Factor completely:
(2x^2)+(50y^2)-20xy
How do I do this? I'm good at factoring but just not this type of factoring. Thanks!
Question about factorization...?
2008-08-15 11:05 am
回答 (5)
2008-08-15 11:16 am
✔ 最佳答案
The first thing you notice is that a factor of 2 can come out:2x² + 50y² - 20xy = 2(x² + 25y² - 10xy)
= 2(x² - 10xy + 25y²)
= 2.(x - 5y)²
At the second line you might not notice the answer...you can continue another way:
2(x² - 10xy + 25y²) = 2(x² - 5xy - 5xy + 25y²)
= 2[ (x² - 5xy) + (-5xy + 25y²) ]
= 2[ x(x - 5y) -5y(x - 5y) ]
= 2[ (x - 5y)(x - 5y) ]
= 2.(x - 5y)²
2008-08-15 6:12 pm
Factor:
2x² + 50y² - 20xy = 2(x² + 25y² - 10xy)
= 2(x² - 10xy + 25y²) = 2(x - 5y)²
2x² + 50y² - 20xy = 2(x² + 25y² - 10xy)
= 2(x² - 10xy + 25y²) = 2(x - 5y)²
2008-08-15 6:15 pm
2x^2 + 50y^2 - 20xy
= 2x^2 - 20xy + 50y^2
= 2x^2 - 10xy - 10xy + 50y^2
= (2x^2 - 10xy) - (10xy - 50y^2)
= 2x(x - 5y) - 10y(x - 5y)
= (x - 5y)(2x - 10y)
= 2(x - 5y)(x - 5y)
= 2x^2 - 20xy + 50y^2
= 2x^2 - 10xy - 10xy + 50y^2
= (2x^2 - 10xy) - (10xy - 50y^2)
= 2x(x - 5y) - 10y(x - 5y)
= (x - 5y)(2x - 10y)
= 2(x - 5y)(x - 5y)
2008-08-15 6:15 pm
2x² - 20xy + 50y²
=> 2x² - 10xy - 10xy + 50y²
=> 2x(x -5y) -10y(x - 5y)
=> (2x - 10y)(x - 5y)
Cheers.
=> 2x² - 10xy - 10xy + 50y²
=> 2x(x -5y) -10y(x - 5y)
=> (2x - 10y)(x - 5y)
Cheers.
參考: Experience
2008-08-15 6:16 pm
Only "intellec..." is correct!
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