Question

1. multiply \\(\frac{x^2+7x+12}{x^2-9} \cdot \frac{x^2-4x+3}{x^2+8x+16}\\). state any restrictions on the variables.

\\(\bigcirc\\) \\(\frac{x-1}{x+4}\\); the variable restrictions are \\(x \
eq -4, x \
eq -3, x \
eq 3\\)
\\(\bigcirc\\) \\(\frac{x+1}{x-4}\\); the variable restrictions are \\(x \
eq -4, x \
eq -3, x \
eq 3\\)
\\(\bigcirc\\) \\(\frac{x-1}{x-4}\\); the variable restrictions are \\(x \
eq -4, x \
eq -3, x \
eq 3\\)
\\(\bigcirc\\) \\(\frac{x+1}{x+4}\\); the variable restrictions are \\(x \
eq -4, x \
eq -3, x \
eq 3\\)

Explanation:

🆕 New Concept Discovered: Multiplying Rational Expressions
Factoring polynomials to simplify products and find domain restrictions.

Step 1: Factor all numerators and denominators

To multiply rational expressions, we first factor each polynomial completely to identify common factors and restrictions.

The given expression is:

$$ \frac{x^2 + 7x + 12}{x^2 - 9} \cdot \frac{x^2 - 4x + 3}{x^2 + 8x + 16} $$

Let's factor each part:

• First numerator: \( x^2 + 7x + 12 \)

We look for two numbers that multiply to \( 12 \) and add to \( 7 \). These are \( 3 \) and \( 4 \).

$$ x^2 + 7x + 12 = (x + 3)(x + 4) $$

• First denominator: \( x^2 - 9 \)

This is a difference of squares: \( a^2 - b^2 = (a - b)(a + b) \).

$$ x^2 - 9 = (x - 3)(x + 3) $$

• Second numerator: \( x^2 - 4x + 3 \)

We look for two numbers that multiply to \( 3 \) and add to \( -4 \). These are \( -1 \) and \( -3 \).

$$ x^2 - 4x + 3 = (x - 1)(x - 3) $$

• Second denominator: \( x^2 + 8x + 16 \)

This is a perfect square trinomial: \( (a + b)^2 = a^2 + 2ab + b^2 \).

$$ x^2 + 8x + 16 = (x + 4)^2 = (x + 4)(x + 4) $$

Step 2: Identify the variable restrictions

Restrictions on a variable exist because we cannot divide by zero. We find these by setting every factor in the original denominators equal to zero before simplifying.

The denominators are:

$$ (x - 3)(x + 3) \quad \text{and} \quad (x + 4)(x + 4) $$

Setting each unique factor to zero:

• \( x - 3 = 0 \implies x

eq 3 \)

• \( x + 3 = 0 \implies x

eq -3 \)

• \( x + 4 = 0 \implies x

eq -4 \)

Thus, the restrictions are:

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

Step 3: Simplify the expression

Now, write the multiplied expression with all factored terms and cancel out common factors that appear in both the numerator and the denominator:

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

Combine into a single fraction:

$$ \frac{(x + 3)(x + 4)(x - 1)(x - 3)}{(x - 3)(x + 3)(x + 4)(x + 4)} $$

Cancel the common factors:

• Cancel \( (x + 3) \) from the top and bottom.
• Cancel \( (x - 3) \) from the top and bottom.
• Cancel one factor of \( (x + 4) \) from the top and bottom.

This leaves:

$$ \frac{x - 1}{x + 4} $$

Answer:

$$ \frac{x - 1}{x + 4}; \text{ the variable restrictions are } x eq -4, z eq -3, z eq 3 $$

*(Note: The options in the image contain a small typographical error where the variable \( x \) is written as \( z \) in the restrictions list: \( z
eq -4, z
eq -3, z
eq 3 \). The correct option is the one showing the simplified expression \( \frac{x-1}{x+4} \).)*