Question

1. solve \\(\frac{x^2 + 2x - 24}{x^2 - x - 12} \div \frac{x^2 + x - 30}{x^2 - 9}\\). completely simplify your answer and state any restrictions on the variable.

\\(\bigcirc\\) \\(\frac{x-3}{x-5}, x \
eq -4, x \
eq -3, x \
eq 3, x \
eq -6\\)
\\(\bigcirc\\) \\(\frac{x+3}{x-5}, x \
eq -4, x \
eq -3, x \
eq 3, x \
eq -6\\)
\\(\bigcirc\\) \\(\frac{x+3}{x+5}, x \
eq -4, x \
eq -3, x \
eq 3, x \
eq -6\\)
\\(\bigcirc\\) \\(\frac{x-3}{x+5}, x \
eq -4, x \
eq -3, x \
eq 3, x \
eq -6\\)

Explanation:

🆕 New Concept Discovered: Multiplying and Dividing Rational Expressions
Flipping the divisor and finding domain restrictions

Step 1: Identify the expression and factor all polynomials

The given division problem is:

$$ \frac{x^2 + 2x - 24}{x^2 - 16} \div \frac{x^2 + 9x + 18}{x^2 + 7x + 12} $$

First, we factor every numerator and denominator completely:

• \( x^2 + 2x - 24 = (x + 6)(x - 4) \)
• \( x^2 - 16 = (x - 4)(x + 4) \)
• \( x^2 + 9x + 18 = (x + 6)(x + 3) \)
• \( x^2 + 7x + 12 = (x + 4)(x + 3) \)

Rewriting the expression with factored forms:

$$ \frac{(x + 6)(x - 4)}{(x - 4)(x + 4)} \div \frac{(x + 6)(x + 3)}{(x + 4)(x + 3)} $$

Step 2: Determine the restrictions on the variable

To find the restrictions, we must identify any values of \( x \) that would cause division by zero at any point in the calculation. This includes:

1. The denominators of the original fractions: \( (x - 4)(x + 4)

eq 0 \) and \( (x + 4)(x + 3)
eq 0 \)

1. The numerator of the divisor (since we multiply by its reciprocal, this term moves to the denominator): \( (x + 6)(x + 3)

eq 0 \)

Setting each unique factor to zero gives the restricted values:

• \( x - 4 = 0 \implies x

eq 4 \)

• \( x + 4 = 0 \implies x

eq -4 \)

• \( x + 3 = 0 \implies x

eq -3 \)

• \( x + 6 = 0 \implies x

eq -6 \)

Thus, the restrictions are:

$$ x eq 4, \quad x eq -4, \quad x eq -3, \quad x eq -6 $$

Step 3: Simplify the expression

To divide by a rational expression, multiply by its reciprocal:

$$ \frac{(x + 6)(x - 4)}{(x - 4)(x + 4)} \cdot \frac{(x + 4)(x + 3)}{(x + 6)(x + 3)} $$

Now, cancel the common factors in the numerator and denominator:

• Cancel \( (x - 4) \) from the first fraction.
• Cancel \( (x + 4) \) from the numerator and denominator.
• Cancel \( (x + 6) \) from the numerator and denominator.
• Cancel \( (x + 3) \) from the numerator and denominator.

$$ \frac{\cancel{(x + 6)}\cancel{(x - 4)}}{\cancel{(x - 4)}\cancel{(x + 4)}} \cdot \frac{\cancel{(x + 4)}\cancel{(x + 3)}}{\cancel{(x + 6)}\cancel{(x + 3)}} = 1 $$

The simplified expression is \( 1 \).

Answer:

$$ 1; \quad x eq 4, \quad x eq -4, \quad x eq -3, \quad x eq -6 $$