help!! math!!?

Question

Reduce the following expression to lowest terms:

w^2 - 7w - 8
-----------------
3w^2 + 27w + 24

Answer 1

(w - 8) (w + 1)
------------------
3(w² + 9w + 8)

(w - 8) (w + 1)
---------------------
3(w + 8) (w + 1)

(w - 8)
--------------
3(w + 8)

Answer 2

Numerator is w^2 - 7w - 8 and can be factored as (w + 1) (w - 8)

Denominator is 3w^2 + 27w + 24 and can be written as

3(w^2 + 9w + 8) and can be further factorised as 3 (w + 1) (w + 8)

So, writing the factors in the numerator and denominator, we get

(w + 1) (w - 8)
-------------------
3(w + 1) (w + 8)

= (w - 8) / 3(w + 8)

Answer 3

Numerator = w^2-7w-8 = w^2 - 8w + w - 8 = w(w-8) +1 (w-8)
= (w-8)(w+1)

Denominator = 3w^2 +27w +24 = 3w^2 +3w +24w +24
=3w(w+1) +24 (w+1) = (w+1) (3w+24) = 3(w+1) (w+8)

Hence the fraction is

(w-8) * (w+1)
-------------------
3(w+1) * (w+8)

(w-8)
= ----------
3(w+8)

= (w-8)/[3*(w+8)]

Answer 4

factor up and down

(w^2 - 7w -8) / (3w^2 + 27w + 24)

[ (w-8)(w+1) ] / [ 3(w^2 + 9w + 8) ]


[ (w-8)(w+1) ] / [ 3 (w+8) (w+1) ], note you can cancel w+1, so


[ (w-8)] / [ 3 (w+8)], done.

Answer 5

(w^2-7w-8)/(3w^2+27w+24)
or (w^2+w-8w-8)/3(w^2+w+8w+8)
or {w(w+1)-8(w+1)}/3{w(w+1)+8(w+1)}
or (w+1)(w-8)/3(w+1)((w+8)
or (w-8)/3(w+8) ans

Answer 6

w^2 - 7w - 8
-----------------
3w^2 + 27w + 24

=w(w-7)-8
-------------
w(3w+27)+24

=(w-7)-8
-------------
(3w+27)+24

(w-7)-8=(3w+27)+24

where w is -7.4

Answer 7

(w-7)(w+1)/3(w+1)(w+8) = (w-7)/3(w+8)

Answer 8

You first need to factorize both top & bottom:

(w + 1) (w - 8)
-------------------
(w + 1) (w + 8)

Then, notice that (w + 1) occurs in both the numerator & denominator, and so cancels:

(w - 8)
---------
(w + 8)

Answer 9

I well decrease the terms by derative the experession so,
w ^ 2 - 7 w - 8
w - 7
1
---------------------------------------------------
3 w ^ 2 + 27 w + 24
6 w + 27
6