Question

find the quotient 3m2−27n2/3m+9n

Answer 1

can someone help ill give you a medel

Answer 2

factor the numerator and denominator

here is the 1st step

\[\frac{3(m^2 - 9n^2)}{3(m + 3n)}\]

remove the common factor...

next the numerator is the difference of 2 squares and can be factored

a^2 - b^2 = (a - b)(a + b)

then remove the common factor you identify

Answer 3

i dont get it

Answer 4

@campbell_st

Answer 5

well when you cancel the 3

its

\[\frac{m^2 - 9n^2}{m + 3n}\]

factor the quadratic in the numerator

Answer 6

i did all that but my answer was wrong @campbell_st

Answer 7

so what's you answer...?

Answer 8

it was 7a +8b @campbell_st

Answer 9

wow.... thats funny given this question is m and n

Answer 10

7m+8n i was typing to fast

Answer 11

ok...

so here is a question...

can you factor

a^2 - b^2

Answer 12

no i cant its new and everytime i tried to learn this we end up leaving class @campbell_st

Answer 13

ok... this quadratic is the difference of 2 squares...

and can be factored as

a^2 - b^2 = (a - b)(a + b)

does that make sense..?

Answer 14

3-27 then 3+9 @campbell_st

Answer 15

nope... you are factoring

m^2 - 9n^2

Answer 16

\[m^2 - 9n^2 = (m - ?)(m + ?)\]

what's the value of ?

so that ? x ? = 9n^2

Answer 17

i just need the answer to past @campbell_st

Answer 18

my grade

Answer 19

whats the square root of 9...?

Answer 20

3

Answer 21

yes... so you problem is

\[\frac{3m^2 - 27n^2}{3m + 9m} = \frac{3(m - 3n)(m + 3n)}{3(m + 3n)}\]

cancel the common factors and you'll have your answer

Answer 22

i dont know how @campbell_st

Answer 23

outside the brackets... whats the same in the top and the bottom...?

Answer 24

3mn