Question

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Folosind descompunerea in factori rezolvati ecuatiile :
a) x^2 -14x + 40
b) x^2 -13x +40
c) x^2 +11x +30
d) x^2 +9x -10
e)x^2 -7x +12
f) x^2 +8x +7

Answer 1

$a) x^2-14x+40 = x^2-10x-4x+40 = x(x-10) - 4(x-10)= \\ = (x-4)(x-10) \Rightarrow\boxed{x_1 = 4, x_2 = 10} \\ \\ b) x^2-13x+40 = x^2-8x-5x+40 = x(x-8)-5(x-8) = \\ = (x-8)(x-5) \Rightarrow \boxed{x_1=8, x_2 = 5 } \\ \\ c) x^2+11x+30=x^2+5x+6x+30 = x(x+5)+6(x+5)= \\ =(x+5)(x+6) \Rightarrow \boxed{x_1 = -5, x_2=-6} \\ \\ d) x^2+9x-10= x^2-x+10x-10 = x(x-1)+10(x-1) =$
$=(x-1)(x+10)\Rightarrow \boxed{x_1 = 1, x_2 = -10} \\ \\ e) x^2-7x+12 = x^2-3x-4x+12 = x(x-3)-4(x-3)= \\ =(x-3)(x-4) \Rightarrow \boxed{x_1 = 3, x_2 = 4} \\ \\ f) x^2+8x+7=x^2+x+7x+7 = x(x+1)+7(x+1) = \\ =(x+1)(x+7) \Rightarrow \boxed{x_1 = -1, x_2 = -7}$

Answer 2

a)x²-14x+40=x²-10x-4x+40
                   =x(x-10)-4(x-10)
                   =(*x-10)(x-4) =>x₁=10 si x₂=4

b)x²-13x+40=x²-8x-5x+40
                   =x(x-8)-5(x-8)
                   =(x-8)(x-5) =>x₁=8 si x₂=5

c)x²+11x+30=x²+5x+6x+30
                   =x(x+5)+6(x+5)
                   =(x+5)(x+6) =>x₁=-5 si x₂=-6

d)x²+9x-10=x²+10x-x-10
                 =x(x+10)-(x+10)
                 =(x+10)(x-1) =>x₁=-10 si x₂=1

e)x²-7x+12=x²-3x-4x+12
                 =x(x-3)-4(x-3)
                 =(x-3)(x-4) =>x₁=3 si x₂=4

f)x²+8x+7=x²+7x+x+7
               =x(x+7)+(x+7)
               =(x+7)(x+1) =>x₁=-7 si x₂=-1