Descompunere în produs de factori ireductibili
[tex]a) = (3x+1)^3 - 4(3x+1)[/tex]
[tex]= (3x+1)\cdot\big[(3x+1)^2 - 2^2\big][/tex]
[tex]= (3x+1)\cdot\big[(3x+1) - 2\big] \cdot\big[(3x+1)+2\big][/tex]
[tex]= (3x+1)\cdot\big(3x+1-2\big) \cdot\big(3x+1+2\big)[/tex]
[tex]= (3x+1)\cdot\big(3x-1\big) \cdot\big(3x+3\big)[/tex]
[tex]b) = (2x+3)^3 - 9(2x+3)[/tex]
[tex]= (2x+3) \cdot \big[(2x+3)^2 - 3^2\big][/tex]
[tex]= (2x+3)(2x+3-3)(2x+3+3)[/tex]
[tex]= 2x(2x+3)(2x+6)[/tex]
[tex]= 4x \cdot (2x+3) \cdot (x+3)[/tex]
[tex]c) = (x^2+2x-1)^2 - 2^2[/tex]
[tex]= (x^2+2x-1-2)(x^2+2x-1+2)[/tex]
[tex]= (x^2+2x-3)(x^2+2x+1)[/tex]
[tex]= (x^2-x+3x-3)(x+1)^2[/tex]
[tex]= \big[x(x-1)+3(x-1)\big] \cdot (x+1)^2[/tex]
[tex]= (x-1) \cdot (x+3) \cdot (x+1)^2[/tex]
[tex]d) = (x^2 + 4x - 3)^2 - 2^2[/tex]
[tex]= (x^2 + 4x - 3 - 2)(x^2 + 4x - 3 + 2)[/tex]
[tex]= (x^2 + 4x - 5)(x^2 + 4x - 1)[/tex]
Pentru x²+4x-5=0 avem Δ=16+20=36, de unde x₁=(-4-6)/2=-5, iar x₂=(-4+6)/2=1
Pentru x₂+4x-1=0 avem Δ=16+4=20, de unde x₁=(-4-2√5)/2=-2-√5, iar x₂=(-4+2√5)/2=-2+√5
[tex]= (x - 1) \cdot (x + 5) \cdot (x + 2 - \sqrt{5}) \cdot (x + 2 + \sqrt{5} )[/tex]
[tex]e) = (x^2 + 4x + 5)^2 - 1^2[/tex]
[tex]= (x^2 + 4x + 5 - 1)(x^2 + 4x + 5 + 1)[/tex]
[tex]= (x^2 + 4x + 4)(x^2 + 4x + 6)[/tex]
[tex]= (x + 2)^2 \cdot (x^2 + 4x + 6)[/tex]
unde x²+4x+6 = x²+4x+4+2 = (x+2)²+2 > 0 ⇒ x ∉ R
[tex]f) = (x^2+4x+1)^2 - 3^2[/tex]
[tex]= (x^2+4x+1-3)(x^2+4x+1+3)[/tex]
[tex]= (x^2+4x-2)(x^2+4x+4)[/tex]
Pentru x₂+4x-2=0 avem Δ=16+8=24, de unde x₁=(-4-2√6)/2=-2-√6, iar x₂=(-4+2√6)/2=-2+√6
[tex]= (x+2+\sqrt{6} ) \cdot (x+2-\sqrt{6} ) \cdot (x+2)^2[/tex]
