При каких a и b многочлен p(x) =(a+b) x^5+abx^2+1 делится на x^2-3x+2

   1. Многочлен P(x) представим в виде произведения двух многочленов:

• ((a + b)x^3 + px^2 + qx + 1/2)(x^2 - 3x + 2) = (a + b)x^5 + abx^2 + 1;
• (a + b)x^5 + (-3(a + b) + p)x^4 + (2(a + b) - 3p + q)x^3 + (2p - 3q + 1/2)x^2 + (2q - 3/2)x + 1 = (a + b)x^5 + abx^2 + 1;
• (-3(a + b) + p)x^4 + (2(a + b) - 3p + q)x^3 + (2p - 3q + 1/2)x^2 + (2q - 3/2)x = abx^2;

• {-3(a + b) + p = 0;{2(a + b) - 3p + q = 0;{2p - 3q + 1/2 = ab;{2q - 3/2 = 0;
• {a + b = p/3;{2p/3 - 3p + 3/4 = 0;{2p - 9/4 + 1/2 = ab;{q = 3/4;
• {a + b = p/3;{8p - 36p + 9 = 0;{2p - 7/4 = ab;{q = 3/4;
• {a + b = p/3;{p = 9/28;{2p - 7/4 = ab;{q = 3/4;
• {p = 9/28;{q = 3/4;{a + b = 3/28;{18/28 - 7/4 = ab;
• {p = 9/28;{q = 3/4;{a + b = 3/28;{ab = -31/28.

   2. По теореме Виета a и b являются корнями уравнения:

• t^2 - 3/28t - 31/28 = 0;
• D = (3/28)^2 + 4 * 31/28 = 1/28^2 * (9 + 3472) = 1/28^2 * 3481;
• √D = 1/28 * 59 = 59/28;

• t = (3/28 ± 59/28)/2 = 3/56 ± 59/56;
• t1 = 3/56 - 59/56 = -56/56 = -1;
• t2 = 3/56 + 59/56 = 62/56 = 31/28;

      (a; b) = (-1; 31/28), (31/28; -1).

   Ответ: (-1; 31/28), (31/28; -1).