Question

add the following fractions. simplify your answer as much as possible.\\(\frac{4z^2 + 15z + 42}{z^2 + 3z - 40} - \frac{3z^2 + 2z + 2}{z^2 + 3z - 40}\\)\
answer

Explanation:

Step1: Subtract numerators (same denominator)

Since the denominators are the same ($z^2 + 3z - 40$), we subtract the numerators:
$(4z^2 + 15z + 42) - (3z^2 + 2z + 2)$

Step2: Simplify the numerator

Distribute the negative sign and combine like terms:
$4z^2 + 15z + 42 - 3z^2 - 2z - 2 = (4z^2 - 3z^2) + (15z - 2z) + (42 - 2) = z^2 + 13z + 40$

Step3: Factor numerator and denominator

Factor the numerator $z^2 + 13z + 40$:
$z^2 + 13z + 40 = (z + 5)(z + 8)$

Factor the denominator $z^2 + 3z - 40$:
$z^2 + 3z - 40 = (z + 8)(z - 5)$

Step4: Cancel common factors

The fraction becomes $\frac{(z + 5)(z + 8)}{(z + 8)(z - 5)}$. Cancel the common factor $(z + 8)$ (note $z
eq -8$ to avoid division by zero):
$\frac{z + 5}{z - 5}$

Answer:

$\frac{z + 5}{z - 5}$ (with the restriction $z
eq -8, 5$)