For how many integers $n$, will the inequality $(n - 5) (n - 10) - 3(n − 2) \leq 0$ be satisfied?

We start with: $(n - 5)(n - 10) - 3(n − 2) \leq 0$

Expanding $(n-5)(n-10)$:

$(n-5)(n-10) = n \cdot n + n \cdot (-10) + (-5) \cdot n + (-5) \cdot (-10)$

$= n^2 - 10n - 5n + 50 = n^2 - 15n + 50$

Expanding $-3(n-2)$:

$-3(n-2) = -3n + 6$

Combining everything:

$(n^2 - 15n + 50) + (-3n + 6) \leq 0$

$n^2 - 15n + 50 - 3n + 6 \leq 0$

$n^2 - 18n + 56 \leq 0$

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We need to factor $n^2 - 18n + 56$.

We're looking for two numbers that:

Multiply to give 56 (the constant term)

Add to give -18 (the coefficient of n)

Let's think: $56 = 4 \times 14$ and $4 + 14 = 18$

Since we need $-18$, we use $-4$ and $-14$:

$n^2 - 18n + 56 = (n - 4)(n - 14)$

So our inequality becomes: $(n - 4)(n - 14) \leq 0$

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For a product of two factors to be $\leq 0$, one factor must be positive and the other negative (or one of them equals zero).

Let's analyze the signs:

$(n - 4) = 0$ when $n = 4$

$(n - 14) = 0$ when $n = 14$

These are the roots of our quadratic.

Sign Analysis:

When $n < 4$: both $(n-4) < 0$ and $(n-14) < 0$, so $(n-4)(n-14) > 0$

When $n = 4$: $(n-4) = 0$, so $(n-4)(n-14) = 0$

When $4 < n < 14$: $(n-4) > 0$ and $(n-14) < 0$, so $(n-4)(n-14) < 0$

When $n = 14$: $(n-14) = 0$, so $(n-4)(n-14) = 0$

When $n > 14$: both $(n-4) > 0$ and $(n-14) > 0$, so $(n-4)(n-14) > 0$

Therefore: $(n-4)(n-14) \leq 0$ when $4 \leq n \leq 14$

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Since we need integer values of $n$ where $4 \leq n \leq 14$:

$n$ can be: $4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14$

Counting: $14 - 4 + 1 = 11$ values

Answer: The inequality is satisfied for 11 integers.

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When solving quadratic inequalities of the form $(x-a)(x-b) \leq 0$ where $a < b$:

The solution is always $a \leq x \leq b$ (between the roots, inclusive)

This is because the parabola opens upward and is negative between its roots