For all x > 2, which of the following is equivalent to the expression \frac{6x^3 - 7x^2 - … — SAT

For all $x > 2$ , which of the following is equivalent to the expression $\frac{6x^3 - 7x^2 - 16x + 12}{2x^2 - 3x - 2}$ ?

The correct answer is the expression that simplifies to

3x + 1 - \frac{7}{2x + 1}

The correct answer is found by factoring the quadratic denominator into

(2x + 1)(x - 2)

. Testing the root

x = 2

in the numerator reveals that

(x - 2)

is also a factor of the numerator, allowing the expression to be rewritten as

(x - 2)(6x^2 + 5x - 6)

. Canceling the common factor

(x - 2)

leaves

\frac{6x^2 + 5x - 6}{2x + 1}

. Performing polynomial long division on this remaining term yields a quotient of

3x + 1

and a remainder of

-7

, which is expressed as the correct answer.

Factor the denominator of the rational expression.

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

Identifying the factors of the denominator helps reveal potential common factors in the numerator.

Factor the numerator

6x^3 - 7x^2 - 16x + 12

by testing

x = 2

as a possible root.

Since

6(2)^3 - 7(2)^2 - 16(2) + 12 = 48 - 28 - 32 + 12 = 0

(x - 2)

is a factor. Dividing the cubic polynomial by

(x - 2)

yields

6x^3 - 7x^2 - 16x + 12 = (x - 2)(6x^2 + 5x - 6)

Finding the common linear factor allows us to simplify the rational expression.

Simplify the rational expression by canceling the common factor

(x - 2)

For

x > 2

\frac{(x - 2)(6x^2 + 5x - 6)}{(x - 2)(2x + 1)} = \frac{6x^2 + 5x - 6}{2x + 1}

Since

x > 2

x - 2 \neq 0

, which mathematically permits dividing out the common factor.

Perform polynomial long division on

\frac{6x^2 + 5x - 6}{2x + 1}

6x^2 + 5x - 6 = (3x + 1)(2x + 1) - 7

Dividing the quadratic expression by the linear expression separates the polynomial into a quotient and a rational remainder.

Rewrite the final expression with the quotient and remainder.

3x + 1 - \frac{7}{2x + 1}

Expressing the result in the standard quotient-remainder form matches the target expression.

Simplification and division of rational algebraic expressions

Question