Разложите на линейные множители квадратный трехчлен:
1) $x^2 - 7x + 12$;
2) $x^2 + 8x + 15$;
3) $x^2 - 3x - 10$;
4) $-x^2 - 5x - 6$;
5) $-x^2 + x + 2$;
6) $6x^2 - 5x - 1$;
7) $4x^2 + 3x - 22$;
8) $-3a^2 + 8a + 3$;
9) $\frac{1}{6}b^2 - \frac{5}{6}b + 1$;
10) $-2x^2 - 0,5x + 1,5$;
11) $0,4x^2 - 2x + 2,5$;
12) $-1,2m^2 + 2,6m - 1$.

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Решение

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ГДЗ учебник по алгебре 8 класс Мерзляк, Полонский, Якир. §22. Упражнения. Номер №753

Решение 1

$x^2 - 7x + 12 = 0$
$D = b^2 - 4ac = (-7)^2 - 4 * 1 * 12 = 49 - 48 = 1 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{7 + \sqrt{1}}{2 * 1} = \frac{7 + 1}{2} = \frac{8}{2} = 4$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{7 - \sqrt{1}}{2 * 1} = \frac{7 - 1}{2} = \frac{6}{2} = 3$
Ответ:$x^2 - 7x + 12 = (x - 3)(x - 4)$

Решение 2

$x^2 + 8x + 15 = 0$
$D = b^2 - 4ac = 8^2 - 4 * 1 * 15 = 64 - 60 = 4 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-8 + \sqrt{4}}{2 * 1} = \frac{-8 + 2}{2} = \frac{-6}{2} = -3$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-8 - \sqrt{4}}{2 * 1} = \frac{-8 -2}{2} = \frac{-10}{2} = -5$
$x^2 + 8x + 15 = (x - (-5))(x - (-3)) = (x + 5)(x + 3)$
Ответ: $x^2 + 8x + 15 = (x + 5)(x + 3)$

Решение 3

$x^2 - 3x - 10 = 0$
$D = b^2 - 4ac = (-3)^2 - 4 * 1 * (-10) = 9 + 40 = 49 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{3 + \sqrt{49}}{2 * 1} = \frac{3 + 7}{2} = \frac{10}{2} = 5$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{3 - \sqrt{49}}{2 * 1} = \frac{3 - 7}{2} = \frac{-4}{2} = -2$
$x^2 - 3x - 10 = (x - 5)(x - (-2)) = (x - 5)(x + 2)$
Ответ: $x^2 - 3x - 10 = (x - 5)(x + 2)$

Решение 4

$-x^2 - 5x - 6 = 0$
$D = b^2 - 4ac = (-5)^2 - 4 * (-1) * (-6) = 25 - 24 = 1 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{5 + \sqrt{1}}{2 * (-1)} = \frac{5 + 1}{-2} = \frac{6}{-2} = -3$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{5 - \sqrt{1}}{2 * (-1)} = \frac{5 - 1}{-2} = \frac{4}{-2} = -2$
$-x^2 - 5x - 6 = -(x - (-3))(x - (-2)) = -(x + 3)(x + 2)$
Ответ: $-x^2 - 5x - 6 = -(x + 3)(x + 2)$

Решение 5

$-x^2 + x + 2 = 0$
$D = b^2 - 4ac = 1^2 - 4 * (-1) * 2 = 1 + 8 = 9 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-1 + \sqrt{9}}{2 * (-1)} = \frac{-1 + 3}{-2} = \frac{2}{-2} = -1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-1 - \sqrt{9}}{2 * (-1)} = \frac{-1 - 3}{-2} = \frac{-4}{-2} = 2$
$-x^2 + x + 2 = -(x - (-1))(x - 2) = -(x + 1)(x - 2) = (x + 1)(2 - x)$
Ответ: $-x^2 + x + 2 = (x + 1)(2 - x)$

Решение 6

$6x^2 - 5x - 1 = 0$
$D = b^2 - 4ac = (-5)^2 - 4 * 6 * (-1) = 25 + 24 = 49 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{5 + \sqrt{49}}{2 * 6} = \frac{5 + 7}{12} = \frac{12}{12} = 1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{5 - \sqrt{49}}{2 * 6} = \frac{5 - 7}{12} = \frac{-2}{12} = -\frac{1}{6}$
$6x^2 - 5x - 1 = 6(x - (-\frac{1}{6}))(x - 1) = 6(x + \frac{1}{6})(x - 1) = (6x + 1)(x - 1)$
Ответ: $6x^2 - 5x - 1 = (6x + 1)(x - 1)$

Решение 7

$4x^2 + 3x - 22 = 0$
$D = b^2 - 4ac = 3^2 - 4 * 4 * (-22) = 9 + 352 = 361 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-3 + \sqrt{361}}{2 * 4} = \frac{-3 + 19}{8} = \frac{16}{8} = 2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-3 - \sqrt{361}}{2 * 4} = \frac{-3 - 19}{8} = \frac{-22}{8} = -\frac{11}{4}$
$4x^2 + 3x - 22 = 4(x - (-\frac{11}{4}))(x - 2) = 4(x + \frac{11}{4})(x - 2) = (4x + 11)(x - 2)$
Ответ: $4x^2 + 3x - 22 = (4x + 11)(x - 2)$

Решение 8

$-3a^2 + 8a + 3 = 0$
$D = b^2 - 4ac = 8^2 - 4 * (-3) * 3 = 64 + 36 = 100 > 0$
$a_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-8 + \sqrt{100}}{2 * (-3)} = \frac{-8 + 10}{-6} = \frac{2}{-6} = -\frac{1}{3}$
$a_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-8 - \sqrt{100}}{2 * (-3)} = \frac{-8 - 10}{-6} = \frac{-18}{-6} = 3$
$-3a^2 + 8a + 3 = -3(a - (-\frac{1}{3}))(a - 3) = -3(a + \frac{1}{3})(a - 3) = (3a + 1)(3 - a)$
Ответ: $-3a^2 + 8a + 3 = (3a + 1)(3 - a)$

Решение 9

$\frac{1}{6}b^2 - \frac{5}{6}b + 1 = 0$
$D = b^2 - 4ac = (-\frac{5}{6})^2 - 4 * \frac{1}{6} * 1 = \frac{25}{36} - \frac{2}{3} = \frac{25 - 24}{36} = \frac{1}{36} > 0$
$b_1 = \frac{-b + \sqrt{D}}{2a} = \frac{\frac{5}{6} + \sqrt{\frac{1}{36}}}{2 * \frac{1}{6}} = \frac{\frac{5}{6} + \frac{1}{6}}{\frac{1}{3}} = \frac{1}{\frac{1}{3}} = 3$
$b_2 = \frac{-b - \sqrt{D}}{2a} = \frac{\frac{5}{6} - \sqrt{\frac{1}{36}}}{2 * \frac{1}{6}} = \frac{\frac{5}{6} - \frac{1}{6}}{\frac{1}{3}} = \frac{\frac{4}{6}}{\frac{1}{3}} = \frac{\frac{2}{3}}{\frac{1}{3}} = 2$
$\frac{1}{6}b^2 - \frac{5}{6}b + 1 = \frac{1}{6}(b - 3)(b - 2) = \frac{1}{3}(b - 3) * \frac{1}{2}(b - 2) = (\frac{1}{3}b - 1)(\frac{1}{2}b - 1)$
Ответ: $\frac{1}{6}b^2 - \frac{5}{6}b + 1 = (\frac{1}{3}b - 1)(\frac{1}{2}b - 1)$

Решение 10

$-2x^2 - 0,5x + 1,5 = 0$
$D = b^2 - 4ac = (-0,5)^2 - 4 * (-2) * 1,5 = 0,25 + 12 = 12,25 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{0,5 + \sqrt{12,25}}{2 * (-2)} = \frac{0,5 + 3,5}{-4} = \frac{4}{-4} = -1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{0,5 - \sqrt{12,25}}{2 * (-2)} = \frac{0,5 - 3,5}{-4} = \frac{3}{4}$
$-2x^2 - 0,5x + 1,5 = -2(x - (-1))(x - \frac{3}{4}) = -2(x + 1)(x - \frac{3}{4}) = (x + 1)(\frac{3}{2} - x) = (x + 1)(1,5 - x)$
Ответ: $-2x^2 - 0,5x + 1,5 = (x + 1)(1,5 - x)$

Решение 11

$0,4x^2 - 2x + 2,5 = 0$
$D = b^2 - 4ac = (-2)^2 - 4 * 0,4 * 2,5 = 4 - 4 = 0$
$x = \frac{-b + \sqrt{D}}{2a} = \frac{2 + \sqrt{0}}{2 * 0,4} = \frac{2}{0,8} = \frac{20}{8} = \frac{5}{2} = 2,5$
$0,4x^2 - 2x + 2,5 = 0,4(x - 2,5)(x - 2,5) = 0,4(x - 2,5)^2$
Ответ: $0,4x^2 - 2x + 2,5 = 0,4(x - 2,5)^2$

Решение 12

$-1,2m^2 + 2,6m - 1 = 0$
$D = b^2 - 4ac = 2,6^2 - 4 * (-1,2) * (-1) = 6,76 - 4,8 = 1,96 > 0$
$m_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-2,6 + \sqrt{1,96}}{2 * (-1,2)} = \frac{-2,6 + 1,4}{-2,4} = \frac{-1,2}{-2,4} = 0,5$
$m_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-2,6 - \sqrt{1,96}}{2 * (-1,2)} = \frac{-2,6 - 1,4}{-2,4} = \frac{-4}{2,4} = \frac{40}{24} = \frac{5}{3}$
$-1,2m^2 + 2,6m - 1 = 0 = -1,2(m - 0,5)(m - \frac{5}{3}) = (m - 0,5)(\frac{12}{10} * \frac{5}{3} -1,2m) = (m - 0,5)(\frac{6}{5} * \frac{5}{3} -1,2m) = (m - 0,5)(2 -1,2m)$
Ответ: $-1,2m^2 + 2,6m - 1 = (m - 0,5)(2 -1,2m)$