Question

can someone explain this process of simplifying in more detail. I'm so confused

Answer 1

especially from the first to second step .. thanks

Answer 2

@hartnn

Answer 3

its just factoring out common terms

Answer 4

but how though

Answer 5

how did the (3x+4)^5 and (3x+4)^4 become (3x+4)^4 and (3x+4)^1

Answer 6

from the first step to the second

Answer 7

\(\Large a^5 +a^4 = a^4 (a+1)\)

Answer 8

thats what is done with 3x+4 terms

a=3x+4

Answer 9

hmm let me think about it for a it and ill get back to u :P

Answer 10

sure :)

Answer 11

so if it was (3x+4)^5 and (3x+4)^3.. how do we factor out the like term

Answer 12

we would have factored OUT (3x+4)^3

so, (3x+4)^5 would have become (3x+4)^2

Answer 13

and (3x+4)^3 would have become 1

(3x+4)^3 [(3x+5)^2 + 1]

Answer 14

(3x+4)^2 i mean***

Answer 15

sorry I'm confused :P

Answer 16

\(\large ☻☺+☻♥= ☻(☺+♥) \\ \text{this is factoring, } \\ \text{☻ was common from both terms and factored out.}\)

Answer 17

you have something like
\(a (3x+4)^5 + b (3x+4)^4\)
right??
a,b are some functions of x which we are not concerned with now..

your doubts is how do we get to here :
\(a (3x+4)^5 + b (3x+4)^4 = (3x+4)^4 [a (3x+4) +b]\)
right ?

Answer 18

yes

Answer 19

\(\Large (3x+4)^5 = (3x+4)^4 \times (3x+4)\)

Answer 20

oh yea because the if u add the exponents 4+1 it is equal to 5

Answer 21

YES, thats correct.

\(a (3x+4)^5 + b (3x+4)^4 = a (3x+4)^4 (3x+4) + b (3x+4)^4\)

n0w (3x+4)^4 is common!
and can be factored out

Answer 22

oh ok and same goes with (x^3-x+2) in the same question i gave

Answer 23

yes, correct

Answer 24

(x^3-x+2)^2=(x^3-x+2)^1+(x^3-x+2)^1

Answer 25

\((x^3-x+2)^2=(x^3-x+2)^1\times (x^3-x+2)^1\)

Answer 26

\(a^2 = a \times a\)
not a+a

Answer 27

ph yes sorry ... thanks a lot

Answer 28

oh*

Answer 29

ask if any more doubts :)
welcome ^_^