The polynomial $p(x)$ is given by $p(x) = x^3 + (b + 2)x^2 + 2(b + 2)x + 8$ where $b$ is a constant - AQA - A-Level Maths Pure - Question 11 - 2022 - Paper 1

The graph of a cubic function can intersect the x-axis at most three times. Since $p(x)$ meets the x-axis at exactly two points, this implies there is one repeated root.

The possible forms of the polynomial are:
$p(x) = k(x - r)^2 (x - s)$
for some constants $k, r,$ and $s$.

Since one of the roots is $x = -2$, we can express the polynomial as:
$p(x) = k(x + 2)^2(x - s)$.
The remaining factor $(x - s)$ needs to ensure the graph meets the x-axis only twice. Hence, if it has only one distinct real root, it must be suggested that $s eq -2$.

The graph is a cubic function with two x-intercepts, meaning it has a local maximum or minimum.
Starting from the left:

• As $x$ approaches -rac{7}{2}, $p(x)$ goes to positive infinity.
• At $x = -2$, the graph touches (but does not cross) the x-axis, indicating a double root.
• As $x$ approaches the second root, $p(x)$ will cross the x-axis and continue decreasing.
• Finally, as $x$ approaches positive infinity, the function $p(x)$ goes to positive infinity.

To find the value of $b$, we first expand $p(x)$:
$p(x) = (x + 2)(x^2 + bx + 4) = x^3 + bx^2 + 4x + 2x^2 + 8$
Combining like terms yields:
$p(x) = x^3 + (b + 2)x^2 + 4x + 8.$

Next, equate the coefficients with the original polynomial formed earlier and determine the condition needed for one root.
Since we established one repeated root and that $b$ must satisfy $b^2 - 16 = 0$, solving for $b$, we have:
$b = 4 ext{ or } b = -4.$
We verify which gives us exactly two real roots. Substituting both values, we find that $b = 4$ provides the correct single point of intersection, confirming the final answer.