Question

разложите на множители выделив квадрат двучлена x^2-8x+12
9x^2-6x-8
cрoчHo, c пoдpoбHыM решением
не по дискриминату и не по теореме виетта

Answer 1

$1) x^2 - 8x + 12 = 0\\x^2 - 8x + 16 - 4 = 0 \\(x - 4)^2 = 4 \\x - 4 = \pm 2 \\x_1 = 6 \\x_2 = 2 \\ x^2 - 8x + 12 = (x - 6)(x - 2) \\ \\ 2) 9x^2-6x-8 = 0 \\x^2 - \frac{2}{3}x - \frac{8}{9} = 0 \\x^2 - \frac{2}{3}x + \frac{1}{9} - 1 = 0 \\(x - \frac{1}{3})^2 = 1 \\x - \frac{1}{3} = \pm 1 \\x_1 = -\frac{2}{3} \\x_2 = \frac{4}{3} \\ 9x^2-6x-8 = 9(x + \frac{2}{3})(x - \frac{4}{3})$

Answer 2

x² - 8x + 12 = x² - 2*x*4 + 4²-4² + 12 = (x-4)² -16+12 = (x-4)² - 2² = (x-4-2)(x-4+2) =

= (x-6)(x-2)

9x² - 6x - 8 = (3x)² - 2*3x*1 + 1-1 - 8 = (3x-1)² - 1-8 = (3x-1)² - 3² = (3x-1-3)(3x-1+3) =

= (3x-4)(3x+2)