Відповіді на вправи: 7.29

7.29. Розкладіть на множники квадратний тричлен:

1) x² – x – 12 = 0; x₁ = 4; x₂ = –3.

Отже, x² – x – 12 = (x – 4)(x + 3).

2) –x² + 2x + 35 = 0; x² – 2x – 35 = 0; x₁ = 7; x₂ = –5.

Отже, –x² + 2x + 35 = –(x – 7)(x + 5) = (7 – x)(x + 5).

3) 6x² + 11x – 2 = 0; D = 11² – 4 • 6 • (–2) = 121 + 48 = 169;

x₁ = -11+1312 = 212 = 16; x₂ = -11-1312 = –2412 = –2.

Отже, 6x² + 11x – 2 = 6(x – 16)(x + 2) = (6x – 1)(x + 2).

4) 23x² + 3x – 6 = 0 |• 3; 2x² + 9x – 18 = 0;

D = 9² – 4 • 2 • (–18) = 81 + 144 = 225;

x₁ = -9+154 = 64 = 32; x₂ = -9-154 = –244 = –6.

Отже, 23 х² + 3х – 6 = 23(х – 32)(х + 6) = (23х – 1)(х + 6).