請教因式分解

1.x^2-(y+z)^2(運用恆等式)
=(x-y-z)(x+y+z)
注意:用a^2-b^2=(a+b)(a-b)
(2x)^2-(5y)^2
=(2x+5y)(2x-5y)
注意:用a^2-b^2=(a+b)(a-b)
3.16s^2-t^2-8s+2t
=(4s)^2-t^2-8s+2t
=(4s-t)(4s+t)-2(4s-t)
=(4s-t)(4s+t-2)
4.x^4-1
=(x^2)^2-(1)^2
=(x^2+1)(x^2-1)
=(x^2+1)(x-1)(x+1)
注意:用a^2-b^2=(a+b)(a-b)

a^2-b^2 =(a+b)(a-b)

1.x^2-(y+z)^2(運用恆等式)
a=x, b=y+z

x^2-(y+z)^2
=[x+(y+z)][x-(y+z)]
=(x+y+z)(x-y-z)



2.(2x)^2-(5y)^2(運用恆等式)
a=2x, b=5y

(2x)^2-(5y)^2
=(2x+5y)(2x-5y)



3.16s^2-t^2-8s+2t

For 16s^2-t^2
a^2=(4s)^2, b^2=t^2

16s^2-t^2-8s+2t
=(4s+t)(4s-t)-2(4s-t)
=(4s-t)(4s+t-2)



4.x^4-1
a^2=x^4=(x^2)^2
b^2 = 1^2

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

For (x^2-1)
a=x, b=1

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

1) (x+y+z)[x-(y+z)]
=(x+y+z)(x-y-z)


2) (2x+5y)(2x-5y)


3) 8s[2s-1]-t[t-2]

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