f.2 factorization

1.4k+1-[5k-4(k+1)]
= 4k + 1 – 5k + 4k + 4
= 3k + 5

2.c²+7c+[5c²-(3c-4c²)]
= c²+7c+5c²-3c+4c²
= 10c2 + 4c
= 2c(5c + 2)

3. a+3b(b-3a)-[2a-b-(a-2b)]
= a+3b2 - 9ab - 2a + b + a - 2b
= 3b2 - 9ab - b
= b(3b – 9a – 1)

4.c[c(c+d)-d(c-d)]
= c[c2 + cd – cd + d2]
= c(c2 + d2)

5. 2x-[3x-x(3-x)]
= 2x-[3x - 3x + x2]
= 2x - x2
= x(2 – x)

6.7a+a(4b-8-a)
= 7a + 4ab – 8a – a2
= 4ab – a – a2
= a(4b – 1 – a)

由第7至10題是不是要展開
7.(n-5)(n-5)
= (n – 5)2
= n2 – 10n + 25

8.(3x-5)(2x+3)
= 6x2 – x – 15


9.(7s-5t)(5s-2t)
= 35s2 – 34st + 10t2

10.(4c-5d)(3x+2d)
= 4c(3x + 2d) – 5d(3x + 2d)
= 12cx + 8cd – 15dx – 10d2

1)4k+1-[5k-4(k+1)]
=4k+1-[5k-4k-1]
=4k+1-[k-1]
=4k+1-k+1
=3k+2

2)c²+7c+[5c²-(3c-4c²)]
=c²+7c+[5c²-3c+4c²]
=c²+7c+[9c²-3c]
=c²+7c+9c²-3c
=10c²+4c
=2c(5c+2)

3)a+3b(b-3a)-[2a-b-(a-2b)]
=a+3b²-9ab-[2a-b-a+2b]
=a+3b²-9ab-[a+b]
=a+3b²-9ab-a-b
=3b²-9ab-b
=b(3b-9a-1)

4)c[c(c+d)-d(c-d)]
=c(c+d)[c-d]
=c(c+d)(c-d)

5)2x-[3x-x(3-x)]
=2x-[3x-3x+x²]
=2x-x²
=x(2-x)

6)7a+a(4b-8-a)
=7a+4ab-8a-a²
=4ab-a-a²
=a(4b-a-1)

7).(n-5)(n-5)
=(n-5)²
=n²-2(n)(5)+5²
=n²-10n+25

8)(3x-5)(2x+3)
=3x(2x+3)-5(2x+3)
=6x²+9x-10x-15
=6x²-x-15

9)(7s-5t)(5s-2t)
=7s(5s-2t)-5t(5s-2t)
=35s²-14st-25st+10t²
=35s²-39st+10t²

10)(4c-5d)(3x+2d)
=4c(3x+2d)-5d(3x+2d)
=12cx+8cd-15xd-10d²