f.3 factorization

2012-09-23 1:43 am

回答 (2)

2012-09-23 11:03 am
✔ 最佳答案
13.
m⁴n + mn⁴
= mn(m³ + n³)
= mn(m + n)(m² - mn + n²)


14.
-2h³k³ + 16
= 16 - 2h³k³
= 2[8 - h³k³]
= 2[2³ - (hk)³]
= 2(2 - hk)[2² + 2hk + (hk)²]
= 2(2 - hk)(4 + 2hk + h²k²)


15.
(a + 3)³ + a³
= [(a + 3) + a] [(a + 3)² - (a + 3)a + a²]
= [a + 3 + a] [a² + 6a + 9 - a² - 3a + a²]
= (2a + 3)(a² + 3a + 9)


16.
p³ - (p - q)³
= [p - (p - q)] [p² + p(p - q) + (p - q)²]
= [p - p + q] [p² + p² - pq + p² - 2pq + q²]
= q(3p² - 3pq + q²)


17.
Method 1 :
(x + 1)³ - (x - 1)³
= [(x + 1) - (x - 1)] [(x + 1)² +(x + 1)(x - 1) + (x - 1)²]
= [x + 1 - x + 1] [x² + 2x + 1 + x² - 1 + x² - 2x + 1]
= 2(3x² + 1)

Method 2 :
(x + 1)³ - (x - 1)³
= (x³ + 3x² + 3x + 1) - (x³ - 3x² + 3x - 1)
= x³ + 3x² + 3x + 1 - x³ + 3x² - 3x + 1
= 6x² + 2
= 2(3x² + 1)


18.
125a³ + b³ - 5a - b
= [125a³ + b³] - (5a + b)
= [(5a)³ + b³] - (5a + b)
= [5a + b] [(5a)² - (5a)b + b²] - (5a+ b)
= (5a + b)(25a² - 5ab + b²) - (5a + b)
= (5a + b) [(25a² - 5ab + b²) - 1]
= (5a + b)(25a² - 5ab + b² - 1)


19.
2h³ - 16k³ + 3h - 6k
= 2(h³ - 8k³) + 3(h - 2k)
= 2[h³ - (2k)³] + 3(h - 2k)
= 2[h - 2k] [h² + h(2k) + (2k)²] + 3(h - 2k)
= 2(h - 2k)(h² + 2hk + 4k²) + 3(h - 2k)
= (h - 2k)(2h² + 4hk + 8k²) + 3(h - 2k)
= (h - 2k) [(2h² + 4hk + 8k²) + 3]
= (h - 2k)(2h² + 4hk + 8k² + 3)


20.
m³ + m² - n³ - n²
= (m³ - n³) + (m² - n²)
= (m - n)(m² + mn + n²) + (m - n)(m + n)
= (m - n)[(m² + mn + n²) + (m +n)]
= (m - n)(m² + mn + n² + m + n)


21.
t⁶ - t³ + t - 1
= t³(t³ - 1) + (t - 1)
= t³(t - 1)(t² + t + 1) + (t- 1)
= (t - 1)(t⁵ + t⁴ + t³) + (t - 1)
= (t - 1) [(t⁵ + t⁴ + t³) + 1]
= (t - 1) [t⁵ + t⁴ + t³ + 1]
= (t - 1) [t⁴(t + 1) + (t³ + 1)]
= (t - 1) [t⁴(t + 1) + (t + 1)(t² - t + 1)]
= (t - 1)(t + 1) [t⁴ + (t² - t + 1)]
= (t - 1)(t + 1)(t⁴ + t² - t + 1)


22.
u³ + 8 - (u + 2)²
= (u³ + 2³) - (u + 2)²
= (u + 2)[u² - 2u + 2²] - (u + 2)²
= (u + 2)(u² - 2u + 4) - (u + 2)²
= (u + 2)[(u² - 2u + 4) - (u + 2)]
= (u + 2)[u² - 2u + 4 - u - 2]
= (u + 2)(u² - 3u + 2)
= (u + 2)(u - 1)(u - 2)
= (u - 1)(u - 2)(u + 2)
參考: micatkie
2012-09-23 7:39 am
13
m^4n+mn^4
=mn(m^3+^3)
=mn(m+n)(m^2-mn+n^2)

14
-2h^3k^3+16
=-2[(hk)^-2^3]
=-2(hk-2)(h^2k^2+2hk+4)

15
(a+3)^3+a^3
=[(a+3)+a][(a+3)^2-a(a+3)+a^2]
=(2a+3)(a^2+6a+9-a^2+3a+a^2)
=(2a+3)(a^2+9a+9)

16
p^3-(p-q)^3
=[p-(p-q)][p^2+p(p-q)+(p-q)^2]
=(-q)(p^2+p^2-pq+p^2-2pq+q^2)
=(-q)(3p^2-3pq+q^2)

17
(x+1)^3-(x-1)^3
=[(x+1)-(x-1)][(x+1)^2+(x+1)(x-1)+(x-1)^2]
=(2)(x^2+2x+1+x^2-1+x^2-2x+1)
=2(3x^2+1)

18
125a^3+b^3-5a-b
=(5a+b)(5a^2-5ab+b)-(5a+b)
=(5a+b)(5a^2-5ab+b+1)

19
2h^3-16k^3+3h-6k
=2(h^2-8k^2)+3(h-2k)
=2(h-2k)(h^2+2hk+4k^2)+3(h-2k)
=(h-2k)[2(h^2+2hk+4k^2)+3]
=(h-2k)(2h^2+4hk+8k^2+3)

20
m^3+m^2-n^3-n^2
=(m-n)(m^2+mn+n^2)+(m-n)(m+n)
=(m-n)(m^2+mn+n^2+m+n)

21
t^6-t^3+t-1
=t^3(t^2-1)+(t-1)
=t^3(t-1)(t+1)+(t-1)
=(t-1)[t^3(t+1)+1]
=(t-1)(t^4+t^2+1)

22
u^3+8-(u+2)^2
=(u+2)(u^2+2u+4)-(u+2)^2
=(u+2)(u^2+2u+4-u-2)
=(u+2)(u^2+u-2)
=(u-1)(u+2)^2


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