For all x > 0, the expression \frac{6x^2 + 13x - 5}{2x + 5} - \frac{3x^2 - 4x}{x} is equiv… — SAT

For all $x > 0$ , the expression $\frac{6x^2 + 13x - 5}{2x + 5} - \frac{3x^2 - 4x}{x}$ is equivalent to a constant $c$ . What is the value of $c$ ?

Cevap: 3

The correct answer is 3.

To find the constant value of the expression, both rational expressions are first simplified. The quadratic in the numerator of the first fraction,

6x^2 + 13x - 5

, can be factored as

(3x - 1)(2x + 5)

. Canceling the common factor of

2x + 5

in the numerator and denominator simplifies the first term to

3x - 1

. The second fraction,

\frac{3x^2 - 4x}{x}

, can be simplified by dividing each term in the numerator by

x

, which results in

3x - 4

. Subtracting the second simplified expression from the first gives

(3x - 1) - (3x - 4) = 3x - 1 - 3x + 4 = 3

. Thus, the constant value of the expression is

3

Factor the numerator of the first rational expression.

6x^2 + 13x - 5 = (3x - 1)(2x + 5)

This allows us to simplify the fraction by canceling common factors.

Simplify the first fraction by canceling the common term

(2x + 5)

\frac{(3x - 1)(2x + 5)}{2x + 5} = 3x - 1

Since

x > 0

2x + 5 \neq 0

, so we can cancel the common factor.

Simplify the second fraction by factoring out

x

and canceling it.

\frac{3x^2 - 4x}{x} = 3x - 4

Since

x > 0

x \neq 0

, so we can divide each term in the numerator by

x

Subtract the two simplified expressions.

(3x - 1) - (3x - 4) = 3

To find the constant value

c

, we subtract the second simplified expression from the first and combine like terms, paying close attention to distributing the negative sign.

Simplifying rational expressions by factoring and performing polynomial subtraction.

Soru