Some questions

2008-07-05 8:48 pm
1. Factorize the following expressions. Steps are necessary.

a) pm - mn - 4p - 4n

b) b + a^3 - a + ba^2

c) (a - b)^2 - 6(a - b) + 9

d) c^2 - (16a^2 - 40ab + 25b^2)

e) a^2 - 4y^2 + 20y - 25

f) x^4 + x^2y^2 + y^4

g) (a - 5b)^2 - (5b + 1)^2

h) 3(4x + y)^2 - 48(x - 2y)^2

i) (100/x^2y^2) - (64/a^2b^2)

回答 (1)

2008-07-06 12:52 am
✔ 最佳答案
a)
The expression cannot be factorized.

If one of the (-) signs is changed to (+) sign, it can be factorized.
For example,
pm - mn - 4p + 4n
= m(p - n) - 4(p - n)
= (m - 4)(p - n)


b)
The expression cannot be factorized.

If one of the (+) signs is changed to (-) sign, it can be factorized.
For example,
b + a3 - a - ba2
= b - a + a3 - ba2
= -(a - b) + a2(a - b)
= (a2 - 1)(a - b)
= (a + 1)(a - 1)(a - b)


c) (a - b)2 - 6(a - b) + 9

= [(a - b) - 3]2

= (a - b - 3)2


d) c2 - (16a2 - 40ab + 25b2)

= c2 - [(4a)2 - 2(4a)(5b) + (5b)2]

= c2 - (4a - 5b)2

= [c + (4a - 5b)][c - (4a - 5b)]

= (c + 4a - 5b)(c - 4a + 5b)


e) a2 - 4y2 + 20y - 25

= a2 - [(2y)2 - 2(2y)(5) + (5)2]

= a2 - (2y - 5)2

= [a + (2y - 5)][a- (2y - 5)]

= (a + 2y - 5)(a - 2y + 5)


f) x4 + x2y2 + y4

= x4 + 2x2y2 + y4 - x2y2

= [(x2)2 + 2(x2)(y2) + (y2)2] - (xy)2

= (x2 + y2)2 - (xy)2

= [(x2 + y2) + xy][(x2 + y2) - xy]

= (x2 + xy + y2)(x2 - xy + y2)


g) (a - 5b)2 - (5b + 1)2

= [(a - 5b) + (5b + 1)][(a - 5b) - (5b + 1)]

= [a - 5b + 5b + 1][a - 5b - 5b - 1]

= (a + 1)(a - 10b -1)


h) 3(4x + y)2 - 48(x - 2y)2

= 3[(4x + y)2 - 16(x - 2y)2]

= 3[(4x + y)2 - 42(x - 2y)2]

= 3[(4x + y) + 4(x - 2y)][(4x + y) - 4(x - 2y)]

= 3[4x + y + 4x - 8y][4x + y - 4x + 8y]

= 3(8x - 7y)(9y)

= 27y(8x - 7y)
or 33y(8x - 7y)


i) (100/x2y2) - (64/a2b2)

= 4[(25/x2y2) - (16/a2b2)]

= 4[(5/xy)2 - (4/ab)2]

= 4[(5/xy) + (4/ab)][(5/xy) - (4/ab)]
or 4(5ab + 4xy)(5ab - 4xy)/x2y2a2b2
=


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