Answer :
To list all possible rational zeros for the given polynomials using the Rational Zero Test, we can follow these steps:
### Step 1: Identify the coefficients
We need to identify the constant term and the leading coefficient for each polynomial.
#### For [tex]\( f(x) = 2x^4 - 17x^3 + 35x^2 + 9x - 45 \)[/tex]:
- Leading coefficient (a): 2
- Constant term (b): -45
#### For [tex]\( f(x) = 4x^5 - 8x^4 - 5x^3 + 10x^2 + x - 2 \)[/tex]:
- Leading coefficient (a): 4
- Constant term (b): -2
### Step 2: Find the factors
Next, we find the factors of the constant term and the leading coefficient.
#### For [tex]\( f(x) = 2x^4 - 17x^3 + 35x^2 + 9x - 45 \)[/tex]:
- Factors of the constant term (-45): [tex]\(\pm 1, \pm 3, \pm 5, \pm 9, \pm 15, \pm 45\)[/tex]
- Factors of the leading coefficient (2): [tex]\(\pm 1, \pm 2\)[/tex]
#### For [tex]\( f(x) = 4x^5 - 8x^4 - 5x^3 + 10x^2 + x - 2 \)[/tex]:
- Factors of the constant term (-2): [tex]\(\pm 1, \pm 2\)[/tex]
- Factors of the leading coefficient (4): [tex]\(\pm 1, \pm 2, \pm 4\)[/tex]
### Step 3: List all possible rational zeros
The possible rational zeros can be found by taking all possible positive and negative combinations of the factors of the constant term divided by the factors of the leading coefficient.
#### For [tex]\( f(x) = 2x^4 - 17x^3 + 35x^2 + 9x - 45 \)[/tex]:
Possible rational zeros are:
[tex]\[ \pm \frac{1}{1}, \pm \frac{3}{1}, \pm \frac{5}{1}, \pm \frac{9}{1}, \pm \frac{15}{1}, \pm \frac{45}{1}, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{5}{2}, \pm \frac{9}{2}, \pm \frac{15}{2}, \pm \frac{45}{2} \][/tex]
This reduces to:
[tex]\[ \{-45.0, -22.5, -9.0, -5.0, -4.5, -3.0, -2.5, -1.5, -1.0, -0.5, 0.5, 1.0, 1.5, 2.5, 3.0, 4.5, 5.0, 9.0, 22.5, 45.0\} \][/tex]
#### For [tex]\( f(x) = 4x^5 - 8x^4 - 5x^3 + 10x^2 + x - 2 \)[/tex]:
Possible rational zeros are:
[tex]\[ \pm \frac{1}{1}, \pm \frac{2}{1}, \pm \frac{1}{2}, \pm \frac{2}{2}, \pm \frac{1}{4}, \pm \frac{2}{4} \][/tex]
This reduces to:
[tex]\[ \{-2.0, -1.0, -0.5, -0.25, 0.25, 0.5, 1.0, 2.0\} \][/tex]
These are the possible rational zeros for each polynomial based on the Rational Zero Test.
### Step 1: Identify the coefficients
We need to identify the constant term and the leading coefficient for each polynomial.
#### For [tex]\( f(x) = 2x^4 - 17x^3 + 35x^2 + 9x - 45 \)[/tex]:
- Leading coefficient (a): 2
- Constant term (b): -45
#### For [tex]\( f(x) = 4x^5 - 8x^4 - 5x^3 + 10x^2 + x - 2 \)[/tex]:
- Leading coefficient (a): 4
- Constant term (b): -2
### Step 2: Find the factors
Next, we find the factors of the constant term and the leading coefficient.
#### For [tex]\( f(x) = 2x^4 - 17x^3 + 35x^2 + 9x - 45 \)[/tex]:
- Factors of the constant term (-45): [tex]\(\pm 1, \pm 3, \pm 5, \pm 9, \pm 15, \pm 45\)[/tex]
- Factors of the leading coefficient (2): [tex]\(\pm 1, \pm 2\)[/tex]
#### For [tex]\( f(x) = 4x^5 - 8x^4 - 5x^3 + 10x^2 + x - 2 \)[/tex]:
- Factors of the constant term (-2): [tex]\(\pm 1, \pm 2\)[/tex]
- Factors of the leading coefficient (4): [tex]\(\pm 1, \pm 2, \pm 4\)[/tex]
### Step 3: List all possible rational zeros
The possible rational zeros can be found by taking all possible positive and negative combinations of the factors of the constant term divided by the factors of the leading coefficient.
#### For [tex]\( f(x) = 2x^4 - 17x^3 + 35x^2 + 9x - 45 \)[/tex]:
Possible rational zeros are:
[tex]\[ \pm \frac{1}{1}, \pm \frac{3}{1}, \pm \frac{5}{1}, \pm \frac{9}{1}, \pm \frac{15}{1}, \pm \frac{45}{1}, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{5}{2}, \pm \frac{9}{2}, \pm \frac{15}{2}, \pm \frac{45}{2} \][/tex]
This reduces to:
[tex]\[ \{-45.0, -22.5, -9.0, -5.0, -4.5, -3.0, -2.5, -1.5, -1.0, -0.5, 0.5, 1.0, 1.5, 2.5, 3.0, 4.5, 5.0, 9.0, 22.5, 45.0\} \][/tex]
#### For [tex]\( f(x) = 4x^5 - 8x^4 - 5x^3 + 10x^2 + x - 2 \)[/tex]:
Possible rational zeros are:
[tex]\[ \pm \frac{1}{1}, \pm \frac{2}{1}, \pm \frac{1}{2}, \pm \frac{2}{2}, \pm \frac{1}{4}, \pm \frac{2}{4} \][/tex]
This reduces to:
[tex]\[ \{-2.0, -1.0, -0.5, -0.25, 0.25, 0.5, 1.0, 2.0\} \][/tex]
These are the possible rational zeros for each polynomial based on the Rational Zero Test.
