Question

Descompuneți in factori folosind .entoda factorului comun:
3x(x-2)+5(x-2)

2a +4b+6c

(x+1)^2-3(x+1)+(x+1)

24ab +12a^2b^3-6^4b^2

Answer 1

○ = (x-2)(3x+5) ○ = 2(a+2b+3c) ○ = (x+1)(x+1-3+1) = (x+1)(x-1) ○ = 6ab(4+2ab² - a^3 b) (rezolvare pt "6^4 b²") Rezolvarea pt ultimul subpunct asa cum este scris de tine este: ○ = b(24a+12a²b²-6^4 b) = b(24a+12a²b²-1296b) = b*12(2a+a²b²-108b)

Answer 2

a)

3x(x-2)+5(x-2)=

3x²-6x+5x-10=

3x²+(-6x+5x)-10=

3x²-x-10

b)

2a+4b+6c=

$2( \frac{2a}{2} + \frac{4b}{2} + \frac{6c}{2}) =$

2(a+2b+3c)

c)

(x+1)²-3(x+1)+(x+1)=

(x+1)²-3(x+1)+x+1=

x²+2x+1-3x-3+x+1=

x²+(2x-3x+x)+(1-3+1)=

x²-1