Вправи
Вправи: 812 (1-2)
Готова відповідь до завдання 812 (1-2) з ГДЗ Алгебра 8 клас Тарасенкова.
Алгебра, 8 клас · Вправи

Текстова відповідь
812. Розкладіть на лінійні множники квадратний тричлен.
1) x² – 3x – 10 = (x + 2)(x – 5);
D = (–3)2 – 4 • 1 • (–10) = 9 + 40 = 49.
x1 = 3-492•1 = 3-72 = -42 = –2;
x2 = 3+492•1 = 3+72 = 102 = 5.
2) x² – 10x + 24 = (x – 4)(x – 6);
D = (–10)2 – 4 • 1 • 24 = 100 – 96 = 4.
x1 = 10-42•1 = 10-22 = 82 = 4;
x2 = 10+42•1 = 10+22 = 122 = 6.
3) –x² + 16x – 15 = –(x – 15)(x – 1);
D = 162 – 4 • (–1) • (–15) = 256 + 60 = 196.
x1 = -16-1962•(-1) = -16-14-2 = -30-2 = 15;
x2 = -16+1962•(-1) = -16+14-2 = -2-2 = 1.
4) x² – 2x – 15 = (x + 3)(x – 5);
D = (–2)2 – 4 • 1 • (–15) = 4 + 60 = 64.
x1 = 2-642•1 = 2-72 = -62 = –3;
x2 = 2+642•1 = 2+82•1 = 102 = 5.
5) –x² + 4x – 3 = –(x – 1)(x – 3);
D = 42 – 4 • (–1) • (–3) = 16 – 12 = 4.
x1 = -4-42•(-1( = -4-2-2 = -6-2 = 3;
x2 = -4+42•(-1) = -4+2-2 = -2-2 = 1.
6) x² – 6x – 7 = (x + 1)(x – 7);
D = (–6)2 – 4 • 1 • (–7) = 36 + 28 = 64.
x1 = 6-642•1 = 6-82 = -22 = –1;
x2 = 6+642•1 = 6+82 = 142 = 7.
7) x² + 4x + 4 = (x + 2)2;
8) –x² + 10x – 25 = –(x² – 10x + 25) = –(x – 5)².