2017-I-14

$\begin{array}{rcl}
f(x) & = & (2x^2+ax+4)(3x+7) + (bx+c) \\
f(x) & = & 6x^2 + (3a+14)x^2 +(12 + 7a + b)x + (28+c)
\end{array}$

By comparing the coefficients $x^2$ of the polynomial, we have

$\begin{array}{rcl}
-13 & = & 3a+14 \\
3a & = & -27 \\
a & = & -9
\end{array}$

1. By the division algorithm, we have

   $\begin{array}{rcl}
   g(x) & = & (2x^2+ax+4)Q(x) + (bx+c)
   \end{array}$,

   for some polynomial $Q(x)$. Therefore, we have

   $\begin{array}{cl}
   & f(x) – g(x) \\
   = & [(2x^2+ax+4)(3x+7)+(bx+c)] – [(2x^2+ax+4)Q(x) + (bx+c)] \\
   = & (2x^2+ax+4)[(3x+7) – Q(x)]
   \end{array}$

   Since $(3x+7) – Q(x)$ is a polynomial, then by the division algorithm $f(x) – g(x)$ is divisible by $2x^2+ax+4$.

2. Consider $f(x)-g(x)=0$. By the results of (a) and (b)(i), we have

   $\begin{array}{rcl}
   (2x^2-9x+4)[(3x+7) – Q(x)] & = & 0 \\
   \end{array}$

   $\therefore 2x^2-9x+4=0$ or $(3x+7) – Q(x)=0$.

   Consider $2x^2-9x+4=0$, we have

   $\begin{array}{rcl}
   2x^2-9x+4 & = & 0 \\
   (2x – 1)(x-4) & = & 0
   \end{array}$

   $\therefore x=\dfrac{1}{2}$ or $x=4$.

   Since $\dfrac{1}{2}$ is not an integer, then not all the roots of the equation $f(x)-g(x)=0$ are integers. I don’t agree.