An algebraic expression raised to different powers might appear in different terms. Factor out this expression raised to the lowest power.
Practice
1.
8(x + 2)3 + 5(x
+ 2)2 =
2.
– 4(x + 16)4 + 9(x
+ 16)2 + x + 16 =
3.
(15xy – 1)(2x – 1)3 –
8(2x – 1)2 =
Solutions
1.
8(x + 2)3 + 5(x
+ 2)2 = [8(x + 2)] (x + 2)2 + 5(x
+ 2)2
= [8(x
+ 2) + 5] (x + 2)2
= (8x
+ 16 + 5) (x + 2)2
= (8x
+ 21) (x + 2)2
2.
– 4(x + 16)4 + 9(x
+ 16)2 + x + 16 =
= [– 4(x
+ 16)3] (x + 16) + 9(x
+ 16) (x + 16) + 1(x
+ 16)
= [– 4(x + 16)3
+ 9(x + 16) + 1] (x + 16)
= [– 4(x + 16)3
+ 9x + 144 + 1] (x + 16)
= – 4(x + 16)3
+ 9x + 145 + 1] (x + 16)
3.
(15xy – 1)(2x – 1)3 –
8(2x – 1)2 =
= (15xy – 1)
(2x – 1) (2x – 1)3 – 8(2x – 1)2
= (15xy – 1)
(2x – 1) – 8] (2x – 1)2
Sumber
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