VIPSolutionsProvide answer7:31 NRB81%a {) 25 app.outlierai/screening?id=68a5cc2 + H J’ Outlier —_— & 6m 54s remainin Determine the real and complex roots of the polynomial x® = 3? +.x = 3 = 0 using any appropriate method. Clearly explain each step of your process and provide the complete set of solutions. © ® 0 & 1 1 vr rt 1 11 1 “111 vr 1 L111 “1 1 111111 238 1 1 1 1 1 1 EEG 1#1 , English(India) . J 9 [1 O Vv
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Provide answer
7:31 NRB81%a
{) 25 app.outlierai/screening?id=68a5cc2 + H
J’ Outlier —_— & 6m 54s remainin
Determine the real and complex roots of the polynomial x® = 3? +.x = 3 = 0 using any appropriate method. Clearly
explain each step of your process and provide the complete set of solutions.
© ® 0 &
1 1 vr rt 1 11 1
“111 vr 1 L111
“1 1 111111
238 1 1 1 1 1 1 EEG
1#1 , English(India) . J
9 [1 O Vv
This is AI generated Response.
Asked by: Gurubindra Singh
Created at: 2025-08-20 19:31:57
Asked by: Gurubindra Singh
Created at: 2025-08-20 19:31:57
Final Answer:
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Let's solve the equation step by step:
Given polynomial:
\[ x^3 - 3x^2 + x - 3 = 0 \]
### Step 1: Try Rational Root Theorem
Possible rational roots are ±1, ±3.
#### Test x = 1:
\[
(1)^3 - 3(1)^2 + 1 - 3 = 1 - 3 + 1 - 3 = -4 \neq 0
\]
#### Test x = -1:
\[
(-1)^3 - 3(-1)^2 + (-1) - 3 = -1 - 3(1) - 1 - 3 = -1 - 3 - 1 - 3 = -8 \neq 0
\]
#### Test x = 3:
\[
(3)^3 - 3(3)^2 + 3 - 3 = 27 - 27 + 3 - 3 = 0
\]
So **x = 3 is a root**.
### Step 2: Polynomial Division (Divide by x - 3)
Divide \( x^3 - 3x^2 + x - 3 \) by \( x - 3 \):
Using synthetic division with root 3:
Coefficients: 1, -3, 1, -3
Carry down 1:
3 × 1 = 3 → -3+3=0
3 × 0 = 0 → 1+0=1
3 × 1 = 3 → -3+3=0
So, quotient is \( x^2 + 1 \).
### Step 3: Solve Quadratic \( x^2 + 1 = 0 \)
\[
x^2 + 1 = 0 \implies x^2 = -1 \implies x = i, x = -i
\]
### **Final Answer:**
**The roots of the polynomial are:**
\[
\boxed{x = 3, \quad x = i, \quad x = -i}
\]
---
**Summary of steps:**
1. Used Rational Root Theorem to find \( x = 3 \) is a real root.
2. Divided the polynomial by \( x-3 \) to get \( x^2 + 1 \).
3. Solved \( x^2 + 1 = 0 \) for complex roots \( x = i \) and \( x = -i \).
4. The complete set of solutions: \( \boxed{3,\,i,\,-i} \).
Question:
Provide answer without latex
This is AI generated Response.
Asked by: Gurubindra Singh
Created at: 2025-08-20 19:32:23
Asked by: Gurubindra Singh
Created at: 2025-08-20 19:32:23
Final Answer:
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Let's solve the polynomial step by step:
Given polynomial:
x^3 - 3x^2 + x - 3 = 0
Step 1: Use Rational Root Theorem to find possible rational roots.
Possible roots are ±1 and ±3.
Test x = 1:
1 - 3 + 1 - 3 = -4 (not zero)
Test x = -1:
-1 - 3(1) - 1 - 3 = -1 -3 -1 -3 = -8 (not zero)
Test x = 3:
27 - 3(9) + 3 - 3 = 27 - 27 + 3 - 3 = 0
Since x = 3 is a root, we can divide the polynomial by (x - 3).
Step 2: Polynomial division of x^3 - 3x^2 + x - 3 by (x - 3):
Using synthetic division with root 3:
Coefficients: 1, -3, 1, -3
Bring down 1:
- Multiply 3 * 1 = 3; add to -3 → 0
- Multiply 3 * 0 = 0; add to 1 → 1
- Multiply 3 * 1 = 3; add to -3 → 0
Resulting quotient: x^2 + 1
Step 3: Solve quadratic x^2 + 1 = 0:
x^2 = -1
x = i, x = -i
Final Roots:
x = 3, i, -i
Summary:
The roots of the polynomial are 3, i, and -i.
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