Question

ajutați maaaaa ex 2 și 3​

Answer 1

Explicație pas cu pas:

descompunere în produs de factori ireductibili:

2.

${(3x + 5)}^{3} - 12x - 20 =$

$= {(3x + 5)}^{3} - 4(x + 5)$

$= (3x + 5)[{(3x + 5)}^{2} - {2}^{2} ]$

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

$= (3x + 5)(3x + 3)(3x + 7)$

$= 3(x + 1)(3x + 5)(3x + 7)$

3.

$( {x}^{2} + x)( {x}^{2} + x - 8) + 12 =$

$= {( {x}^{2} + x)}^{2} - 8( {x}^{2} + x) + 12$

$= {( {x}^{2} + x)}^{2} - 2 \cdot 4( {x}^{2} + x) + {4}^{2} - 4$

$= {({x}^{2} + x - 4)}^{2} - {2}^{2}$

$= ({x}^{2} + x - 4 - 2)({x}^{2} + x - 4 + 2)$

$= ({x}^{2} + x - 6)({x}^{2} + x - 2)$

$= (x - 2)(x + 3)(x - 1)(x + 2)$

Answer 2

a)

$E(x)=(3x+5)^3-12x-20=\\(3x+5)^3-4(3x+5)=\\(3x+5)((3x+5)^2-4)=\\(3x+5)((3x+5)^2-2^2)=\\(3x+5)(3x+5-2)(3x+5+2)=\\(3x+5)(3x+3)(3x+7)=\\3(x+1)(3x+5)(3x+7)$

b)

Notăm $x^2+x=t$

$t(t-8)+12=t^2-8t+12=(t-2)(t-6)$

$(x^2+x)(x^2+x-8)+12=(t-2)(t-6)=(x^2+x-2)(x^2+x-6)=(x-1)(x+2)(x-2)(x+3)$