### Instruction: Find the largest value of $n$ such that $3x^2 +nx + 72$ can be factored as the product of two linear factors with integer coefficients.### Response: I notice that the leading coefficient is 3, which is prime, so the only possible linear factors are of the form $(3x \pm a)$ and $(x \pm b)$, where $a$ and $b$ are integers.
I also notice that the constant term is 72, which has many factors, so I have to consider different combinations of $a$ and $b$ that satisfy $ab = 72$.
To maximize $n$, I want to maximize the sum of the coefficients of $x$ in the linear factors, which is $(3b \pm a)$.
Since $a$ and $b$ have the same sign, the sum is maximized when they are both positive.
So I can write $3x^2 + nx + 72 = (3x + a)(x + b)$, where $a$ and $b$ are positive factors of 72.
Expanding the right-hand side, I get $3x^2 + (3b + a)x + ab$, which means that $n = 3b + a$.
Now I need to find the largest possible values of $a$ and $b$.
One way to do this is to list all the factors of 72 in pairs, starting from the largest ones: $(72, 1), (36, 2), (24, 3), (18, 4), (12, 6), (9, 8)$.
For each pair, I can calculate the value of $n$ by plugging in $a$ and $b$: $n = 3b + a = 3(1) + 72 = 75, 3(2) + 36 = 42, 3(3) + 24 = 33, 3(4) + 18 = 30, 3(6) + 12 = 30, 3(8) + 9 = 33$.
The largest value of $n$ among these is 75, which occurs when $a = 72$ and $b = 1$.
So the largest value of $n$ such that $3x^2 + nx + 72$ can be factored as the product of two linear factors with integer coefficients is 75.