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We want all real values of $x$ such that

$x^2 + 7x + 6 \ge 0.$

This is a quadratic inequality, so the standard method is:

1. Factor the quadratic.
2. Find its zeros.
3. Use the sign of the quadratic on each interval determined by those zeros.

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1. Factor the quadratic

We look for two numbers whose product is $6$ and whose sum is $7$.
Those numbers are $1$ and $6$, because $1 \cdot 6 = 6$ and $1 + 6 = 7$.

So we can factor:

$x^2 + 7x + 6 = (x + 1)(x + 6).$

Thus our inequality becomes

$(x + 1)(x + 6) \ge 0.$

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2. Find the zeros (where it equals 0)

Set the expression equal to zero:

$(x + 1)(x + 6) = 0.$

This happens when either factor is zero:

• $x + 1 = 0 \Rightarrow x = -1,$
• $x + 6 = 0 \Rightarrow x = -6.$

So the quadratic is zero at $x = -6$ and $x = -1$.

These two points divide the real line into three intervals:

1. $(-\infty, -6)$
2. $[-6, -1]$
3. $(-1, \infty)$

We will study the sign of $(x + 1)(x + 6)$ on each interval.

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3. Sign analysis of the product

Because the quadratic has positive leading coefficient (the coefficient of $x^2$ is $1 > 0$), its graph is a parabola opening upwards. That means:

• It is negative between its roots.
• It is positive outside its roots.
• It is zero exactly at the roots.

We can also confirm this with a sign chart.

Interval 1: $x < -6$

Take a test value, for example $x = -7$:

• $x + 6 = -7 + 6 = -1 < 0,$
• $x + 1 = -7 + 1 = -6 < 0.$

Both factors are negative, so their product is positive:

$(x + 1)(x + 6) > 0 \quad \text{for } x < -6.$

Interval 2: $-6 < x < -1$

Take a test value, for example $x = -3$:

• $x + 6 = -3 + 6 = 3 > 0,$
• $x + 1 = -3 + 1 = -2 < 0.$

One factor is positive and the other negative, so the product is negative:

$(x + 1)(x + 6) < 0 \quad \text{for } -6 < x < -1.$

Interval 3: $x > -1$

Take a test value, for example $x = 0$:

• $x + 6 = 0 + 6 = 6 > 0,$
• $x + 1 = 0 + 1 = 1 > 0.$

Both factors are positive, so the product is positive:

$(x + 1)(x + 6) > 0 \quad \text{for } x > -1.$

At the roots

At $x = -6$ and $x = -1$, we have

$(x + 1)(x + 6) = 0.$

Since our inequality is $\ge 0$, those points are included in the solution set.

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4. Solution set

Putting all this together:

• The expression is positive for $x < -6$ and $x > -1$.
• The expression is zero at $x = -6$ and $x = -1$.
• The expression is negative for $-6 < x < -1$.

Therefore the set of all real $x$ such that

$x^2 + 7x + 6 \ge 0$

is

$x \le -6 \quad \text{or} \quad x \ge -1.$

In interval notation: $(-\infty, -6] \cup [-1, \infty).$

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5. Match with the answer choices

Now compare with the options:

• A. $-6 \le x \le -1$ → Only the interval between the roots. That is where the expression is negative, except at the endpoints. So A is incorrect.
• B. $-6 \le x \le 1$ → Includes points between $-6$ and $-1$, where the expression is negative, and misses all $x > 1$ and $x < -6$. Incorrect.
• C. $1 \le x \le 6$ → Only a small part of where the expression is positive; it excludes $x \le -6$ and $-1 \le x < 1$ and $x > 6$. Incorrect.
• D. $x \le -6 \text{ or } x \ge -1$ → Exactly matches our solution set. This is correct.
• E. $x \le 1 \text{ or } x \ge 6$ → Includes some values (like $0$) where the expression is positive, but also includes values (like $-3$) where the expression is negative; and it misses values less than $-6$. Incorrect.

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Correct answer: D. $x \le -6 \text{ or } x \ge -1.$