2007-02-18 01:30:25 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
You typed the problem incorrect. It should be x^2-2xy+3y^2
First: multiply the 1st & 3rd coefficient to get "3." Find two numbers that give you "3" when multiplied & "-2" (2nd coefficient) when added/subtracted. The numbers are (-3 & 1).
Sec: rewrite the expression with the new middle coefficients.
x^2 - 3xy + xy + 3y^2
Thord: when there are 4 terms - group "like" terms & factor both sets of parenthesis.
(x^2 - 3xy) + (xy + 3y^2)
x(x - 3y) + y(x - 3y)
(x - 3y)(x + y) Or, (x + y)(x - 3y)
2007-02-18 05:04:33 · answer #1 · answered by ♪♥Annie♥♪ 6 · 1⤊ 1⤋
i assume the question is x^2-2xy-3y^2
then
=x^2 +xy -3xy -3y^2
=x(x+y) -3y(x+y)
=(x+y)(x-3y)
2007-02-18 09:37:32 · answer #2 · answered by san 3 · 1⤊ 0⤋
x² - 2xy + 3x² =
x (x-2y+3x) =
2x ( 2x - y)
2007-02-18 09:43:39 · answer #3 · answered by M. Abuhelwa 5 · 0⤊ 1⤋
hey dude first check the question properly
2007-02-18 11:44:09 · answer #4 · answered by aparna k 1 · 0⤊ 0⤋
x² - 2xy + 3x²
4x² - 2xy
2x(2x - y)
- - - - - - - - -s-
2007-02-18 10:15:23 · answer #5 · answered by SAMUEL D 7 · 0⤊ 1⤋