2. จงแยกตัวประกอบของพหุนามต่อไปนี้
วิธีคิด
สูตรการแยกตัวประกอบของพหุนามดีกรีสองที่เป็นผลต่างของกำลังสอง
\(\mathsf{A^2 \, – \, B^2 = (A + B)(A \, – \, B)}\)
เมื่อ A แทนพจน์หน้า และ B แทนพจน์หลังของพหุนามดีกรีสอง
1) \(\small\mathsf{{(a \; – \, 2)}^2 \, – \, 1}\)
วิธีทำ
\(\small\mathsf{{(a \; – \, 2)}^2 \, – \, 1}\)
= \(\small\mathsf{{(a \; – \, 2)}^2 \, – \, 1^2}\)
= [(a – 2) + 1][(a – 2) – 1]
= (a – 1)(a – 3)
= (a – 1)(a – 3)
ตอบ (a – 1)(a – 3)
2) \(\small\mathsf{25 \; – \, {(y + 1)}^2}\)
วิธีทำ
\(\small\mathsf{25 \; – \, {(y + 1)}^2}\)
= \(\small\mathsf{5^2 \; – \, {(y + 1)}^2}\)
= [5 + (y + 1)][5 – (y + 1)]
= (6 + y)(4 – y)
= (6 + y)(4 – y)
ตอบ (6 + y)(4 – y)
3) \(\small\mathsf{{(x + 2)}^2 \, – \, 4}\)
วิธีทำ
\(\small\mathsf{{(x + 2)}^2 \, – \, 4}\)
= \(\small\mathsf{{(x + 2)}^2 \, – \, 2^2}\)
= [(x + 2) + 2][(x + 2) – 2]
= (x + 4)(x)
= x(x + 4)
= (x + 4)(x)
= x(x + 4)
ตอบ x(x + 4)
4) \(\small\mathsf{{(x \; – \, 3)}^2 \, – \, 36}\)
วิธีทำ
\(\small\mathsf{{(x \; – \, 3)}^2 \, – \, 36}\)
= \(\small\mathsf{{(x \; – \, 3)}^2 \, – \, 6^2}\)
= [(x – 3) + 6][(x – 3) – 6]
= (x + 3)(x – 9)
= (x + 3)(x – 9)
ตอบ (x + 3)(x – 9)
5) \(\small\mathsf{81 \; – \, {(x + 5)}^2}\)
วิธีทำ
\(\small\mathsf{81 \; – \, {(x + 5)}^2}\)
= \(\small\mathsf{9^2 \; – \, {(x + 5)}^2}\)
= [9 + (x + 5)][9 – (x + 5)]
= (14 + x)(4 – x)
= (14 + x)(4 – x)
ตอบ (14 + x)(4 – x)
6) \(\small\mathsf{x^2 \; – \, {(2x + 1)}^2}\)
วิธีทำ
\(\small\mathsf{x^2 \; – \, {(2x + 1)}^2}\)
= [x + (2x + 1)][x – (2x + 1)]
= (3x + 1)(-x – 1)
= (3x + 1)(-1)(x + 1)
= (-1)(3x + 1)(x + 1)
= (3x + 1)(-1)(x + 1)
= (-1)(3x + 1)(x + 1)
ตอบ (-1)(3x + 1)(x + 1)
7) \(\small\mathsf{4x^2 \; – \, {(x \; – \, 2)}^2}\)
วิธีทำ
\(\small\mathsf{4x^2 \; – \, {(x \; – \, 2)}^2}\)
= \(\small\mathsf{{(2x)}^2 \; – \, {(x \; – \, 2)}^2}\)
= [2x + (x – 2)][2x – (x – 2)]
= (3x – 2)(x + 2)
= (3x – 2)(x + 2)
ตอบ (3x – 2)(x + 2)
8) \(\small\mathsf{{(2x + 3)}^2 \, – \, 25x^2}\)
วิธีทำ
\(\small\mathsf{{(2x + 3)}^2 \, – \, 25x^2}\)
= \(\small\mathsf{{(2x + 3)}^2 \, – \, {(5x)}^2}\)
= [(2x + 3) + 5x][(2x + 3) – 5x]
= (7x + 3)(-3x + 3)
= (7x + 3)(-3x + 3)
ตอบ (7x + 3)(-3x + 3)
9) \(\small\mathsf{{(x + 6)}^2 \, – \, {(x + 4)}^2}\)
วิธีทำ
\(\small\mathsf{{(x + 6)}^2 \, – \, {(x + 4)}^2}\)
= [(x + 6) + (x + 4)][(x + 6) – (x + 4)]
= (x + 6 + x + 4)(x + 6 – x – 4)
= (2x + 10)(2)
= 2(x + 5)(2)
= 4(x + 5)
= (2x + 10)(2)
= 2(x + 5)(2)
= 4(x + 5)
ตอบ 4(x + 5)
10) \(\small\mathsf{{(x \; – \, 8)}^2 \, – \, {(x \; – \, 5)}^2}\)
วิธีทำ
\(\small\mathsf{{(x \; – \, 8)}^2 \, – \, {(x \; – \, 5)}^2}\)
= [(x – 8) + (x – 5)][(x – 8) – (x – 5)]
= (x – 8 + x – 5)(x – 8 – x + 5)
= (2x – 13)(-3)
= -3(2x – 13)
= (2x – 13)(-3)
= -3(2x – 13)
ตอบ -3(2x – 13)
11) \(\small\mathsf{{(3x + 2)}^2 \, – \, {(x \; – \, 1)}^2}\)
วิธีทำ
\(\small\mathsf{{(3x + 2)}^2 \, – \, {(x \; – \, 1)}^2}\)
= [(3x + 2) + (x – 1)][(3x + 2) – (x – 1)]
= (3x + 2 + x – 1)(3x + 2 – x + 1)
= (4x + 1)(2x + 3)
= (4x + 1)(2x + 3)
ตอบ (4x + 1)(2x + 3)
12) \(\small\mathsf{{(4x \; – \, 3)}^2 \, – \, {(5x + 2)}^2}\)
วิธีทำ
\(\small\mathsf{{(4x \; – \, 3)}^2 \, – \, {(5x + 2)}^2}\)
= [(4x – 3) + (5x + 2)][(4x – 3) – (5x + 2)]
= (4x – 3 + 5x + 2)(4x – 3 – 5x – 2)
= (9x – 1)(-x – 5)
= (9x – 1)(-1)(x + 5)
= (-1)(9x – 1)(x + 5)
= (9x – 1)(-x – 5)
= (9x – 1)(-1)(x + 5)
= (-1)(9x – 1)(x + 5)
ตอบ (-1)(9x – 1)(x + 5)
13) \(\small\mathsf{9{(x \; – \, 7)}^2 \, – \, 100x^2}\)
วิธีทำ
\(\small\mathsf{9{(x \; – \, 7)}^2 \, – \, 100x^2}\)
= \(\small\mathsf{3^2{(x \; – \, 7)}^2 \, – \, {(10x)}^2}\)
= \(\small\mathsf{{[3(x \; – \, 7)]}^2 \, – \, {(10x)}^2}\)
= \(\small\mathsf{{(3x \; – \, 21)}^2 \, – \, {(10x)}^2}\)
= [(3x – 21) + 10x][(3x – 21) – 10x]
= (13x – 21)(-7x – 21)
= (13x – 21)(-7)(x + 3)
= (-7)(13x – 21)(x + 3)
= \(\small\mathsf{{(3x \; – \, 21)}^2 \, – \, {(10x)}^2}\)
= [(3x – 21) + 10x][(3x – 21) – 10x]
= (13x – 21)(-7x – 21)
= (13x – 21)(-7)(x + 3)
= (-7)(13x – 21)(x + 3)
ตอบ (-7)(13x – 21)(x + 3)
14) \(\small\mathsf{144x^2 \; – \, {(2x \; – \, 3)}^2}\)
วิธีทำ
\(\small\mathsf{144x^2 \; – \, {(2x \; – \, 3)}^2}\)
= \(\small\mathsf{{(12x)}^2 \; – \, {(2x \; – \, 3)}^2}\)
= [12x + (2x – 3)][12x – (2x – 3)]
= (12x + 2x – 3)(12x – 2x + 3)
= (14x – 3)(10x + 3)
= (12x + 2x – 3)(12x – 2x + 3)
= (14x – 3)(10x + 3)
ตอบ (14x – 3)(10x + 3)
15) \(\small\mathsf{25x^2 \; – \, 16{(x \; – \, 5)}^2}\)
วิธีทำ
\(\small\mathsf{25x^2 \; – \, 16{(x \; – \, 5)}^2}\)
= \(\small\mathsf{{(5x)}^2 \; – \, 4^2{(x \; – \, 5)}^2}\)
= \(\small\mathsf{{(5x)}^2 \; – \, {[4(x \; – \, 5)]}^2}\)
= \(\small\mathsf{{(5x)}^2 \; – \, {(4x \; – \, 20)}^2}\)
= [5x + (4x – 20)][5x – (4x – 20)]
= (5x + 4x – 20)(5x – 4x + 20)
= (9x – 20)(x + 20)
= \(\small\mathsf{{(5x)}^2 \; – \, {(4x \; – \, 20)}^2}\)
= [5x + (4x – 20)][5x – (4x – 20)]
= (5x + 4x – 20)(5x – 4x + 20)
= (9x – 20)(x + 20)
ตอบ (9x – 20)(x + 20)
16) \(\small\mathsf{{(5x + 3)}^2 \, – \, 121x^2}\)
วิธีทำ
\(\small\mathsf{{(5x + 3)}^2 \, – \, 121x^2}\)
= \(\small\mathsf{{(5x + 3)}^2 \, – \, {(11x)}^2}\)
= [(5x + 3) + 11x][(5x + 3) – 11x]
= (16x + 3)(-6x + 3)
= (16x + 3)(-3)(2x – 1)
= (-3)(16x + 3)(2x – 1)
= (16x + 3)(-6x + 3)
= (16x + 3)(-3)(2x – 1)
= (-3)(16x + 3)(2x – 1)
ตอบ (-3)(16x + 3)(2x – 1)