 3 + 1 = 0 \)[/tex]
Therefore, [tex]\( x = \frac{1}{3} \)[/tex] is a root.
3. Polynomial Division: Now that we have found one root, [tex]\( x = \frac{1}{3} \)[/tex], we can perform synthetic or polynomial division to simplify the cubic polynomial.
Divide [tex]\( -3x^3 + 19x^2 - 9x + 1 \)[/tex] by [tex]\( x - \frac{1}{3} \)[/tex]. After division, we get a quadratic polynomial.
4. Solve the Quadratic: The division gives us a quadratic, let's assume it is of the form [tex]\( ax^2 + bx + c \)[/tex]. Solve this quadratic equation using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
5. Final Roots: Combining the solution [tex]\( x = \frac{1}{3} \)[/tex] with the solutions from the quadratic equation, you will get all the roots of the polynomial.
By performing these steps, you will have found all the roots of [tex]\( f(x) \)[/tex]. Each root will be either rational or irrational, depending on the discriminant of the quadratic obtained through division.
