Factor the difference of squares / Reverse Foil help?

2015-09-01 8:35 am

(1/9)y² - (1/3)yx + (1/4)x²

Can I Divide by their LCD (1/36) ?
4y² - 12yx + 8x² = 0 ?
Can I reduce to
4(y² - 3yx + 2x²) = 0
Then I get confused. Is there another method besides reverse foil ?

更新1:

+ 2x², = addition. Factors of 1 = 1*1 = 1 Factors of 2 = 2*1 = 2 We want -3yx 1 xy + 2 xy DOES NOT = -3xy. Do I only need 3yx, not -3yx ?

更新2:

9x² + 30xy + 25y² = 0 I get (3x + 5y)(3x + 5y). Fractions are killing me.

回答 (3)

胡雪8°

2015-09-01 9:00 am

In the first step may be multiply the expression by 36/36 (i.e 1)

The expression
= (1/9)y² - (1/3)yx + (1/4)x²
= (36/36) [(1/9)y² - (1/3)yx + (1/4)x²]
= (1/36)(4y² - 12yx + 9x²)
= (1/36)[(2y)² - 2(2y)(3x) + (3x)²]

(2y)² - 2(2y)(3x) + (3x)² is in the form of a² - 2ab + b² which is equal to (a - b)²,
where a = 2y and b = 3x

Then, the expression :
= (1/36)[(2y)² - 2(2y)(3x) + (3x)²]
= (1/36)(2y - 3x)²

Alternatively, the expression
= (1/9)y² - (1/3)yx + (1/4)x²
= (y²/9) - (yx/3) + (x²/4)
= (y/3)² - 2(y/3)(x/2) (x/2)²
= [(y/3) - (x/2)]²

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[匿名]

2016-12-15 4:27 pm

Solving Difference Of Squares

參考: https://shrinkurl.im/a9zfW

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[匿名]

2015-09-01 9:38 am

(1/9)y² - (1/3)yx + (1/4)x²
=(1/36)(4y^2-12yx+9x^2)
=(1/36)(2y-3x)(2y-3x)

You can only equate to 0 if you are solving an equation.
Factoring the difference of squares is a little different.

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收錄日期: 2021-05-02 11:16:35
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