Solve polynomial for x?? (Pre-Cal)?

Question

http://www2.murrieta.k12.ca.us/vmhs/staff/mfitzpatrick/xc/xcprob1.jpg

There's the link to it.... please help!

Answer 1

simplify the complex fraction at begining by multiplying all parts by LCD = (x-1)x-3)(x+1)
I will show the result after cancelling the denominator with matching factor

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

in the numerator you have a common (x-3) and in the denominator a common (x+1) . Factoring those out:
[(x+3)(3(x-3) + 2(x+1))]/[(x+1)(2(x-3) - 1(x-1))]

[(x+3)(5x -1)] / [(x+1)(x - 5)]

multiplying this by the reciprocal of the second part of equation, the (5x - 1) factors will cancel, leaving you with

[(x+3)(x^3 - 9x^2 + 15x + 25)] /[x^2-4x-5] = 3

simplifying the numerator and multiplying denom over to other side

x^4 -12x^3 + 42x^2 -20x - 75 = 3x^2 - 12x - 15

x^4 - 12x^3 +39x^2 -8x - 60 = 0

using synthetic division, I divided out 2 and found a zero with the result of
x^3 - 10x^2 +19x + 30

dividing this by 6, I got a zero remainder with quotient now of

x^2 - 4x - 5
which factors as (x - 5)(x + 1)
so final two answers are 5 and -1

answers are 2,6,5,-1

but cannot use the 5 or -1 as they make original denominators equal 0

Answer 2

Steps
1. rewrite the numerator in the first term as
(5x - 1) / [(x + 1) ( x- 1)]

2. rewrite the denominator of the first terms as
(x-5) / [(x-1) (x-3)]

3. the first term reduces to
[ (5x - 1) (x-3) ] / [ (x + 1) (x - 5)]

4. multiple this by the reciprocal of the second term
the cubic part is (x+1) (x -5) (x-5)

5. the expresion will reduce to

(x - 3) (x - 5) = 3

x^2 - 8x + 15 = 3
x^2 - 8x + 12 = 0
(x - 6) (x - 2) = 0

x = 2 or 6

Answer 3

I tried this when you posted before (or maybe it was someone else) and got nowhere. In fact, I tried dividing that polynomial by all the factors of the problem and, alas, no luck. Sorry! :{

Answer 4

x=2
Did it by inspection.