For all x > 3, which of the following expressions is equivalent to \frac{x^2 - 9}{x^2 - 2x… — SAT

For all $x > 3$ , which of the following expressions is equivalent to $\frac{x^2 - 9}{x^2 - 2x - 3} \cdot \frac{x + 1}{x}$ ?

\frac{x+3}{x}

Factoring the numerator of the first fraction as

(x-3)(x+3)

and its denominator as

(x-3)(x+1)

allows the common factor

(x-3)

to cancel out, leaving

\frac{x+3}{x+1}

. Multiplying this result by the second fraction,

\frac{x+1}{x}

, permits the cancellation of the common factor

(x+1)

, which simplifies the entire expression to the equivalent form

\frac{x+3}{x}

Factor the numerator and the denominator of the first rational expression:

\frac{x^2 - 9}{x^2 - 2x - 3}

The numerator factors as

(x - 3)(x + 3)

using the difference of squares identity. The denominator factors as

(x - 3)(x + 1)

by finding two numbers that multiply to

-3

and add to

-2

Factoring polynomials is necessary to identify and cancel common factors.

Simplify the first fraction by canceling the common factor

(x-3)

from both the numerator and denominator.

The first fraction simplifies to

\frac{x+3}{x+1}

Since

x > 3

, the term

x-3

is non-zero, making division by it valid.

Multiply the simplified first fraction by the second fraction:

\frac{x+3}{x+1} \cdot \frac{x+1}{x}

The common factor

(x+1)

in the numerator and denominator cancels out, resulting in

\frac{x+3}{x}

Canceling the common factor

(x+1)

yields the simplest equivalent expression.

Simplifying rational expressions by factoring polynomials and canceling common factors.

Soru