F.3 Fatorization(:

1) 4x³ - 39xy² + 35y³= 4x³ - 4y³ + 39y³ - 39xy²= 4(x³ - y³) + 39y²(y - x)= 4(x - y)(x² + xy + y²) - 39y²(x - y)= (x - y) (4x² + 4xy + 4y² - 39y²)= (x - y) (4x² + 4xy - 35y²)= (x - y) (2x - 5y) (2x + 7y)
2)x(y² - z²) + y(z² - x²) + z(x² - y²)= xy² - xz² + yz² - yx² + zx² - zy²= (z - y)x² + (y² - z²)x + yz² - zy²= (z - y)x² + (y - z)(y + z)x + yz(z - y)= (z - y) [x² - (y + z)x + yz]= (z - y) (x - y) (x - z)
3) x^4-3x^2+9y^4I think the question is
x^4 - 3x²y² + 9y^4= x^4 + 6x²y² + 9y^4 - 9x² y²= (x²+ 3y²)² - (3xy)²= (x² - 3xy + 3y²) (x² + 3xy + 3y²)
4)x^4 + 4= x^4 + 4x² + 4 - 4x²= (x² + 2)² - (2x)²= (x² - 2x + 2)(x² + 2x + 2)
5)x^4 - 6x²y² + y^4= x^4 - 2x²y² + y^4 - 4x²y²= (x² - y²)^2 - (2xy)²= (x² - 2xy - y²) (x² + 2xy - y²)
A) - 10p² - 11pq + 6q²= - 10p² - 15pq + 4pq + 6q² = - 5p(2p + 3q) + 2q(2p + 3q)= (2p + 3q) (2q - 5p)
B)20a² + 30ab - 10ab² = 10a (2a + 3b - b²)
C)64x^4 - 81= (8x²)² - 9²= (8x² - 9) (8x² + 9)= (2√2 x - 3) (2√2 x + 3) (8x² + 9)
D)(3/4)m² - 3= 3 [(1/4)m² - 1]= 3 (m/2 - 1) (m/2 + 1)
E)x² - 5x + 4= x² - x - 4x + 4= x(x - 1) - 4(x - 1)= (x - 1) (x - 4)

(3/4)m^2-3
(3/4)(m^2-4)
(3/4)(m-2)(m+2)
Fraction can be expressed in Factorization,but it is not a good way.

2010-11-27 12:43:09 補充：
64x^4 - 81
(8x^2-9)(8x^2+9)----------(It is sufficent)