\( \left. \begin{array} { l } { ( x - 2 ) ( 4 x ^ { 2 } - m x + n ) \leq 0 \rightarrow ( - \infty , 6 ] } \\ { m = ? \quad n = ? } \end{array} \right. \)

We start with the inequality \[ (x-2)(4x^2 - m x + n) \le 0 \] and the information that its solution set is \[ (-\infty, 6]. \] This means that the inequality holds (i.e. the expression is nonpositive) for every \(x \le 6\) and is positive for \(x > 6\). Because the polynomial is cubic (i.e. degree 3) and its leading coefficient is positive (since multiplying out gives a leading term \(4x^3\)), its typical sign pattern when it has three distinct roots \(r_1 < r_2 < r_3\) is: - \(f(x) < 0\) for \(x < r_1\), - \(f(x) > 0\) for \(r_1 < x < r_2\), - \(f(x) < 0\) for \(r_2 < x < r_3\), - \(f(x) > 0\) for \(x > r_3\). For the inequality to hold for every \(x\le6\) (i.e. no intervals of positivity in that range), the sign-switch from negative to positive that normally occurs between the first and second roots must be removed. One way to guarantee this is if one of the roots has multiplicity 2 so that the polynomial does not change sign at that point. Since the factor \((x-2)\) is already present, a natural choice is to take \(x=2\) as a double root, and let the third (simple) root be \(x=6\). Then the factorization of the cubic polynomial would be \[ f(x)= (x-2)^2 \, k(x-6), \] where \(k\) is a constant chosen so that the leading coefficient is \(4\). Expanding the factor \((x-2)^2(x-6)\) first, we have: \[ (x-2)^2 = x^2 - 4x+4. \] Then, \[ (x-2)^2 (x-6)= (x^2-4x+4)(x-6). \] Multiplying term‐by‐term: \[ \begin{aligned} (x^2-4x+4)(x-6) &= x^3 - 6x^2 - 4x^2 + 24x+4x -24\\[1mm] &= x^3 - 10x^2 + 28x - 24. \end{aligned} \] Thus, \[ f(x)= k(x^3 - 10x^2+28x-24). \] Since the original cubic has leading coefficient \(4\) (because \((x-2)(4x^2- m x+n)\) expands with leading term \(4x^3\)), we choose \(k=4\). Hence, \[ f(x)= 4(x^3 - 10x^2+28x-24)= 4x^3 - 40x^2+112x-96. \] Now, we also have the original expression written as: \[ f(x)= (x-2)(4x^2 - m x+n). \] Let us expand this: \[ \begin{aligned} (x-2)(4x^2- m x+n) &= x(4x^2-mx+n) -2(4x^2-mx+n)\\[1mm] &= 4x^3 - m x^2+nx -8x^2+2m x-2n\\[1mm] &= 4x^3 + (-m-8)x^2+ (n+2m)x -2n. \end{aligned} \] We now match the coefficients of like powers of \(x\) from the two forms of \(f(x)\): 1. Coefficient of \(x^3\): \[ 4 = 4. \] 2. Coefficient of \(x^2\): \[ -m-8=-40. \] Solving for \(m\): \[ -m-8=-40\quad\Longrightarrow\quad m+8=40\quad\Longrightarrow\quad m=32. \] 3. Coefficient of \(x\): \[ n+2m=112. \] Substituting \(m=32\): \[ n+64=112\quad\Longrightarrow\quad n=48. \] 4. Constant term: \[ -2n = -96. \] This yields \[ n=48, \] which is consistent. Thus, the values are given by: \[ m=32 \quad \text{and} \quad n=48. \] It is also useful to check the sign of \(f(x)=4(x-2)^2(x-6)\). Notice that \((x-2)^2\) is always nonnegative. Therefore, the sign of \(f(x)\) is determined solely by the factor \((x-6)\). This means: - For \(x<6\), we have \((x-6)<0\) so \(f(x)<0\). - At \(x=6\), \(f(x)=0\). - For \(x>6\), \((x-6)>0\) so \(f(x)>0\). Thus, the inequality \(f(x)\le0\) is satisfied for \(x\le6\), which is exactly what was given. The final answer is: \[ \boxed{m=32,\quad n=48.} \]