Question

Descompuneți în factori :

a) bx-ax+a-b=?
b)x^2-4x-77=?
c)x^6-x^4+x^2-1= ?

Answer 1

a) bx-ax+a-b = b(x-1) - a(x-1) = (x-1)(b-a)

b) x²-4x-77 = 0 => x₁ = [4 +√(16+308)] /2 = (4+18)/2 = 11  ; x₂ = (4-18)/2 = -7

x²-4x-77 = (x+7)(x-11)

c) x⁶-x⁴+x²-1 = x²(x⁴+1) - (x⁴+1) = (x²-1)(x⁴+1) = (x-1)(x+1)(x⁴+1)

Answer 2

$a)bx - ax + a - b = \\ \\ x( \frac{bx}{x} - \frac{ax}{x}) + (a - b) = \\ \\ x(b - a) + (a - b) = \\ \\ (b - a)( \frac{x(b - a)}{b - a} + \frac{a - b}{b - a}) = \\ \\ (b - a)(x - 1)$

$b) {x}^{2} - 4x - 77 = \\ \\ {x}^{2} - 11x + 7x - 77 = \\ \\ x( \frac{ {x}^{2} }{x} - \frac{11x}{x}) + 7( \frac{7x}{7} - \frac{7 \times 11}{7}) = \\ \\ x(x ^{2 - 1} - 11) + 7(x - 11) = \\ \\ x(x - 11) + 7(x - 11) = \\ \\ (x - 11)( \frac{x(x - 11)}{x - 11} + \frac{7(x - 11)}{x - 11}) = \\ \\ (x - 11)(x + 7)$

$c) {x}^{6} - {x}^{4} + {x}^{2} - 1 = \\ \\ {x}^{4}( \frac{ {x}^{6} }{ {x}^{4} } - \frac{ {x}^{4} }{ {x}^{4} }) + (x - 1)(x + 1) = \\ \\ {x}^{4}( {x}^{6 - 4} - 1) + (x - 1)(x + 1) = \\ \\ {x}^{4}( {x}^{2} - 1) + (x - 1)(x + 1) = \\ \\ {x}^{4}( {x}^{2} - {1}^{2}) + (x - 1)(x + 1) = \\ \\ {x}^{4}((x - 1)(x + 1)) + (x - 1)(x + 1) = \\ \\ {x}^{4}(x - 1)(x + 1) + (x - 1)(x + 1) = \\ \\ (x - 1)(x + 1)( \frac{ {x}^{4}(x - 1)(x + 1) }{(x - 1)(x + 1)} + \frac{(x - 1)(x + 1)}{(x - 1)(x + 1)}) = \\ \\ (x - 1)(x + 1)( {x}^{4} + 1)$