Answer :
Sure! Let's find the roots of the polynomial [tex]\( f(x) = x^5 - 3x^4 + 8x^3 - 8x^2 + 7x - 5 \)[/tex] given that some of the roots include 1 and [tex]\( i \)[/tex].
### Step-by-Step Solution:
1. Given roots:
- [tex]\( x = 1 \)[/tex]
- [tex]\( x = i \)[/tex]
Since polynomial coefficients are real numbers, the complex conjugate of any non-real roots must also be a root. Therefore, [tex]\( x = -i \)[/tex] is also a root.
2. Polynomial and its Roots:
- We know the given polynomial has [tex]\( x = 1 \)[/tex], [tex]\( x = i \)[/tex], and [tex]\( x = -i \)[/tex] as roots.
3. Construct the corresponding factors:
- For [tex]\( x = 1 \)[/tex]: [tex]\( (x - 1) \)[/tex]
- For [tex]\( x = i \)[/tex] and [tex]\( x = -i \)[/tex]: The corresponding factor is [tex]\( (x - i)(x + i) = x^2 + 1 \)[/tex]
4. Form the partially factored polynomial:
- Combine the factors we have: [tex]\( (x - 1)(x^2 + 1) \)[/tex]
5. Perform polynomial division:
To find the remaining polynomial after factoring out [tex]\( (x - 1)(x^2 + 1) \)[/tex] from [tex]\( x^5 - 3x^4 + 8x^3 - 8x^2 + 7x - 5 \)[/tex], we need to perform polynomial division on [tex]\( f(x) = x^5 - 3x^4 + 8x^3 - 8x^2 + 7x - 5 \)[/tex]:
- First, divide [tex]\( f(x) \)[/tex] by [tex]\( (x - 1) \)[/tex]:
[tex]\[ f(x) \div (x - 1) \][/tex]
This will give us a quotient polynomial.
Let's denote this quotient by [tex]\( Q_1(x) \)[/tex]:
[tex]\[ Q_1(x) = x^4 - 2x^3 + 6x^2 - 2x + 5 \][/tex]
- Next, divide the obtained quotient [tex]\( Q_1(x) \)[/tex] by [tex]\( (x^2 + 1) \)[/tex]:
[tex]\[ Q_1(x) \div (x^2 + 1) \][/tex]
This will give us another quotient polynomial.
Let's denote the new quotient by [tex]\( Q_2(x) \)[/tex]:
[tex]\[ Q_2(x) = x^2 - 2x + 5 \][/tex]
6. Solving the remaining polynomial [tex]\( Q_2(x) = x^2 - 2x + 5 \)[/tex]:
[tex]\[ x^2 - 2x + 5 = 0 \][/tex]
Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i \][/tex]
The remaining roots are [tex]\( x = 1 + 2i \)[/tex] and [tex]\( x = 1 - 2i \)[/tex].
### In summary:
The roots of the polynomial [tex]\( x^5 - 3x^4 + 8x^3 - 8x^2 + 7x - 5 \)[/tex] are:
1. [tex]\( x = 1 \)[/tex]
2. [tex]\( x = i \)[/tex]
3. [tex]\( x = -i \)[/tex]
4. [tex]\( x = 1 + 2i \)[/tex]
5. [tex]\( x = 1 - 2i \)[/tex]
### Step-by-Step Solution:
1. Given roots:
- [tex]\( x = 1 \)[/tex]
- [tex]\( x = i \)[/tex]
Since polynomial coefficients are real numbers, the complex conjugate of any non-real roots must also be a root. Therefore, [tex]\( x = -i \)[/tex] is also a root.
2. Polynomial and its Roots:
- We know the given polynomial has [tex]\( x = 1 \)[/tex], [tex]\( x = i \)[/tex], and [tex]\( x = -i \)[/tex] as roots.
3. Construct the corresponding factors:
- For [tex]\( x = 1 \)[/tex]: [tex]\( (x - 1) \)[/tex]
- For [tex]\( x = i \)[/tex] and [tex]\( x = -i \)[/tex]: The corresponding factor is [tex]\( (x - i)(x + i) = x^2 + 1 \)[/tex]
4. Form the partially factored polynomial:
- Combine the factors we have: [tex]\( (x - 1)(x^2 + 1) \)[/tex]
5. Perform polynomial division:
To find the remaining polynomial after factoring out [tex]\( (x - 1)(x^2 + 1) \)[/tex] from [tex]\( x^5 - 3x^4 + 8x^3 - 8x^2 + 7x - 5 \)[/tex], we need to perform polynomial division on [tex]\( f(x) = x^5 - 3x^4 + 8x^3 - 8x^2 + 7x - 5 \)[/tex]:
- First, divide [tex]\( f(x) \)[/tex] by [tex]\( (x - 1) \)[/tex]:
[tex]\[ f(x) \div (x - 1) \][/tex]
This will give us a quotient polynomial.
Let's denote this quotient by [tex]\( Q_1(x) \)[/tex]:
[tex]\[ Q_1(x) = x^4 - 2x^3 + 6x^2 - 2x + 5 \][/tex]
- Next, divide the obtained quotient [tex]\( Q_1(x) \)[/tex] by [tex]\( (x^2 + 1) \)[/tex]:
[tex]\[ Q_1(x) \div (x^2 + 1) \][/tex]
This will give us another quotient polynomial.
Let's denote the new quotient by [tex]\( Q_2(x) \)[/tex]:
[tex]\[ Q_2(x) = x^2 - 2x + 5 \][/tex]
6. Solving the remaining polynomial [tex]\( Q_2(x) = x^2 - 2x + 5 \)[/tex]:
[tex]\[ x^2 - 2x + 5 = 0 \][/tex]
Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i \][/tex]
The remaining roots are [tex]\( x = 1 + 2i \)[/tex] and [tex]\( x = 1 - 2i \)[/tex].
### In summary:
The roots of the polynomial [tex]\( x^5 - 3x^4 + 8x^3 - 8x^2 + 7x - 5 \)[/tex] are:
1. [tex]\( x = 1 \)[/tex]
2. [tex]\( x = i \)[/tex]
3. [tex]\( x = -i \)[/tex]
4. [tex]\( x = 1 + 2i \)[/tex]
5. [tex]\( x = 1 - 2i \)[/tex]
