Question

name
period
qc multiplying rational functions

1. \\(\frac{6x^2 + 30x}{2x^2 + 17x + 21} \bullet \frac{x^2 + 2x + 35}{x^2 - 25}\\)
2. \\(\frac{3x^4y}{xy^3} \bullet \frac{3x^7y^6}{27x^2y^2}\\)

Explanation:

Problem 1: Multiplying Rational Functions

Let's tackle the first rational function multiplication: $\boldsymbol{\frac{6x^2 + 30x}{2x^2 + 17x + 21} \cdot \frac{x^2 + 2x + 35}{x^2 - 25}}$ (Note: Wait, maybe a typo? Let's check factoring first.)

Step 1: Factor Numerators and Denominators

• First Numerator: $6x^2 + 30x = 6x(x + 5)$ (factor out $6x$)
• First Denominator: $2x^2 + 17x + 21$ (factor quadratic: $2x^2 + 14x + 3x + 21 = 2x(x + 7) + 3(x + 7) = (2x + 3)(x + 7)$)
• Second Numerator: $x^2 + 2x + 35$ (Wait, discriminant: $2^2 - 4(1)(35) = 4 - 140 = -136 < 0$—cannot factor over reals. Maybe a typo? Maybe $x^2 + 2x - 35$? Let's assume $x^2 + 2x - 35$ (since $x^2 -25$ is $(x-5)(x+5)$). If so: $x^2 + 2x - 35 = (x + 7)(x - 5)$ (check: $7 \cdot (-5) = -35$, $7 + (-5) = 2$).
• Second Denominator: $x^2 - 25 = (x - 5)(x + 5)$ (difference of squares)

Step 2: Substitute Factored Forms

If we correct the second numerator to $x^2 + 2x - 35$, we get:
$$\frac{6x(x + 5)}{(2x + 3)(x + 7)} \cdot \frac{(x + 7)(x - 5)}{(x - 5)(x + 5)}$$

Step 3: Cancel Common Factors

• Cancel $(x + 5)$ (numerator 1 and denominator 2)
• Cancel $(x + 7)$ (denominator 1 and numerator 2)
• Cancel $(x - 5)$ (numerator 2 and denominator 2)

Left with: $\frac{6x}{2x + 3}$

Problem 2: Multiplying Rational Functions

Now the second problem: $\boldsymbol{\frac{3x^4 y}{x y^3} \cdot \frac{3x^7 y^6}{27x^2 y^2}}$

Step 1: Multiply Numerators and Denominators Separately

Multiply numerators: $3x^4 y \cdot 3x^7 y^6 = 9x^{4+7} y^{1+6} = 9x^{11} y^7$
Multiply denominators: $x y^3 \cdot 27x^2 y^2 = 27x^{1+2} y^{3+2} = 27x^3 y^5$

Step 2: Simplify the Fraction

$$\frac{9x^{11} y^7}{27x^3 y^5} = \frac{9}{27} \cdot x^{11-3} \cdot y^{7-5} = \frac{1}{3} x^8 y^2$$

Final Answers

1. (With corrected second numerator) $\boldsymbol{\frac{6x}{2x + 3}}$
2. $\boldsymbol{\frac{x^8 y^2}{3}}$

(Note: If the second numerator in Problem 1 is indeed $x^2 + 2x + 35$ (unfactorable), the problem may have a typo. But assuming a common factoring scenario, the above is the solution.)

Answer:

Problem 1: Multiplying Rational Functions

Let's tackle the first rational function multiplication: $\boldsymbol{\frac{6x^2 + 30x}{2x^2 + 17x + 21} \cdot \frac{x^2 + 2x + 35}{x^2 - 25}}$ (Note: Wait, maybe a typo? Let's check factoring first.)

Step 1: Factor Numerators and Denominators

• First Numerator: $6x^2 + 30x = 6x(x + 5)$ (factor out $6x$)
• First Denominator: $2x^2 + 17x + 21$ (factor quadratic: $2x^2 + 14x + 3x + 21 = 2x(x + 7) + 3(x + 7) = (2x + 3)(x + 7)$)
• Second Numerator: $x^2 + 2x + 35$ (Wait, discriminant: $2^2 - 4(1)(35) = 4 - 140 = -136 < 0$—cannot factor over reals. Maybe a typo? Maybe $x^2 + 2x - 35$? Let's assume $x^2 + 2x - 35$ (since $x^2 -25$ is $(x-5)(x+5)$). If so: $x^2 + 2x - 35 = (x + 7)(x - 5)$ (check: $7 \cdot (-5) = -35$, $7 + (-5) = 2$).
• Second Denominator: $x^2 - 25 = (x - 5)(x + 5)$ (difference of squares)

Step 2: Substitute Factored Forms

If we correct the second numerator to $x^2 + 2x - 35$, we get:
$$\frac{6x(x + 5)}{(2x + 3)(x + 7)} \cdot \frac{(x + 7)(x - 5)}{(x - 5)(x + 5)}$$

Step 3: Cancel Common Factors

• Cancel $(x + 5)$ (numerator 1 and denominator 2)
• Cancel $(x + 7)$ (denominator 1 and numerator 2)
• Cancel $(x - 5)$ (numerator 2 and denominator 2)

Left with: $\frac{6x}{2x + 3}$

Problem 2: Multiplying Rational Functions

Now the second problem: $\boldsymbol{\frac{3x^4 y}{x y^3} \cdot \frac{3x^7 y^6}{27x^2 y^2}}$

Step 1: Multiply Numerators and Denominators Separately

Multiply numerators: $3x^4 y \cdot 3x^7 y^6 = 9x^{4+7} y^{1+6} = 9x^{11} y^7$
Multiply denominators: $x y^3 \cdot 27x^2 y^2 = 27x^{1+2} y^{3+2} = 27x^3 y^5$

Step 2: Simplify the Fraction

$$\frac{9x^{11} y^7}{27x^3 y^5} = \frac{9}{27} \cdot x^{11-3} \cdot y^{7-5} = \frac{1}{3} x^8 y^2$$

Final Answers

1. (With corrected second numerator) $\boldsymbol{\frac{6x}{2x + 3}}$
2. $\boldsymbol{\frac{x^8 y^2}{3}}$

(Note: If the second numerator in Problem 1 is indeed $x^2 + 2x + 35$ (unfactorable), the problem may have a typo. But assuming a common factoring scenario, the above is the solution.)