Can someone factor this and explain to me how it's done please??
12x^2 + 32xy - 35y^2
Algebra 2 help?
2008-07-29 5:59 pm
回答 (6)
2008-07-29 6:27 pm
✔ 最佳答案
One way is to treat it as a quadratic in 'x'.12x^2 + 32xy - 35y^2 = 0
-35 X 12 = -70 X 6 = 42 X -10.
Since 42 + (-10) = 32, required middle nomial is split
12x^2 + 42xy - 10xy + 35y^2 = 0
6x(2x + 7y) - 5y(2x + 7y) = 0
(6x -5y)(2x + 7y) = 0
2008-07-30 1:22 am
The following are the pairs of factors of 12 1*12 2*6 3*4
The following are the pairs of factors of 35 1*35 5*7
With the equation
(a*x + b*y) * (c*x - d*y) (- because the -35 is negative)
we know that b * d = 35
a * c = 12
and a*d - b*y = 32
We look for the proper pairings
It takes time and arithmetic, but their aren't that many choices
The winner is:
(2x + 7y) * (6x - 5y) (my fourth guess)
The following are the pairs of factors of 35 1*35 5*7
With the equation
(a*x + b*y) * (c*x - d*y) (- because the -35 is negative)
we know that b * d = 35
a * c = 12
and a*d - b*y = 32
We look for the proper pairings
It takes time and arithmetic, but their aren't that many choices
The winner is:
(2x + 7y) * (6x - 5y) (my fourth guess)
2008-07-30 1:15 am
First, look for a GCF. There is none in this problem. Then, factor. I use the guess & check method.
The answer is
(6x-5y)(2x+7y)
The answer is
(6x-5y)(2x+7y)
2008-07-30 1:11 am
12XY((X/Y)+8/3-(35/12)(Y/X))
2008-07-30 1:06 am
(6x -5y)(2x+7y)
2008-07-30 1:06 am
12x^2 + 32xy - 35y^2
= 12x^2 + 42xy - 10xy - 35y^2
= (12x^2 + 42xy) - (10xy + 35y^2)
= 6x(2x + 7y) - 5y(2x + 7y)
= (2x + 7y)(6x - 5y)
= 12x^2 + 42xy - 10xy - 35y^2
= (12x^2 + 42xy) - (10xy + 35y^2)
= 6x(2x + 7y) - 5y(2x + 7y)
= (2x + 7y)(6x - 5y)
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