Question

find the excluded value pls...medal

Answer 1

(3x^2+5x-8)/(6x+16)*(4x^2+x)/(3x^2-3x)

Answer 2

@Australopithecus @jdoe0001 @pooja195 @TheSmartOne @Awolflover1

Answer 3

2/(x^2-1)+x/(x^2-2x+1)

Answer 4

@Astrophysics

Answer 5

let me rewrite that \(\Large \frac{(3x^2+5x-8)}{(6x+16)} \times \frac{(4x^2+x)}{(3x^2-3x)}\)

Answer 6

that's what you have?

Answer 7

for excluded values you look at what makes the denominator equal to zero

Answer 8

but before doing so we would like to simplify first factoring if possible

Answer 9

the answer was (3x^33-8)(4x+1)/6(3x+8)(x-1) @xapproachesinfinity

Answer 10

can we factor \(3x^2+5x-8\)

Answer 11

answer for what?

Answer 12

yes we can factor that @xapproachesinfinity

Answer 13

so what are the factors

Answer 14

(3x+8)(x-1) @xapproachesinfinity

Answer 15

yes!

Answer 16

so we have \(\Large \frac{(3x+8)(x-1)}{(6x+16)}\times \frac{(4x+1)}{3(x-1)}\)
\(\Large \frac{(3x+8)(4x+1)}{3(6x+16)}\)

Answer 17

so what make denominator now equal to 0

Answer 18

6x+16=0 ===> x=-16/6
x=-8/3

Answer 19

excluded value -8/3

Answer 20

ok thanks can you help with the secondequation @xapproachesinfinity

Answer 21

actually i have to go for some time
i fasting i need to go eat :)