[15] Factorization

1.(1+x)^2-9 = 1+2x+x^2-9 = x^2+2x-8 = (x-2)(x+4)

2.(p+q)^2 -144 = (p+q)^2 - (12)^2 = (p+q+12)(p+q-12)

3.16a^2-(4a-b)^2 = (4a)^2-(4a-b)^2 = [4a-(4a-b)][4a+(4a-b)] = b(8a-b)

4.9(h+k)^2 - 16k^2 = (3h+3k)^2 - (4k)^2 = [(3h+3k)+(4k)][(3h+3k)-(4k)] = (3h+7k)(3h-k)

5. (a-1)^2 - (b-1)^2 = [(a-1)+(b-1)][(a-1)-(b-1)] = (a+b-2)(a-b)

6.9(x+y)^2 - (x-y)^2 = (3x+3y)^2 - (x-y)^2 = [(3x+3y)+(x-y)][(3x+3y)-(x-y)] =
(4x+2y)(2x+4y) = 4(2x+y)(x+2y)

7.p^2 - q^2 + 4p + 4q = (p+q)(p-q)+4(p+q) = (p+q)(p-q+4)

8.a^2 - b^2 + bc - ac = (a+b)(a-b)-ac+bc = (a+b)(a-b)-c(a-b) = (a-b)(a+b-c)

9.75 (m-3n)^2 - 27(2m - 5n)^2 = 3[-(m-3n)^2 - 9(2m-5n)^2] = 3[(5m-15n)^2-(6m-15n)^2]
= 3[(5m-15n)+(6m-15n)][(5m-15n)-(6m-15n)] = 3(11m-30n)(-m)=3m(30n-11m)

10.3u^2 - 12v^2 - 6u + 12v = 3[u^2 - (2v)^2 - 2u + 4v] = 3[(u+2v)(u-2v)+2(2v-u)]
= 3[(u+2v)(u-2v)-2(u-2v)] = 3[(u-2v)(u+2v-2)]

1. (x+4)(x-2)

2. (p+q+12)(p+q-12)

3. b(8a-b)

4. (3h+7k)(3h-k)

5. (a+b-2)(a-b)

6. (4x+2y)(2x+4y)

7. (p+q)(p-q+4)

8. (a-b)(a+b-c)

9. 75 (m-3n)^2 - 27(2m - 5n)^2
=3{[5(m-3n)]^2-[3(2m-5n)^2]}
=3[(5m-15n)^2-(6m-15n)^2]
=-3m(11m-30n)

10. 3u^2 - 12v^2 - 6u + 12v
=3[u^2-(2v)^2]-6(u-2v)
=3(u+2v)(u-2v)-6(u-2v)
=3(u-2v)(u+2v-2)

1.(1+x)2﹣9
= (1+x)2﹣32
= [(1+x)﹣3][(1+x)+3]
= (1+x﹣3)(1+x+3)
= (x﹣2)(x+4)

2.(p+q)2﹣144
= (p+q)2﹣122
= [(p+q)﹣12][(p+q)+12]
= (p+q﹣12)(p+q+12)

3.16a2﹣(4a﹣b)2
= (4a)2﹣(4a﹣b)2
= [(4a)﹣(4a﹣b)][(4a)+(4a﹣b)]
= (4a﹣4a+b)(4a+4a﹣b)
= b(8a﹣b)

4.9(h+k)2﹣16k2
= [3(h+k)]2﹣(4k)2
= [3(h+k)﹣4k][3(h+k)+4k]
= (3h+3k﹣4k)(3h+3k+4k)
= (3h﹣k)(3h+7k)

5.(a﹣1)2﹣(b﹣1)2
= [(a﹣1)﹣(b﹣1)][(a﹣1)+(b﹣1)]
= (a﹣1﹣b+1)(a﹣1+b﹣1)
= (a﹣b)(a+b﹣2)

6.9(x+y)2﹣(x﹣y)2
= [3(x+y)]2﹣(x﹣y)2
= [3(x+y)﹣(x﹣y)][3(x+y)+(x﹣y)]
= (3x+3y﹣x+y)(3x+3y+x﹣y)
= (2x+4y)(4x+2y)
= 2(x+y)2(2x+y)
= 4(x+y)(2x+y)

7.p2﹣q2 + 4p + 4q
= (p2﹣q2)+4(p+q)
=(p+q)(p﹣q)+4(p+q)
=(p+q)(p﹣q+4)


a2﹣b2 + bc﹣ac
=(a+b)(a﹣b)+c(b﹣a)
=-(a+b)(a﹣b)﹣c(a﹣b)
=(a﹣b)(a+b﹣c)

75(m﹣3n)2﹣27(2m﹣5n)2
=3[25(m﹣3n)2-9(2m﹣5n)2]
=3[(5m﹣15n)2-(6m﹣15n)2]
=3[(5m﹣15n)+(6m﹣15n)][(5m﹣15n)﹣(6m﹣15n)]
=3[5m﹣15n+6m﹣15n][5m﹣15n﹣6m+15n]
=3[11m﹣30n][-m]
=3s(30n﹣11m)

3u2﹣12v2﹣6u + 12v
= 3[(u2﹣4v2)﹣2(u﹣2v)]
= 3[(u+2v)(u﹣2v)﹣2(u﹣2v)]
= 3(u﹣2v)[(u+2v)﹣2]
= 3(u﹣2v)(u+2v﹣2)

應用恆等式a2﹣b2 = (a﹣b)(a+b)

2007-12-31 14:21:38 補充：
第9題的答案是= 3[11m﹣30n][-m]= 3m(30n﹣11m)