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descompuneti in factori (x-3)(x-1)(x+1)(x+3)+15

x(x^2+3x+4)(x+3)+4

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


Răspuns :

Explicație pas cu pas:

(x-3)(x-1)(x+1)(x+3)+15=(x²-1)(x²-9)+15=x⁴-x²-9x²+9+15=

x⁴-10x²+24=x⁴-6x²-4x²+24=x²(x²-6)-4(x²-6)=(x²-6)(x²-4)=

(x+2)(x-2)(x²-6)

x(x²+3x+4)(x+3)+4=(x²+3x)(x²+3x+4)+4

Notam t=x²+3x

t(t+4)+4=t²+4t+4=(t+2)²=(x²+3x+2)²=(x²+x+2x+2)²=

[x(x+1)+2(x+1)]²=[(x+1)(x+2)]²=(x+1)²*(x+2)²

(x²+x-1)+3(x²+x-1)+2=(x²+x-1)(1+3)+2=4(x²+x-1)+2=

4x²+4x-4+2=4x²+4x+2