Question

Descumpuneti, aplicati formulele, factorul comun si artifici:

5x³-45x=
7x³-28x=
x5-16x=
x³+x²+x+1=
x³+3x²-x-3=
x³-2x³-x+2=
x³-4x²-4x+16=
x³+3x²-4x-12=
x³+3x²-9x-27=
x³+2x²-16(x+2)=
9(2x-5)²-4(4x-3)²=
4x²-9-(2x+3)²=
9(x-2)²-4(3x-7)²=
4(5x+3)²-25(x-2)²=
(2x²-2x+5)²-(x²+2x+2)²=
x²+3x+2=
x²+5x+4=
x²+7x+12=
x²+8x+15=
x²+9x+14=
x2+11x=30=
x²+15x+56=
x2+12x+35=
x²-5x+6=
x²-9x+20=

Answer 1

1) 5x(x la puterea 2-9) 
2) 7x(x la puterea 2-4) 
3) x(5-16) 
4) x la puterea 3(x+1) la puterea 2
de la 5-15 nu stiu sa fac 
16) (x+1)(x+2) 
17) (x+1)(x+4) 
18) (x+3)(x+4) 
19) (x+5)(x+3) 
20) (x+2)(x+7) 
21) - 
22)  
23) (x+5)(x+7) 
24) (x-5)(x+1) 
25) -

Answer 2

5x³-45x=
7x³-28x=7x(x²-4)=7x(x-2)(x+2)
x⁵-16x=x(x⁴-16)=x(x²-4)(x²+4)=x(x+2)(x-2)(x²+4)
x³+x²+x+1=x²(x+1)+(x+1)=(x+1)(x²+1)
x³+3x²-x-3=x²(x+3)-(x+3)=(x+3)(x²-1)=(x+3)(x-1)(x+1)
x³-2x²-x+2=x²(x-2)-(x-2)=(x-2)(x²-1)=(x-2)(x-1)(x+1)
x³-4x²-4x+16=x²(x-4)-4(x-4)=(x-4)(x²-4)=(x-4)(x-2)(x+2)
x³+3x²-4x-12=x²(x+3)-4(x+3)=(x+3)(x²-4)=(x+3)(x-2)(x+2)
x³+3x²-9x-27=x²(x+3)-9(x+3)=(x+3)(x²-9)=(x+3)(x-3)(x+3)
x³+2x²-16(x+2)=x²(x+2)-16(x+2)=(x+2)(x²-16)=(x+2)(x-4)(x+4)
9(2x-5)²-4(4x-3)²=[3(2x-5)-2(4x-3)][3(2x-5)+2(4x-3)]=
=(6x-15-8x+6)(6x-15+8x-6)=(-2x-9)(14x-21)=7(-2x-9)(2x-3)
4x²-9-(2x+3)²=(2x-3)(2x+3)-(2x+3)²=(2x+3)(2x-3-2x-3)=-6(2x+3)
9(x-2)²-4(3x-7)²=[3(x-2)-2(3x-7)][3(x-2)+2(3x-7)]=
=(3x-6-6x+14)(3x-6+6x-14)=(-3x+8)(9x-20)
4(5x+3)²-25(x-2)²=[2(5x+3)-5(x-2)][2(5x+3)+5(x-2)]=
=(10x+6-5x+10)(10x+6+5x-10)=(5x+16)(15x-4)
(2x²-2x+5)²-(x²+2x+2)²=(2x²-2x+5-x²-2x-2)(2x²-2x+5+x²+2x+2)=
=(2x²-2x+5-x²-2x-2)(2x²-2x+5+x²+2x+2)=(x²-4x+3)(3x²+7)=(x²-x-3x+3)(3x²+7)=
=[x(x-1)-3(x-1)](3x²+7)=(x-1)(x-3)(3x²+7)
x²+3x+2=x²+2x+x+2=x(x+2)+(x+2)=(x+2)(x+1)
x²+5x+4=x²+4x+x+4=x(x+4)+(x+4)=(x+4)(x+1)
x²+7x+12=x²+4x+3x+12=x(x+4)+3(x+4)=(x+4)(x+3)
x²+8x+15=x²+5x+3x+15=x(x+5)+3(x+5)=(x+5)(x+3)
x²+9x+14=x²+7x+2x+14=x(x+7)+2(x+7)=(x+7)(x+2)
x+11x+30=x²+6x+5x+30=x(x+6)+5(x+6)=(x+6)(x+5)
x²+15x+56=x²+8x+7x+56=x(x+8)+7(x+8)=(x+8)(x+7)
x²+12x+35=x²+7x+5x+35=x(x+7)+5(x+7)=(x+7)(x+5)
x²-5x+6=x²-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)
x²-9x+20=x²-4x-5x+20=x(x-4)-5(x-4)=(x-4)(x-5)