Question

AJUTOR!!!!! DAU COROANA!!!

1) Descompuneti in factori:

a) 3x^2 +5x -2
b) 3x^2 - 22x +7
c) 3x^2 - 10x - 8

Answer 1

$a) {3x}^{2} + 5x - 2 = \\ \\ {3x}^{2} - x + 6x - 2 = \\ \\ x( \frac{ {3x}^{2} }{x} - \frac{x}{x}) + 2( \frac{2 \times 3x}{2} - \frac{2}{2}) = \\ \\ x( {3x}^{2 - 1} - 1) + 2(3x - 1) = \\ \\ x(3x - 1) + 2(3x - 1) = \\ \\ (3x - 1)( \frac{x(3x - 1)}{3x - 1} + \frac{2(3x - 1)}{3x - 1}) = \\ \\ (3x - 1)(x + 2)$

$b) {3x}^{2} - 22x + 7 = \\ \\ 3x - 21x - x + 7 = \\ \\ 3x( \frac{ {3x}^{2} }{3x} - \frac{3 \times 7x}{3x}) + ( - x + 7) = \\ \\ 3x( {x}^{2 - 1} - 7) + ( - x + 7) = \\ \\ 3x(x - 7) + ( - x + 7) = \\ \\ (x - 7)( \frac{3x(x - 7)}{x - 7} + \frac{ - x + 7}{x - 7}) = \\ \\ (x - 7)(3x - 1)$

$c) {3x}^{2} - 10x - 8 = \\ \\ {3x}^{2} - 12x + 2x - 8 = \\ \\ 3x( \frac{ {3x}^{2} }{3x} - \frac{ {2}^{2} \times 3x }{3x}) + 2( \frac{2x}{2} - \frac{ {2}^{3} }{2}) = \\ \\ 3x( {x}^{2 - 1} - ( {2}^{2})) + 2(x - ( {2}^{3 - 1})) = \\ \\ 3x(x - 4) + 2(x - ( {2}^{2})) = \\ \\ 3x(x - 4) + 2(x - 4) = \\ \\ (x - 4)( \frac{3x(x - 4)}{x - 4} + \frac{2(x - 4)}{x - 4}) = \\ \\ (x - 4)(3x + 2)$