Can I get help factoring 18m^3n + 3m^2n^2 – 6mn^3?
2009-06-18 7:37 am
I was able to find the answer on the internet, but I need help understanding how to do it. If someone could go step by step and explain it to me, I would appreciate it. thanks
回答 (6)
2009-06-18 7:52 am
✔ 最佳答案
first of all take out the terms common in each termthis wud give u.
3mn (6m"2 + mn - 2n"2)
now there are two possible ways.........
simpler one:
again take n'2 common
this gives
3mn'3 ( 6 (m/n)"2 + (m/n)" - 2)
let m/n = y
then we get
3mn'3 (6y'2 + y - 2)
factorising we get
3mn'3 (3y + 2)(2y - 1)
substituting back the value of y,
we get final answer as
3mn (3m +2n)(2m - n)
2009-06-18 11:18 am
take the common factor
which is 3mn
so 3mn(6m^2+mn-2n^2) and deal with the brackets as a quadratic
3mn(3m+2n)(2m-n)
which is 3mn
so 3mn(6m^2+mn-2n^2) and deal with the brackets as a quadratic
3mn(3m+2n)(2m-n)
2009-06-18 8:15 am
18m^3n + 3m^2n^2 - 6mn^3
= 3(6m^3n + m^2n^2 - 2mn^3)
= 3mn(6m^2 + mn - 2n^2)
= 3mn(6m^2 + 4mn - 3mn - 2n^2)
= 3mn[(6m^2 + 4mn) - (3mn + 2n^2)]
= 3mn[2m(3m + 2n) - n(3m + 2n)]
= 2mn(3m + 2n)(2m - n)
= 3(6m^3n + m^2n^2 - 2mn^3)
= 3mn(6m^2 + mn - 2n^2)
= 3mn(6m^2 + 4mn - 3mn - 2n^2)
= 3mn[(6m^2 + 4mn) - (3mn + 2n^2)]
= 3mn[2m(3m + 2n) - n(3m + 2n)]
= 2mn(3m + 2n)(2m - n)
2009-06-18 7:50 am
18m^3n + 3m^2n^2 – 6mn^3
(3mn is the GCF.)So,
= 3mn(6m^2+mn-2n^2). <-----Answer.
(3mn is the GCF.)So,
= 3mn(6m^2+mn-2n^2). <-----Answer.
2009-06-18 7:45 am
find whats common in all 3 parts. i see that all coefficients are multiples of 3 and all have 1 n and 1 m
so:
(3nm)(6m^2+mn+2n^2) i hope thats right
so:
(3nm)(6m^2+mn+2n^2) i hope thats right
2009-06-18 7:44 am
3mn(6m^2+mn+2n^2)
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