2. จงแยกตัวประกอบของพหุนามต่อไปนี้
1) m(n + 3) + 5(n +3)
วิธีทำ
m(n + 3) + 5(n +3) = (n +3)(m + 5)
ตอบ (n +3)(m + 5)
2) (x + y)z – (x + y)
วิธีทำ
(x + y)z – (x + y) = (x + y)(z – 1)
ตอบ (x + y)(z – 1)
3) 4t(a + b) – s(a + b)
วิธีทำ
4t(a + b) – s(a + b) = (a + b)(4t – s)
ตอบ (a + b)(4t – s)
4) \(\mathsf{(4y^2 + 3)y + 6(4y^2 + 3)}\)
วิธีทำ
\(\mathsf{(4y^2 + 3)y + 6(4y^2 + 3) = (4y^2 + 3)(y + 6)}\)
ตอบ \(\mathsf{(4y^2 + 3)(y + 6)}\)
5) a(b – 3c) + x(b – 3c)
วิธีทำ
a(b – 3c) + x(b – 3c) = (b – 3c)(a + x)
ตอบ (b – 3c)(a + x)
6) ax + by + bx + ay
วิธีทำ
ax + by + bx + ay
= (ax + bx) + (ay + by)
= x(a + b) + y(a + b)
= (a + b)(x + y)
= (a + b)(x + y)
ตอบ (a + b)(x + y)
7) 5a – 10x + ab – 2bx
วิธีทำ
5a – 10x + ab – 2bx
= (5a + ab) – (10x + 2bx)
= a(5 + b) – 2x(5 + b)
= (5 + b)(a – 2x)
= (5 + b)(a – 2x)
ตอบ (5 + b)(a – 2x)
8) na + 3b + nb + 3a
วิธีทำ
na + 3b + nb + 3a
= (na + nb) + (3a + 3b)
= n(a + b) + 3(a + b)
= (a + b)(n + 3)
= (a + b)(n + 3)
ตอบ (a + b)(n + 3)
9) xy – st – xt + sy
วิธีทำ
xy – st – xt + sy
= (xy + sy) – (xt + st)
= y(x + s) – t(x + s)
= (x + s)(y – t)
= (x + s)(y – t)
ตอบ (x + s)(y – t)
10) \(\small{\mathsf{n^2m + n^2p \, – \, 8m \, – \, 8p}}\)
วิธีทำ
\(\small{\mathsf{n^2m + n^2p \, – \, 8m \, – \, 8p}}\)
= \(\small{\mathsf{(n^2m + n^2p) \, – \, (8m + 8p)}}\)
= \(\small{\mathsf{n^2(m + p) \, – \, 8(m + p)}}\)
= \(\small{\mathsf{(m + p)(n^2 \, – \, 8)}}\)
= \(\small{\mathsf{(m + p)(n^2 \, – \, 8)}}\)
ตอบ \(\small{\mathsf{(m + p)(n^2 \, – \, 8)}}\)
11) \(\small{\mathsf{ab^2 \, – \, cb^2 \, – \, 6a + 6c}}\)
วิธีทำ
\(\small{\mathsf{ab^2 \, – \, cb^2 \, – \, 6a + 6c}}\)
= \(\small{\mathsf{(ab^2 \, – \, cb^2) \, – \, (6a \, – \, 6c)}}\)
= \(\small{\mathsf{b^2(a \, – \, c) \, – \, 6(a \, – \, c)}}\)
= \(\small{\mathsf{(a \, – \, c)(b^2 \, – \, 6)}}\)
= \(\small{\mathsf{(a \, – \, c)(b^2 \, – \, 6)}}\)
ตอบ \(\small{\mathsf{(a \, – \, c)(b^2 \, – \, 6)}}\)
12) \(\small{\mathsf{2x^3 \, – \, x + 14x^2 \, – \, 7}}\)
วิธีทำ
\(\small{\mathsf{2x^3 \, – \, x + 14x^2 \, – \, 7}}\)
= \(\small{\mathsf{(2x^3 \, – \, x) + (14x^2 \, – \, 7)}}\)
= \(\small{\mathsf{x(2x^2 \, – \, 1) + 7(2x^2 \, – \, 1)}}\)
= \(\small{\mathsf{(2x^2 \, – \, 1)(x + 7)}}\)
= \(\small{\mathsf{(2x^2 \, – \, 1)(x + 7)}}\)
ตอบ \(\small{\mathsf{(2x^2 \, – \, 1)(x + 7)}}\)
13) \(\small{\mathsf{a^2 \, – \, 2b \, – \, 5a^3 + 10ab}}\)
วิธีทำ
\(\small{\mathsf{a^2 \, – \, 2b \, – \, 5a^3 + 10ab}}\)
= \(\small{\mathsf{(a^2 \, – \, 5a^3) \, – \, (2b \, – \, 10ab)}}\)
= \(\small{\mathsf{a^2(1 \, – \, 5a) \, – \, 2b(1 \, – \, 5a)}}\)
= \(\small{\mathsf{(1 \, – \, 5a)(a^2 \, – \, 2b)}}\)
= \(\small{\mathsf{(1 \, – \, 5a)(a^2 \, – \, 2b)}}\)
ตอบ \(\small{\mathsf{(1 \, – \, 5a)(a^2 \, – \, 2b)}}\)
14) \(\small{\mathsf{x^3 \, – \, x^3z + y^2z \, – \, y^2}}\)
วิธีทำ
\(\small{\mathsf{x^3 \, – \, x^3z + y^2z \, – \, y^2}}\)
= \(\small{\mathsf{(x^3 \, – \, x^3z) \, – \, (y^2 \, – \, y^2z)}}\)
= \(\small{\mathsf{x^3(1 \, – \, z) \, – \, y^2(1 \, – \, z)}}\)
= \(\small{\mathsf{(1 \, – \, z)(x^3 \, – \, y^2)}}\)
= \(\small{\mathsf{(1 \, – \, z)(x^3 \, – \, y^2)}}\)
ตอบ \(\small{\mathsf{(1 \, – \, z)(x^3 \, – \, y^2)}}\)