2007-01-26 08:54:51 · 8 answers · asked by chris 2 in Science & Mathematics ➔ Mathematics
First: multiply the 1st & 3rd coefficient to get "30." Find two numbers that give you "30" when multiplied and "11" (2nd/middle coefficient) when added/subtracted. The numbers are: 5 and 6
Sec: rewrite the expression with the new middle coefficients...
x^2 + 5xy + 6xy + 30y^2
Third: when you have 4 terms, group "like" terms & factor...
(x^2 + 5xy) + (6xy + 30y^2)
x(x+5y) + 6y(x+5y)
(x+5y)(x+6y)
2007-01-26 11:21:28 · answer #1 · answered by ♪♥Annie♥♪ 6 · 0⤊ 0⤋
x^2 + 11xy + 30y^2
(x+5y)(x+6y)
To check your answer do expand back out your problem...
x^2 + 6xy + 5xy +30y^2....which is the exact as your original problem when you add the like (xy) terms together.
x^2 + 11xy + 30y^2
2007-01-26 17:16:01 · answer #2 · answered by CC 2 · 0⤊ 0⤋
x² + 11xy + 30y² = (x + 5y)(x + 6y).
2007-01-26 17:03:40 · answer #3 · answered by S. B. 6 · 0⤊ 0⤋
Answer is (x+5y)(x+6y)
2007-01-26 17:00:53 · answer #4 · answered by Anonymous · 1⤊ 0⤋
(x+6y)(x+5y)
2007-01-26 17:01:36 · answer #5 · answered by bruinfan 7 · 0⤊ 0⤋
i have no idea
2007-01-26 17:09:24 · answer #6 · answered by heatmizer323 1 · 0⤊ 1⤋
(x-6y)(x-5y)
2007-01-26 17:08:04 · answer #7 · answered by Anonymous · 0⤊ 1⤋
(x-6y)(x-5y)
2007-01-26 17:00:23 · answer #8 · answered by Dave aka Spider Monkey 7 · 0⤊ 2⤋