factoriser les expressions suivantes
(2x+3)²-(x+1)²
16(x-1)²-(2x-5)²
(4x+3)²-2(4x+3)(x+2)
49-(x+2)²
x²-25+2(x-5)(x+3)
et expliquer moi sil vous plait
(2x+3)²-(x+1)²=(2x+3-x-1)(2x+3+x+1)=(x+2)(3x+4)
16(x-1)²-(2x-5)²=(4(x-1))²-(2x-5)²=(4x-4-2x+5)(4x-4+2x-5)=(2x+1)(6x-9)=3(2x+1)(2x-3)
(4x+3)²-2(4x+3)(x+2)=(4x+3)(4x+3-2(x+2))=(4x+3)(4x+3-2x-4)=(4x+3)(2x-1)
49-(x+2)²=7²-(x+2)²=(7-x-2)(7+x+2)=(5-x)(x+9)
x²-25+2(x-5)(x+3)=(x²-5²)+2(x-5)(x+3)=(x-5)(x+5)+2(x-5)(x+3)=(x-5)(x+5+2(x+3))
=(x-5)(x+5+2x+6)=(x-5)(3x+11)
(2x+3)²-(x+1)²= (4x²+9+12x)-(x²+1+2x)=4x²+9+12x-x²-1-2x=3x²+8+10x
16(x-1)²-(2x-5)²=16(x²+1+2x)-(4x²+25-20x)=...
(4x+3)²-2(4x+3)(x+2)=
49-(x+2)²=7²-(x+2)²=
x²-25+2(x-5)(x+3)=