Question

Descompuneți in factori:

A)x²+x-2=
B)x2-x-2=
C)x²+2x-3=
D)x²+2x-3=
E)x²-12x+32=
F)(x²-x)(x²-x+4)+4=
G)(x²+x)(x²+x+2)+1=

Help pls!

Answer 1

$a) {x}^{2} + x - 2$

$= {x}^{2} +2x - x - 2$

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

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

$b) {x}^{2} - x - 2$

$= {x}^{2} - 2x + x - 2$

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

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

$c) {x}^{2} + 2x - 3$

$= {x}^{2} + 3x - x - 3$

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

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

$e) {x}^{2} - 12x + 32$

$= {x}^{2} - 4x - 8x + 32$

$= x(x - 4) - 8(x - 4)$

$= (x - 8)(x - 4)$

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

${x}^{2} - x = t$

$t(t + 4) + 4 = {t}^{2} + 4t + 4$

$= {t}^{2} + 2t + 2t + 4$

$= t(t + 2) + 2(t + 2)$

$= (t + 2)(t + 2) = {(t + 2)}^{2}$

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

$g)( {x}^{2} + x)( {x}^{2} + x + 2) + 1$

${x}^{2} + x = t$

$t(t + 2) + 1 = {t}^{2} + 2t + 1$

$= {t}^{2} + t + t + 1$

$= t(t + 1) + 1(t + 1)$

$= (t + 1)(t + 1)$

$= {(t + 1)}^{2}$

$= {( {x}^{2} + x + 1)}^{2}$

Answer 2

A)x²+x-2=x²+2x-x-2=x(x+2)-(x+2)=(x+2)(x-1)

B)x²-x-2=x²+x-2x-2=x(x+1)-2(x+1)=(x+1)(x-2)

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

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

E)x²-12x+32=x²-4x-8x+32=x(x-4)-8(x-4)=(x-4)(x-8)

F)(x²-x)(x²-x+4)+4=(x²-x)(x²-x)+4(x²-x)+4=(x²-x)²+2×2×(x²-x)+2²=(x²-x+2)²

G)(x²+x)(x²+x+2)+1=(x²+x)(x²+x)+2×(x²+x)+1=(x²+x)²+2×(x²+x)+1=(x²+x+1)²