F.3 to F.4 Maths暑期作業 Factorization step by step thxx!?

1.(-28)
6xxy-2xxxy-2xy=(2xy)(3x-xx-1)=-(2xy)(xx-3x+1)=-2*7*2=-28

2a.(m+1)(n+1)

2b.(m+1)(n+1)(p+1)

3a,(2a+2)

3b.(x^2+3x+1)(x+1)(x+2)=(x+1)(x+2)(x+(3-5^(1/2))/2)(x+(3+5^(1/2))/2)
let A=(x^2+3x)=>(x^2+3x)^2+3(x^2+3x)+2
=AA+3A+2
=(A+1)(A+2)
=(x^2+3x+1)(x^2+3x+2)
=(x+(3-5^(1/2))/2)(x+(3+5^(1/2))/2)(x+1)(x+2)

4a.x^3+3x^2+x-2
(x+2)(x^2+x-1)
=xxx+xx-x+2xx+2x-2
=x^3+3x^2+x-2

4b.(2x-1)(x+2)^2
2x^3+7x^2+4x-4
=(x+2)(2xx+3x-2)<-這裡用x=-2代進去可以知道含(x+2)
=(2x-1)(x+2)^2

4c.2,3,19(但是我不知道他要怎麼算)

5a.2x^3-x^2-x
x(x-1)(2x+1)
=x(2xx-2x+x-1)
=x(2xx-x-1)
=2x^3-x^2-x

5b.(2x+1)(x-1)^2
2x^3-3x^2+1
=(x-1)(2xx-x-1)
=(2x+1)(x-1)^2

1.
6x^2 y -2x^3 y-2xy
=-(2xy)(x^2-3x+1)
=-2(7)(2)
=-28

2a.
1+m+n+mn
=(1+m)+n(1+m)
=(1+m)(1+n)

2b
1+m+n+p+mn+mp+np+mnp
=(1+m)+(n+p)+m(n+p)+np(1+m)
=(1+m)+np(1+m)+(n+p)(1+m)
=(1+m)(1+np+n+p)
=(1+m)[(1+n)+p(n+1)]
=(1+m)(1+n)(1+p)

3a
(a+1)(a+2)
=a^2 +3a+2

3b
(x^2+3x)^2+3(x^2+3x)+2
=[(x^2+3x)+1][(x^2+3x)+2]
=(x^2+3x+1)(x+1)(x+2)

4a
(x+2)(x^2+x-1)
=x^3+x^2-x+2x^2+2x-2
=x^3+3x^2+x-2

4b
2x^3+6x^2+2x-4
=2(x^3+3x^2+x-2)
=2(x+2)(x^2+x-1)

4c
2x^3+6x^2+2x-4=2736
x=10.16374192