F2 Mathematics (Factorization)

Factorize the following algebraicexpressions.
1. 4u^2－5u－16v^2+10v
=(4u^2－16v^2)－(5u－10v)
=4(u^2－4v^2)－5(u－2v)
=4(u－2v)(u+2v)－5(u－2v)
=(u－2v)(4u+8v－5

2. a) Factorize 2a^2－12a+18=2a^2－12a+18
=2(a^2－6a+9)
=2(a－3)^2
b) Using result of part (a) , factorize 2(a^2－1)^2+18－12(a^2－1)
2(a^2－1)^2+18－12(a^2－1)
=2(a^2－1)^2－12(a^2－1)+18
=2(a^2－1－3)^2
=2(a^2－4)^2
=2(a－2)^2(a+2)^2
c) Hence , factorize 2(a^2－1)^2+28－12a^2
2(a^2－1)^2+28－12a^2
=2(a^2－1)^2 – 12(a^2－1)+18－2
=2(a－2)^2(a+2)^2－2
=2[(a－2)^2(a+2)^2－1]
=2[(a^2－4)^2－1^2]
=2(a^2－3)(a^2－5)

1. 4u^2-5u-16v^2+10v
=(4u^2-16v^2)-(5u-10v)
=(u-2v)(4u+8v-5)

2.a
=2(a^2-6a+9)
=2(a-3)^2

2(a^2-1)^2+18-12(a^2-1)
=2(a^2-1-3)^2
=2(a-2)^2(a+2)^2

2(a^2-1)^2+28-12a^2
=2[(a^2-4)^2-1^2]
=2(a^2-3)(a^2-5)

1. 4u^2-5u-16v^2+10v
=(4u^2-16v^2)-(5u-10v)
=4(u-2v)(u+2v)-5(u-2v)
=(u-2v)(4u+8v-5)

2.a
=2(a^2-6a+9)
=2(a-3)^2

2(a^2-1)^2+18-12(a^2-1)
=2(a^2-1)^2-12(a^2-1)+18
=2(a^2-1-3)^2
=2(a-2)^2(a+2)^2

2(a^2-1)^2+28-12a^2
=2(a^2-1)^2-12(a^2-1)+18-2
=2[(a^2-4)^2-1^2]
=2(a^2-3)(a^2-5)
Quick steps .......sorry

1 (4u^2-16v^2)+(10v-5u)
=4(u-2v)(u+2v)-5(u-2v)
=(u-2v)(2u+8v-5)

2a. 2a^2-12a+18
=2(a^2-6a+9)
=2(a-3)^2


2b let a^2-1 be u
2u^2+18-12u
=2(u-3)^2
=2(a^2-1-3)^2
=2(a^2-4)^2
=2(a-2)^2 *(a+2)^2

2c 2( a² - 1 )² + 28 - 12a²
= 2( a² -1 )² + 18 - 12( a² - 1 ) -2
=2(a^2-4)^2-2
=2[(a^2-4)^2-1}
=2(a^2-4-1)(a^2-4+1)
=2(a^2-5)(a^2-3)

2013-10-24 18:52:05 補充：
更改
1 (4u^2-16v^2)+(10v-5u)
=4(u-2v)(u+2v)-5(u-2v)
=(u-2v)(4u+8v-5)