Question

Разложить на множители квадратный трехчлен,

Answer 1

$1) x^2+3x-88=0\\D=\sqrt{3^2-4*1*(-88)} =\sqrt{9+352} =\sqrt{361} =б19\\\\x_1=\frac{-3+19}{2} =8\\\\x_2=\frac{-3-19}{2} =-11\\\\(x-8)(x+11)\\\\\\2)2x^2-3x+1=0\\D=\sqrt{(-3)^2-4*2*1} =\sqrt{9-8} =\sqrt{1} =б1\\\\x_1=\frac{3+1}{4} =1\\\\x_2=\frac{3-1}{4} =\frac{2}{4} =\frac{1}{2} \\\\2(x-\frac{1}{2} )(x-1)=(2x-1)(x-1)\\\\\\3)3x^2+11x-14=0\\D=\sqrt{11^2-4*3*(-14)} =\sqrt{121+168} =\sqrt{289} =б17\\\\x_1=\frac{-11+17}{6} =1\\\\x_2=\frac{-11-17}{6}=-\frac{28}{6} =-\frac{14}{3} \\\\3(x+\frac{14}{3})(x-1)=(3x+14)(x-1)$

Answer 2

1)

  x²+3x-88 =

=  x²+11x - 8x - 88 =

= (x²+11x) - (8x + 88) =

= x·(x+11) - 8·(x + 11) =

= (x+11)·(x -8)

2)

    2x² - 3x + 1 =

=  2x²- 2x - x + 1 =

=  (2x²- 2x) - (x - 1) =

=  2x·(x- 1) - 1·(x - 1) =

= (x- 1)·(2x - 1)

3)

    3x² + 11x - 14 =

=  3x² + 14x - 3x - 14 =

=  (3x² - 3x)+ (14x - 14) =

=  3x·(x - 1)+ 14·(x - 1) =

=  (x - 1)·(3x + 14)