Question

a X la a doua +4x+3
b. X la a doua+11x+18

Answer 1

$a) {x}^{2} + 4x + 3 = {x}^{2} + x + 3x + 3 = x( \frac{ {x}^{2} }{x} + \frac{x}{x}) + 3( \frac{3x}{3} + \frac{3}{3}) = x( {x}^{2 - 1} + 1) + 3(x + 1) = x(x + 1) + 3(x + 1) = (x + 1)( \frac{x(x + 1)}{x + 1} + \frac{3(x + 1)}{x + 1}) = (x + 1)(x + 3)$

$b) {x}^{2} + 11x + 18 = {x}^{2} + 2x + 9x + 18 = x( \frac{ {x}^{2} }{x} + \frac{2x}{x}) + {3}^{2}( \frac{ {3}^{2} \times x }{ {3}^{2} } + \frac{2 \times {3}^{2} }{ {3}^{2} }) = x( {x}^{2 - 1} + 2) + 9(x + 2) = x(x + 2) + 9(x + 2) = x(x + 2) + {3}^{2}(x + 2) = (x + 2)( \frac{x(x + 2)}{x + 2} + \frac{ {3}^{2}(x + 2) }{x + 2}) = (x + 2)(x + {3}^{2}) = (x + 2)(x + 9)$