Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 292

\[\boxed{\mathbf{292}.}\]

\[1)\ \left( 6x^{2} + 7x + a \right)\ :(2x + 3) =\]

\[= 3x - 1\ (ост.\ a + 3)\]

\[a + 3 = 0\]

\[a = - 3.\]

\[P(x) = (2x + 3)(bx + c) =\]

\[= 2bx^{2} + 3bx + 2xc + 3c =\]

\[= 2bx^{2} + (3b + 1c)x + 3c\]

\[\left\{ \begin{matrix} 2b = 6\ \ \ \ \ \ \ \ \ \ \\ 3b + 2c = 7 \\ 3c = a\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ }\left\{ \begin{matrix} b = 3\ \ \ \\ c = - 1 \\ a = - 3 \\ \end{matrix} \right.\ \]

\[Ответ:при\ a = - 3.\]

\[= bx^{6} + (b + c)x^{5} + (c + d)x^{4} +\]

\[+ (d + k)x^{3} + (k + l)x^{2} +\]

\[+ (l + m)x + m\]

\[b = 1:\]

\[b + c = 1\]

\[c = 0.\]

\[c = 0:\]

\[c + d = - 4\]

\[d = - 4.\]

\[d + k = - 4\]

\[k = 0.\]

\[a + m = 4;\ \ \ a = m\]

\[2a = 4\]

\[a = 2.\]

\[Ответ:при\ a = 2.\]

\[3)\ P(x) =\]

\[= (x - 3)\left( bx^{2} + cx + d \right) =\]

\[= bx^{3} + (c - 3b)x^{2} +\]

\[+ (d - 3c)x - 3d\]

\[\left\{ \begin{matrix} b = 1\ \ \ \ \ \ \ \ \ \ \ \\ c - 3b = a \\ d - 3c = a \\ - 3d = - 15 \\ \end{matrix} \right.\ \rightarrow d = 5;\ \]

\[\ c = a + 3\]

\[d - 3c = a\]

\[5 - 3 \cdot (a + 3) = a\]

\[5 - 3a - 9 = a\]

\[4a = - 4\]

\[a = - 1,\]

\[Ответ:при\ a = - 1.\]

\[4)\ P(x) = (4x + 5)(bx + c) =\]

\[= 4bx^{2} + (5b + 4c)x + 5c\]

\[\left\{ \begin{matrix} \begin{matrix} 4b = - 4\ \ \ \ \ \ \\ 5b + 4c = a \\ \end{matrix} \\ 5c = 5\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} b = - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ c = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ a = - 5 + 4 = - 1 \\ \end{matrix} \right.\ \]

\[Ответ:при\ a = - 1.\]