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

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

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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.} $$ $$ m = 32 $$ and $$ n = 48 $$.

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

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