URGENT: Algebra Help 1/z^2-3z-4 + 3z/z^2-6z-7?

Question

(1/z^2-3z-4) + (3z/z^2-6z-7)

Add Express in simplest form:

It reads 1 OVER z^2-3z-4 PLUS 3z OVER z^2-6z-7

It either:
a. 3z^2-11z-7/(z+1)(z-4)(z-7)

OR.

b. 2z^2-4z+1/(z-1)(z+4)(z+7)

Answer 1

(1/(z^2 - 3z - 4)) + ((3z)/(z^2 - 6z - 7))
(1/((z - 4)(z + 1))) + ((3z)/((z - 7)(z + 1)))

Multiply everything by (z - 4)(z + 1)(z - 7)

((z - 7) + (3z(z - 4)))/((z - 4)(z + 1)(z - 7))

(z - 7 + 3z^2 - 12z)/((z - 4)(z + 1)(z - 7))

(3z^2 - 11z - 7)/((z - 4)(z + 1)(z - 7))

ANS : a.) (3z^2 - 11z - 7)/((z - 4)(z + 1)(z - 7))

Answer 2

If you are looking for details:

Factoring the denominators, the expression becomes

1/[(z - 4)(z + 1)] + 3z/[(z - 7)(z + 1)]

so you can see that the common denominator is

(z - 4)(z + 1)(z - 7)

To combine the fractions you multiply the left fraction by 1 in the form of (z - 7)/(z - 7) and multiply the right fraction by 1 in the form of (z - 4)/(z - 4). When we multiply by 1 we do not change the actual value of the fraction.

If you do the above and simplify the resulting numerator you get

(2z^2 - 11z - 7)/D

where D is the common denominator developed above.

Answer 3

x ( x + 2 ) - 2 ( x + 2 ) so take out the ( )s first = x ^ 2 (x to the second one potential) + 2x - 2x - 4 (its -4 because once you distribute its easily -2 * 2) + 2x and - 2x cancel out = x ^ 2 - 4 wish this helped! XD

Answer 4

first one (A)

Answer 5

i got the first answer, a.

Answer 6

Do your own homework.