Answer :
To find all the zeros and their multiplicities of the function [tex]\( f(x) = x^5 + x^4 - 19x^3 + 17x^2 + 48x - 60 \)[/tex], we can follow these steps:
### Step 1: Apply the Rational Zero Theorem
The Rational Zero Theorem states that any rational zero of the polynomial is of the form [tex]\( \frac{p}{q} \)[/tex], where [tex]\( p \)[/tex] is a factor of the constant term, and [tex]\( q \)[/tex] is a factor of the leading coefficient.
- Constant term: [tex]\(-60\)[/tex]
- Leading coefficient: [tex]\(1\)[/tex]
Factors of [tex]\(-60\)[/tex]: [tex]\(\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 10, \pm 12, \pm 15, \pm 20, \pm 30, \pm 60\)[/tex]
Factors of 1: [tex]\(\pm 1\)[/tex]
Possible rational zeros: [tex]\( \pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 10, \pm 12, \pm 15, \pm 20, \pm 30, \pm 60 \)[/tex]
### Step 2: Use a Graphing Calculator
Graph the function to identify any visible rational zeros. This step helps narrow down the potential candidates for rational zeros to test.
### Step 3: Test Potential Rational Zeros with Synthetic Division
We'll test some of these potential zeros using synthetic division to see which ones result in a remainder of 0.
1. Test [tex]\( x = 2 \)[/tex]:
- Using synthetic division, when you divide [tex]\( f(x) \)[/tex] by [tex]\( x - 2 \)[/tex], you find that the remainder is 0. Thus, [tex]\( x = 2 \)[/tex] is a zero.
2. Divide [tex]\( f(x) \)[/tex] by [tex]\( x - 2 \)[/tex]:
- After dividing, the quotient is [tex]\( x^4 + 3x^3 - 13x^2 - 9x + 30 \)[/tex].
3. Test [tex]\( x = -2 \)[/tex] on the new polynomial:
- Similarly, [tex]\( x = -2 \)[/tex] is found to be a zero.
4. Divide [tex]\( x^4 + 3x^3 - 13x^2 - 9x + 30 \)[/tex] by [tex]\( x + 2 \)[/tex]:
- After dividing, the quotient is [tex]\( x^3 + x^2 - 15x + 15 \)[/tex].
Continuing with synthetic division for the smaller polynomial:
5. Test [tex]\( x = 3 \)[/tex] on this cubic polynomial:
- Performing synthetic division again shows [tex]\( x = 3 \)[/tex] is a zero.
6. Divide [tex]\( x^3 + x^2 - 15x + 15 \)[/tex] by [tex]\( x - 3 \)[/tex]:
- The quotient is [tex]\( x^2 + 4x - 5 \)[/tex].
### Step 4: Solve the Remaining Quadratic
For [tex]\( x^2 + 4x - 5 \)[/tex], use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], [tex]\( c = -5 \)[/tex]:
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{16 + 20}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{36}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm 6}{2} \][/tex]
The solutions are [tex]\( x = 1 \)[/tex] and [tex]\( x = -5 \)[/tex].
### Final Zeros and Their Multiplicities
- [tex]\( x = 2 \)[/tex] (Multiplicity 1)
- [tex]\( x = -2 \)[/tex] (Multiplicity 1)
- [tex]\( x = 3 \)[/tex] (Multiplicity 1)
- [tex]\( x = 1 \)[/tex] (Multiplicity 1)
- [tex]\( x = -5 \)[/tex] (Multiplicity 1)
Thus, the zeros of the polynomial [tex]\( f(x) = x^5 + x^4 - 19x^3 + 17x^2 + 48x - 60 \)[/tex] are [tex]\( x = 2, -2, 3, 1, \)[/tex] and [tex]\(-5\)[/tex], each with multiplicity 1.
### Step 1: Apply the Rational Zero Theorem
The Rational Zero Theorem states that any rational zero of the polynomial is of the form [tex]\( \frac{p}{q} \)[/tex], where [tex]\( p \)[/tex] is a factor of the constant term, and [tex]\( q \)[/tex] is a factor of the leading coefficient.
- Constant term: [tex]\(-60\)[/tex]
- Leading coefficient: [tex]\(1\)[/tex]
Factors of [tex]\(-60\)[/tex]: [tex]\(\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 10, \pm 12, \pm 15, \pm 20, \pm 30, \pm 60\)[/tex]
Factors of 1: [tex]\(\pm 1\)[/tex]
Possible rational zeros: [tex]\( \pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 10, \pm 12, \pm 15, \pm 20, \pm 30, \pm 60 \)[/tex]
### Step 2: Use a Graphing Calculator
Graph the function to identify any visible rational zeros. This step helps narrow down the potential candidates for rational zeros to test.
### Step 3: Test Potential Rational Zeros with Synthetic Division
We'll test some of these potential zeros using synthetic division to see which ones result in a remainder of 0.
1. Test [tex]\( x = 2 \)[/tex]:
- Using synthetic division, when you divide [tex]\( f(x) \)[/tex] by [tex]\( x - 2 \)[/tex], you find that the remainder is 0. Thus, [tex]\( x = 2 \)[/tex] is a zero.
2. Divide [tex]\( f(x) \)[/tex] by [tex]\( x - 2 \)[/tex]:
- After dividing, the quotient is [tex]\( x^4 + 3x^3 - 13x^2 - 9x + 30 \)[/tex].
3. Test [tex]\( x = -2 \)[/tex] on the new polynomial:
- Similarly, [tex]\( x = -2 \)[/tex] is found to be a zero.
4. Divide [tex]\( x^4 + 3x^3 - 13x^2 - 9x + 30 \)[/tex] by [tex]\( x + 2 \)[/tex]:
- After dividing, the quotient is [tex]\( x^3 + x^2 - 15x + 15 \)[/tex].
Continuing with synthetic division for the smaller polynomial:
5. Test [tex]\( x = 3 \)[/tex] on this cubic polynomial:
- Performing synthetic division again shows [tex]\( x = 3 \)[/tex] is a zero.
6. Divide [tex]\( x^3 + x^2 - 15x + 15 \)[/tex] by [tex]\( x - 3 \)[/tex]:
- The quotient is [tex]\( x^2 + 4x - 5 \)[/tex].
### Step 4: Solve the Remaining Quadratic
For [tex]\( x^2 + 4x - 5 \)[/tex], use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], [tex]\( c = -5 \)[/tex]:
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{16 + 20}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{36}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm 6}{2} \][/tex]
The solutions are [tex]\( x = 1 \)[/tex] and [tex]\( x = -5 \)[/tex].
### Final Zeros and Their Multiplicities
- [tex]\( x = 2 \)[/tex] (Multiplicity 1)
- [tex]\( x = -2 \)[/tex] (Multiplicity 1)
- [tex]\( x = 3 \)[/tex] (Multiplicity 1)
- [tex]\( x = 1 \)[/tex] (Multiplicity 1)
- [tex]\( x = -5 \)[/tex] (Multiplicity 1)
Thus, the zeros of the polynomial [tex]\( f(x) = x^5 + x^4 - 19x^3 + 17x^2 + 48x - 60 \)[/tex] are [tex]\( x = 2, -2, 3, 1, \)[/tex] and [tex]\(-5\)[/tex], each with multiplicity 1.
