Question

add and/or subtract, then simplify. list all domain restrictions. (\frac{4}{3x^2 - 5x + 2}-\frac{x + 2}{x^3 - x^2 + 5x - 5}=) restrictions: select an answer (y =) (y
eq) (x
eq) (x =) submit que

Explanation:

Step1: Factor the first denominator

Factor $3x^2-5x+2$:
$3x^2-5x+2=(3x-2)(x-1)$

Step2: Factor the second denominator

Factor $x^3-x^2+5x-5$ by grouping:
$x^3-x^2+5x-5=x^2(x-1)+5(x-1)=(x^2+5)(x-1)$

Step3: Find common denominator

The least common denominator is $(3x-2)(x-1)(x^2+5)$

Step4: Rewrite fractions with LCD

$$\frac{4(x^2+5)}{(3x-2)(x-1)(x^2+5)} - \frac{(x+2)(3x-2)}{(3x-2)(x-1)(x^2+5)}$$

Step5: Expand numerators

Expand $4(x^2+5)=4x^2+20$
Expand $(x+2)(3x-2)=3x^2-2x+6x-4=3x^2+4x-4$

Step6: Subtract numerators

$$\frac{(4x^2+20)-(3x^2+4x-4)}{(3x-2)(x-1)(x^2+5)}$$
$$=\frac{4x^2+20-3x^2-4x+4}{(3x-2)(x-1)(x^2+5)}$$
$$=\frac{x^2-4x+24}{(3x-2)(x-1)(x^2+5)}$$

Step7: Find domain restrictions

Denominators cannot be 0:

1. $3x-2=0 \implies x=\frac{2}{3}$
2. $x-1=0 \implies x=1$
3. $x^2+5=0$ has no real solutions, since $x^2\geq0$ so $x^2+5\geq5>0$

Answer:

Simplified expression: $\frac{x^2-4x+24}{(3x-2)(x-1)(x^2+5)}$
Domain restrictions: $x
eq \frac{2}{3}$, $x
eq 1$