Question

factoring quadratic expressions

Answer 1

\[18n^2 - 8\]

Answer 2

@phi

Answer 3

take out the common factor what is GCF(greatest common factor ) ?

Answer 4

2?

Answer 5

right take out 2 from 18n^2 -8
or in other words divide both terms by common factor \[2(\frac{ 18n^2 }{ 2 }-\frac{ 8 }{ 2 })\]

Answer 6

2(9n^2 - 4)

Answer 7

you should always be on the look out for "difference of squares"

Answer 8

@phi I don't understand...

Answer 9

9 and 4 are perfect square roots and the negative between both terms so you can apply the difference of squares \[\huge\rm a^2-b^2 =(a-b)(a+b)\]

Answer 10

so how would i implement it on 9^2 - 4?

Answer 11

this is how i do it take square roots of both terms
write in two parentheses (sqrt of 1st term `+` sqrt of 2nd term) (sqrt of 1st term `-` sqrt of 2nd term ) lolol

Answer 12

okay.. so would it be

(3+4)(3-4)?

Answer 13

what about n it's 9n^2 - 4

Answer 14

(3n+4)(3n-4)

Answer 15

looks good don't forget the common factor
2(3n+4)(3n-4)

Answer 16

ohh wait no

Answer 17

take square root of both termz