Question

subtract.
\\(\dfrac{6}{2x^2 + x - 28} - \dfrac{2}{2x^2 - 9x + 7}\\)
simplify your answer as much as

Explanation:

Step1: Factor denominators

Factor $2x^2+x-28$: find two numbers that multiply to $2*(-28)=-56$ and add to 1, which are 8 and -7.
$2x^2+x-28=(2x-7)(x+4)$

Factor $2x^2-9x+7$: find two numbers that multiply to $2*7=14$ and add to -9, which are -7 and -2.
$2x^2-9x+7=(2x-7)(x-1)$

Step2: Find common denominator

The common denominator is $(2x-7)(x+4)(x-1)$

Step3: Rewrite fractions with common denominator

$\frac{6}{(2x-7)(x+4)} = \frac{6(x-1)}{(2x-7)(x+4)(x-1)}$
$\frac{2}{(2x-7)(x-1)} = \frac{2(x+4)}{(2x-7)(x+4)(x-1)}$

Step4: Subtract the fractions

$\frac{6(x-1) - 2(x+4)}{(2x-7)(x+4)(x-1)}$

Step5: Expand and simplify numerator

Expand: $6x-6-2x-8$
Combine like terms: $4x-14=2(2x-7)$

Step6: Cancel common factors

Cancel $(2x-7)$ from numerator and denominator:
$\frac{2(2x-7)}{(2x-7)(x+4)(x-1)} = \frac{2}{(x+4)(x-1)}$

Step7: Expand denominator (optional, simplified form)

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

Answer:

$\frac{2}{x^2+3x-4}$ (or $\frac{2}{(x+4)(x-1)}$)