6)
7)
8)
9)
10)
2. จงแยกตัวประกอบของพหุ นามต่อไปนี้
1)
x2 – 25x
2)
x3 – 4x2
3)
x4 – 4x
4)
15x2 – 25x
5)
81x2 – x
6)
7x2 + 49x
7)
88x3 + 8 x2
8)
13x4 + x2
9)
5x3 + 15x2
10) 100x4 + 10x3
11) x2 + 3x – 4
12) x2 + 10x + 25
13) x2 + 6x + 9
14) x2 + 4x + 4
15) x2 + 8x – 20
16) x2 – 10x + 25
17) x2 – 14x + 49
18) x2 + 6x – 16
19) x2 + 6x + 8
20) x2 + x – 30
21)
22)
23)
24)
25)
26)
27)
28)
29)
30)
31)
32)
33)
34)
35)
36)
37)
38)
39)
40)
ho
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(x – 5)(x – 5)
(5x – 4)(5x – 4)
(3x – 1)(3x – 1)
(2x + 1)(3x + 2)
(4x + 2)(x + 4)
x2 + 13x + 30
x2 + 8x + 7
x2 + 10x + 21
x2 – 5x – 50
x2 + 9x + 20
x2 – 10x – 11
x2 + 14x + 13
3x2 + 10x + 3
2x2 + x – 6
2x2 – x – 1
8x2 – 2x – 3
25x2 + 15x + 2
4x2 + 5x – 9
3x2 + 4x – 15
4x2 – 1
25x2 – 1
9x2 – 4
x4 – x2
x3 – 25x
x4 – 4x2
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1. จงหาผลคูณต่อไปนี้
1)
(x + 1)(x – 1)
2)
(x + 3)(x – 3)
3)
(2x + 3)(2x – 3)
4)
(5x + 4)(5x – 4)
5)
(3x + 1)(3x – 1)
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แบบฝึ กหัด 3.3.1 (จากหนังสื อเรียน)
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3. จงแยกตัวประกอบของพหุ นามต่อไปนี้ โดยการทาให้เป็ นกาลังสองสมบูรณ์
1)
x2 + 4x – 32
2)
x2 – 2x – 3
3)
x2 – 4x + 2
4)
x2 + 8x – 5
5)
x2 + 6x + 2
6)
x2 + 8x + 14
7)
x2 – 10x + 7
8)
x2 + 7x + 11
9)
x2 – 2x
10) x2 + 4x
11) –2x2 – 8x + 8
12) 8 + 4x – x2
13) –3x2 + 6x + 4
14) 4x2 – 4x – 9
15) –3x2 + 6x + 2
16) –2x2 + 2x + 1
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เฉลยแบบฝึ กหัด 3.3.1 (จากหนังสื อเรียน)
(x + 1)(x – 1) =
(x + 3)(x – 3) =
(2x + 3)(2x – 3) =
(5x + 4)(5x – 4) =
(3x + 1)(3x – 1) =
(x – 5)(x – 5) =
(5x – 4)(5x – 4) =
(3x – 1)(3x – 1) =
(2x + 1)(3x + 2) =
(4x + 2)(x + 4) =
x2 + (–x) + x + (–1)
x2 + (–3x) + 3x+ (–9)
4x2 + (–6x) + (6x) + (–9)
25x2 + (–20x) + 20x + (–16)
9x2+ (–3x) + 3x + (–1)
x2 + (–5x) + (–5x) + 25
25x2 + (–20x) + (–20x) + 16
9x2 + (–3x) + (–3x) + 1
6x2 + 4x + 3x + 2
4x2 + 16x + 2x + 8
ข้อ 2. 1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
12)
13)
14)
15)
16)
17)
18)
x2 – 25x
x3 – 4x2
x4 – 4x
15x2 – 25x
81x2 – x
7x2 + 49x
88x3 + 8 x2
13x4 + x2
5x3 + 15x2
100x4 + 10x3
x2 + 3x – 4
x2 + 10x + 25
x2 + 6x + 9
x2 + 4x + 4
x2 + 8x – 20
x2 – 10x + 25
x2 – 14x + 49
x2 + 6x – 16
x(x – 25)
x2(x – 4)
x(x3 – 4)
5x(3x – 5)
x(81x – 1)
7x(x + 7)
8x2 (11x + 1)
x2 (13x2 + 1)
5x2(x + 3)
10x3(10x + 1)
(x – 1)(x + 4)
(x + 5)(x + 5)
(x + 3)(x + 3)
(x + 2)(x + 2)
(x – 2)(x + 10)
(x – 5)(x – 5)
(x – 7)(x – 7)
(x – 2)(x + 8)
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ข้อ 1. 1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
=
=
=
(x + 5)2
(x + 3)2
(x + 2)2
=
=
(x – 5)2
(x – 7)2
x2 – 1
x2 – 9
4x2 – 9
25x2 – 16
9 x2 – 1
x2 – 10x + 25
25x2 – 40x + 16
9x2 – 6x + 1
6x2 + 7x + 2
4x2 + 18x + 8
2)
x2 – 2x – 3
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x2 + 4x – 32
(x + 2)(x + 4)
(x – 5)(x + 6)
(x + 3)(x + 10)
(x + 1)(x + 7)
(x + 3)(x + 7)
(x + 5)(x – 10)
(x + 5)(x + 4)
(x + 1)(x – 11)
(x + 1)(x + 13)
(3x + 1)(x + 3)
(2x – 3)(x + 2)
(2x + 1)(x – 1)
(4x – 3)(2x + 1)
(5x + 2)(5x + 1)
(4x + 9)(x – 1)
(3x – 5)(x + 3)
(2x)2 – 12
=
(5x)2 – 12
=
(3x)2 – 22
=
x2(x2 – 1)
=
x(x2 – 25)
=
2 2
x (x – 4)
=
ho
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ข้อ 3. 1)
x2 + 6x + 8
x2 + x – 30
x2 + 13x + 30
x2 + 8x + 7
x2 + 10x + 21
x2 – 5x – 50
x2 + 9x + 20
x2 – 10x – 11
x2 + 14x + 13
3x2 + 10x + 3
2x2 + x – 6
2x2 – x – 1
8x2 – 2x – 3
25x2 + 15x + 2
4x2 + 5x – 9
3x2 + 4x – 15
4x2 – 1
25x2 – 1
9x2 – 4
x4 – x2
x3 – 25x
x4 – 4x2
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19)
20)
21)
22)
23)
24)
25)
26)
27)
28)
29)
30)
31)
32)
33)
34)
35)
36)
37)
38)
39)
40)
(x2 + 4x + 4) – 32 – 4
(x + 2)2 – 36
((x + 2) – 6)((x + 2) + 6)
(x – 4)(x + 8)
(x2 – 2x + 1) – 3 – 1
(x – 1)2 – 4
((x – 1) – 2)((x – 1) + 2)
(x – 3)(x + 1)
(2x – 1)(2x + 1)
(5x – 1)(5x + 1)
(3x – 2)(3x + 2)
x2(x – 1)(x + 1)
x(x – 5)(x + 5)
x2(x – 2)(x + 2)
x2 – 4x + 2
4)
x2 + 8x – 5
5)
x2 + 6x + 2
6)
x2 + 8x + 14
7)
x2 – 10x + 7
8)
nC
x2 + 7x + 11
=
Pib
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=
9)
x2 – 2x
=
=
=
=
=
=
=
(x2 – 4x + 4) + 2 – 4
(x – 2)2 – 2
[(x – 2) – 2 ][(x – 2) + 2 ]
[(x – (2 + 2 )][x – (2 – 2 )]
(x2 + 8x + 16) – 5 – 16
(x + 4)2 – 21
[(x + 4) – 21 ][(x + 4) + 21 ]
[x + (4 – 21 )][x + (4 + 21 )]
(x2 + 6x + 9) + 2 – 9
(x + 3)2 – 7
[(x + 3) – 7 ][(x + 3) + 7 ]
[x + (3 – 7 )][x + (3 + 7 )]
(x2 + 8x + 16) + 14 – 16
(x + 4)2 – 2
[(x + 4) – 2 ][(x + 4) + 2 ]
[x + (4 – 2 )][x + (4 + 2 )]
(x2 – 10x + 25) – 25 + 7
(x – 5)2 – 18
[(x – 5) – 18 ][(x – 5) + 18 )
[x – (5 + 18 )][x – (5 – 18 )]
(x2 + 7x + 49 ) – 49 +11
4 4
7
5
(x + )2 –
2 4
[(x + 7 ) – 5 ][(x + 7 ) + 5 )
2
2
2
2
[x + ( 7  5 )][x + ( 7  5 )]
2
2
2
(x – 2x + 1) – 1
(x – 1)2 – 1
[(x – 1) – 1][(x – 1) + 1]
[x – (1 + 1)][(x – (1 – 1)]
(x – 2)(x)
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3)
12)
=
Pib
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14)
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(x2 + 4x + 4) – 4
(x + 2)2 – 4
[(x + 2) – 2][(x + 2) + 2]
[x + (2 – 2)][x + (2 + 2)]
(x)(x + 4)
–2(x2 + 4x – 4)
–2[(x2 + 4x + 4) – 4 – 4]
–2(x + 2)2 + 16
–2[(x + 2)2 – 8]
–2[((x + 2) – 8 )((x + 2) + 8 )]
–2[(x + (2 – 8 ))(x + (2 + 8 ))]
–(x2 – 4x – 8)
–[(x2 – 4x + 4) – 8 – 4]
– [(x – 2)2 – 12]
–[((x – 2) – 12 )((x – 2 )+ 12 )]
–[(x – (2 + 12 ))(x +(– 2 + 12 )]
–3(x2 – 2x) + 4
–3[(x2 – 2x + 1) – 1] + 4
–3[(x – 1)2 – 1] + 4
–3(x – 1)2 + 7
–3[(x – 1)2 – 7 ]
3
–3[((x – 1) – 7 )((x – 1 )+ 7 )]
3
3
–3[(x – (1 + 7 ))(x – (1 – 7 )]
3
3
2
4(x – x) – 9
4[(x2 – x + 1 ) – 1 ] – 9
4 4
1
2 1
4[(x – ) – ] – 9
2 4
4(x – 1 )2 – 1– 9
2
4(x – 1 )2 – 10
2
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13)
=
=
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–2x2 – 8x + 8 =
=
=
=
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2
8 + 4x – x
=
=
=
=
=
2
–3x + 6x + 4 =
=
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11)
x2 + 4x
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10)
4x2 – 4x – 9
=
=
=
=
=
=
=
–3x2 + 6x + 2 =
=
=
=
=
=
=
=
–2x2 + 2x + 1
=
=
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16)
ms
=
15)
4[(x – 1 )2 – 10 ]
2
4
10
10
4[((x – 1 ) –
)((x – 1 )+
)]
2
2
2
2
4[(x – ( 1  10 ))(x – ( 1  10 )]
2
2
2
–3(x – 2x) + 2
–3[(x2 – 2x + 1) – 1] + 2
–3[(x – 1)2 – 1] + 2
–3(x – 1)2 + 3 + 2
–3(x – 1)2 + 5
–3[(x – 1)2 – 5 ]
3
–3[((x – 1) – 5 )((x – 1 )+ 5 )]
3
3
–3[(x – (1 + 5 ))(x – (1 – 5 )]
3
3
2
–2(x – x) + 1
–2[(x2 – x + 1 ) – 1 ] + 1
4 4
1
–2[(x – )2 – 1 ] + 1
2 4
1
–2 (x – )2 + 1 + 1
2 2
–2 (x – 1 )2 + 3
2 2
–2 [(x – 1 )2 – 3 ]
2 4
3
3
–2 [((x – 1 ) – )((x – 1 )+
)]
2 2
2
2
–2 [(x – ( 1  3 ))(x – ( 1  3 )]
2
2
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