Question

Descompuneți în factori:
a) (2x + 3)^2 - (x^2 + 2x + 1)
b) (x^2 + x + 2)^2 – x^2 + 4x – 4;
c)x^2+8x+16-(x^2-2x+5)^2
d) (x + 2)^4 - (x + 1)^4
f) (2x-1)^8 – (2x – 1)^8
e) (2x - 1)^4-(x + 3)^4​


Rapid va rog!!!!!

Answer 1

Exercițiul f) nu pare scris corect

Nu trebuie ultimul pas, poți sa lași și sub forma de înmulțire de paranteze

Answer 2

a)(2x+3)²-(x²+2x+1)=

(2x+3)²-(x+1)²=

[2x+3-(x+1)]×[2x+3+(x+1)]=

(2x+3-x-1)×(2x+3+x+1)=

(x+2)×(3x+4)

b)(x²+x+2)²-x²+4x-4=

x⁴+x²+4+2x³+4x²+4x-x²+4x-4=

x⁴+2x³+4x²18x

c)x²+8x+16-(x²-2x+5)²=

x²+8x+16-(x⁴+14x²+25-4x³-20x)=

x²+8x+16-x⁴-14x²-25+4x³+20x=

-13x²+28x-9-x⁴+4x³=

-x⁴+4x³-13x²+28x-9

d)(x+2)⁴-(x+1)⁴=

[(x+2)²-(x+1)²]×[(x+2)²+(x+1)²]=

[x+2-(x+1)]×[x+2+(x+1)]×(x²+4x+4+x²+2x+1)

(x+2-x-1)×(x+2+x+1)×(2x²+6x+5)=

1(2x+3)×(2x²+6x+5)

f)(2x-1)⁸-(2x-1)⁸=

0

e)(2x-1)⁴-(x+3)⁴=

[(2x-1)²-(x+3)²]×[(2x-1)²+(x+3)²]=

[2x-1-(x+3)]×[2x-1+(x+3)]×(4x²-4x+1+x²+6x+9)=

(2x-1-x-3)×(2x-1+x+3)×(5x²+2x+10)=

(x-4)×(3x+2)×(5x²+2x+10)