This equation is not quadratic, because it contains a term involving x³. However, we can solve it by using factoring. First we get 0 on the right side by subtracting 17x2 from both sides. Then we factor the polynomial on the left side and use an extension of the zero-factor property.
This equation is not quadratic, because it contains a term involving x³. However, we can solve it by using factoring. First we get 0 on the right side by subtracting 17x2 from both sides. Then we factor the polynomial on the left side and use an extension of the zero-factor property.
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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I need help with all of them plz.
![## Solving Polynomial Equations by Factoring
### Example Problem:
**Solve:** \( 6x^3 + 12x = 17x^2 \)
This equation is not quadratic because it contains a term involving \( x^3 \). However, we can solve it by using factoring. First, we get 0 on the right side by subtracting \( 17x^2 \) from both sides. Then, we factor the polynomial on the left side and use an extension of the zero-factor property.
**Step-by-Step Solution:**
1. **Rewrite the equation to set one side to 0:**
\[
6x^3 + 12x - 17x^2 = 0
\]
This is achieved by subtracting \( 17x^2 \) from both sides:
\[
6x^3 + 12x - 17x^2 = 17x^2 - 17x^2 \implies 6x^3 - 17x^2 + 12x = 0
\]
2. **Factor out the greatest common factor (GCF):**
\[
x(6x^2 - 17x + 12) = 0
\]
3. **Factor the trinomial:**
\[
x(2x - 3)(3x - 4) = 0
\]
4. **Apply the zero-factor property:**
If \( x(2x - 3)(3x - 4) = 0 \), then at least one of the factors must be zero.
\[
x = 0 \quad \text{or} \quad 2x - 3 = 0 \quad \text{or} \quad 3x - 4 = 0
\]
5. **Solve each equation:**
- \( x = 0 \)
- \( 2x - 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2} \)
- \( 3x - 4 = 0 \implies 3x = 4 \implies x = \frac{4}{3} \)
**Solution Set:**
The solutions are \( 0 \), \( 3/](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9896e9f0-21a4-41f5-9cfe-a9d2b9678883%2Ffa6f3d44-597e-4309-811d-55996939c184%2Fnoqj0qk_processed.png&w=3840&q=75)
Transcribed Image Text:## Solving Polynomial Equations by Factoring
### Example Problem:
**Solve:** \( 6x^3 + 12x = 17x^2 \)
This equation is not quadratic because it contains a term involving \( x^3 \). However, we can solve it by using factoring. First, we get 0 on the right side by subtracting \( 17x^2 \) from both sides. Then, we factor the polynomial on the left side and use an extension of the zero-factor property.
**Step-by-Step Solution:**
1. **Rewrite the equation to set one side to 0:**
\[
6x^3 + 12x - 17x^2 = 0
\]
This is achieved by subtracting \( 17x^2 \) from both sides:
\[
6x^3 + 12x - 17x^2 = 17x^2 - 17x^2 \implies 6x^3 - 17x^2 + 12x = 0
\]
2. **Factor out the greatest common factor (GCF):**
\[
x(6x^2 - 17x + 12) = 0
\]
3. **Factor the trinomial:**
\[
x(2x - 3)(3x - 4) = 0
\]
4. **Apply the zero-factor property:**
If \( x(2x - 3)(3x - 4) = 0 \), then at least one of the factors must be zero.
\[
x = 0 \quad \text{or} \quad 2x - 3 = 0 \quad \text{or} \quad 3x - 4 = 0
\]
5. **Solve each equation:**
- \( x = 0 \)
- \( 2x - 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2} \)
- \( 3x - 4 = 0 \implies 3x = 4 \implies x = \frac{4}{3} \)
**Solution Set:**
The solutions are \( 0 \), \( 3/
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