Simplify complex expressions...?

Question

[(x^2-9x+20) / (2x^2 - 7x -15)] * [(2x^2 - x - 6) / (-x^2 + 2x + 8)]

and

(12 - x - x^2) / (5x^2 +30x +40)

and

[(x-4) / (2x^2 - 3x - 2)] - [(3) / (2x +1)]

Answer 1

[(x^2-9x+20) / (2x^2 - 7x -15)] * [(2x^2 - x - 6) / (-x^2 + 2x + 8)]

=[(x-5)(x-4)/(2x+3)(x-5)] * [(2x+3)(x-2)/(-x-2)(x-4)]

=[(x-4)/(2x+3)] * [(2x+3)(x-2)/(-x-2)(x-4)]

=(x-2)/(-x-2)

= (2-x)/(x+2)
--------------------------------------------------------------------

(12 - x - x^2) / (5x^2 +30x +40)

=[(4+x)(3-x)] / [5(x^2 + 6x + 8)]

=[(4+x)(3-x)] / [5(x+4)(x+2)]

= (3-x)/[5(x+2)]

Answer 2

Factor by FOIL if possible, and/or by Distributive property; cancel if multiplying and get common denominators if adding

(12 - x - x^2) / (5x^2 +30x +40)

(4 + x)(3 - x)/5(x^2 + 6x + 8)

(4+x)(3-x)/5(x+4)(x+2)

Since x+4 = 4+x, they cancel, assuming x is not -4

(3-x) / 5(x+2) is the answer


[(x-4) / (2x^2 - 3x - 2)] - [(3) / (2x +1)]
becomes [(x-4) / (2x+1)(x-2) ] - [3 / (2x + 1)]

The LCD is (2x+1)(x-2) so multiply top & bottom of the second fraction by the missing factor (x-2)

[(x-4) / (2x+1)(x-2) ] - [3(x-2) / (2x + 1)(x-2)]

only subtract the tops: (x - 4) - 3(x - 2) = x - 4 - 3x + 6 = -2x + 2

= 2(-x + 1) over the LCD of (2x+1)(x-2)

Answer 3

[(x^2 - 9x + 20) / (2x^2 - 7x -15)] * [(2x^2 - x - 6) / (-x^2 + 2x + 8)]
- [(x - 5)(x - 4) / (2x + 3)(x - 5)] * [(2x + 3)(x - 2) / (x + 2)(x - 4)]
- (x - 2) / (x + 2)

and

(12 - x - x^2) / (5x^2 + 30x + 40)
- (x^2 + x - 12 ) / [5((x^2 + 6x + 8)]
- (x + 4)(x - 3) / [5((x + 2)(x + 4)]
- (x - 3) / [5((x + 2)]

and

[(x - 4) / (2x^2 - 3x - 2)] - [(3) / (2x +1)]
[(x - 4) / (2x + 1)(x - 2)] - [(3) / (2x +1)]
[(x - 4) / (2x + 1)(x - 2)] - [(3)(x - 2) / (2x +1)(x - 2)]
(x - 4 - 3x + 6) / (x - 2)
(2 - 2x) / (x - 2)
2(1 - x) / (x - 2)