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Divide using long division. x2 +7x + 12 14) - X + 4

Divide using long division. x2 +7x + 12 14) - X + 4

Algebra and Trigonometry (6th Edition)
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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### Polynomial Division and Factoring #### 1. Polynomial Division **Problem Statement:** Divide using long division. 14) \(\frac{x^2 + 7x + 12}{x + 4}\) **Solution:** To divide the polynomial \(x^2 + 7x + 12\) by \(x + 4\) using long division, follow these steps: - Write the divisor \(x + 4\) outside the division bracket and the dividend \(x^2 + 7x + 12\) inside the bracket. Here is the visual representation of the long division setup: ``` x + 3 ______________ x + 4 | x^2 + 7x + 12 - (x^2 + 4x) ______________ 3x + 12 - (3x + 12) ______________ 0 ``` 1. Divide the first term of the dividend \(x^2\) by the first term of the divisor \(x\) to get \(x\). 2. Multiply \(x\) by the divisor \(x + 4\) to get \(x^2 + 4x\). 3. Subtract \(x^2 + 4x\) from \(x^2 + 7x + 12\) to get \(3x + 12\). 4. Repeat with the new dividend \(3x + 12\): - Divide \(3x\) by \(x\) to get \(3\). - Multiply \(3\) by the divisor \(x + 4\) to get \(3x + 12\). - Subtract \(3x + 12\) from \(3x + 12\) to get a remainder of 0. Hence, the quotient is \(x + 3\), and the remainder is 0. #### 2. Polynomial Factoring **Problem Statement:** Factor the polynomial completely. If the polynomial is prime, so state. 15) \(x^2 + 6x + 8\) **Solution:** To factor \(x^2 + 6x + 8\): 1. Find two numbers that multiply to \(8\) (the constant term) and add to \(6\) (the coefficient of \(x\)).
Transcribed Image Text:### Polynomial Division and Factoring #### 1. Polynomial Division **Problem Statement:** Divide using long division. 14) \(\frac{x^2 + 7x + 12}{x + 4}\) **Solution:** To divide the polynomial \(x^2 + 7x + 12\) by \(x + 4\) using long division, follow these steps: - Write the divisor \(x + 4\) outside the division bracket and the dividend \(x^2 + 7x + 12\) inside the bracket. Here is the visual representation of the long division setup: ``` x + 3 ______________ x + 4 | x^2 + 7x + 12 - (x^2 + 4x) ______________ 3x + 12 - (3x + 12) ______________ 0 ``` 1. Divide the first term of the dividend \(x^2\) by the first term of the divisor \(x\) to get \(x\). 2. Multiply \(x\) by the divisor \(x + 4\) to get \(x^2 + 4x\). 3. Subtract \(x^2 + 4x\) from \(x^2 + 7x + 12\) to get \(3x + 12\). 4. Repeat with the new dividend \(3x + 12\): - Divide \(3x\) by \(x\) to get \(3\). - Multiply \(3\) by the divisor \(x + 4\) to get \(3x + 12\). - Subtract \(3x + 12\) from \(3x + 12\) to get a remainder of 0. Hence, the quotient is \(x + 3\), and the remainder is 0. #### 2. Polynomial Factoring **Problem Statement:** Factor the polynomial completely. If the polynomial is prime, so state. 15) \(x^2 + 6x + 8\) **Solution:** To factor \(x^2 + 6x + 8\): 1. Find two numbers that multiply to \(8\) (the constant term) and add to \(6\) (the coefficient of \(x\)).
### Multiplying Rational Expressions #### Problem: \[ \text{Multiply.} \] \[ 17) \quad \frac{2y}{4y+2} \cdot \frac{6y+3}{2} \] #### Explanation: To solve the above multiplication of rational expressions, follow these steps: 1. **Factor where possible:** Factor any common factors in both the numerators and denominators. - The first fraction \(\frac{2y}{4y+2}\) can be simplified. The denominator can be factored: \[ 4y + 2 = 2(2y + 1) \] Thus, the fraction becomes: \[ \frac{2y}{2(2y+1)} = \frac{y}{2y+1} \] - The second fraction \(\frac{6y+3}{2}\) can also be simplified. The numerator can be factored: \[ 6y + 3 = 3(2y + 1) \] Thus, the fraction becomes: \[ \frac{3(2y+1)}{2} \] 2. **Multiply the fractions:** \[ \left(\frac{y}{2y+1}\right) \cdot \left( \frac{3(2y+1)}{2} \right) \] 3. **Simplify:** Notice that \(2y + 1\) in the numerator of the second fraction and the denominator of the first fraction will cancel each other out: \[ \frac{y \cdot 3(2y+1)}{(2y+1) \cdot 2} = \frac{3y}{2} \] Therefore, the simplified answer to the multiplication of the given rational expressions is: \[ \frac{3y}{2} \] ### Solving the Equation #### Problem: \[ \text{Solve the equation and check your solution.} \] \[ 18) \quad \frac{1}{3x} + \frac{1}{4x} = \frac{x+9}{8} \] #### Explanation: To solve the given equation, we first obtain a common denominator for the fractions involving \(x\). We'll then solve for \(x\) through the following steps:
Transcribed Image Text:### Multiplying Rational Expressions #### Problem: \[ \text{Multiply.} \] \[ 17) \quad \frac{2y}{4y+2} \cdot \frac{6y+3}{2} \] #### Explanation: To solve the above multiplication of rational expressions, follow these steps: 1. **Factor where possible:** Factor any common factors in both the numerators and denominators. - The first fraction \(\frac{2y}{4y+2}\) can be simplified. The denominator can be factored: \[ 4y + 2 = 2(2y + 1) \] Thus, the fraction becomes: \[ \frac{2y}{2(2y+1)} = \frac{y}{2y+1} \] - The second fraction \(\frac{6y+3}{2}\) can also be simplified. The numerator can be factored: \[ 6y + 3 = 3(2y + 1) \] Thus, the fraction becomes: \[ \frac{3(2y+1)}{2} \] 2. **Multiply the fractions:** \[ \left(\frac{y}{2y+1}\right) \cdot \left( \frac{3(2y+1)}{2} \right) \] 3. **Simplify:** Notice that \(2y + 1\) in the numerator of the second fraction and the denominator of the first fraction will cancel each other out: \[ \frac{y \cdot 3(2y+1)}{(2y+1) \cdot 2} = \frac{3y}{2} \] Therefore, the simplified answer to the multiplication of the given rational expressions is: \[ \frac{3y}{2} \] ### Solving the Equation #### Problem: \[ \text{Solve the equation and check your solution.} \] \[ 18) \quad \frac{1}{3x} + \frac{1}{4x} = \frac{x+9}{8} \] #### Explanation: To solve the given equation, we first obtain a common denominator for the fractions involving \(x\). We'll then solve for \(x\) through the following steps:
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