Question

Factor this pellet :|
3(a-4)^5 (2a +3)^-2 -1/2(a-4)^4 (2a+3)^-1

Answer 1

Is this your expression?\[\frac{3(a-4)^5 (2a +3)^{-2} -1}{2(a-4)^4 (2a+3)^{-1}}\]

Answer 2

\[3(a-4)^{5}(2a+3)^{-2}-1/2(a-4)^{4}(2a+3)^{-1}\]

Answer 3

Thats my expression

Answer 4

Remember\[x^{-a}=\frac{1}{x^a}\]and \[x^ax^b=x^{a+b}\]

Answer 5

this is the textbook step that I don't get:

Answer 6

factor out (a-4)^4 and (2a+3)^-1

Answer 7

\[1/2(a-4)^{4}(2a+3)^{-2} [6(a-4) -(2a+3)]\]

Answer 8

ohh ok

Answer 9

did you understand my instructions? pfenn1 thought it was another type of problem

Answer 10

yeah but shouldn't it be (2a+3)^-1 that gets factored out

Answer 11

ok lol you said that but in the book, it has

Answer 12

(2a+3)^-2 is factored out :S

Answer 13

follow the book, I forget if it is -1 or -2

Answer 14

but it should be -1 because then: -1 - -2 = 1 which shut leave you with (2a+3)

Answer 15

but they got (2a+3) by factoring out (2a+3)^-2

Answer 16

from (2a+3)^1 = (2a +3)

Answer 17

but -1 -(-2) =1 not -2 - (-1)

Answer 18

I see you are getting this.

Answer 19

ok tell me what the book said and I will work it out for you

Answer 20

ok give me a sec, I'll type it out which will take me a few mins:

Answer 21

\[The question itss 1/2(a−4)4(2a+3)−2[6(a−4)−(2a+3)] Thats their first step \]

Answer 22

ops wait something went wrong

Answer 23

Ok so the question is 3(a−4)5(2a+3)−2−1/2(a−4)4(2a+3)−1

Answer 24

Their first step is 1/2(a−4)4(2a+3)−2[6(a−4)−(2a+3)]

Answer 25

then they just expand and isimmpify which I get

Answer 26

crap those are supposed to be exponents

Answer 27

the 4 and the -2 are exponents

Answer 28

klj lol thesis doesn't let me backspace for some reason..