Let $P(x) = x^3 + 3x^2 - 13x + 6.$ (i) Show that $P(2) = 0.$ (ii) Hence, factor the polynomial $P(x)$ as $A(x)B(x)$, where $B(x)$ is a quadratic polynomial - HSC - SSCE Mathematics Extension 1 - Question 11 - 2020 - Paper 1

Since $P(2) = 0$, we know that $(x - 2)$ is a factor of $P(x)$. We can perform polynomial long division of $P(x)$ by $(x - 2)$:

1. Divide the leading term: $x^3 \div x = x^2$
2. Multiply $(x - 2)$ by $x^2$: $x^3 - 2x^2$
3. Subtract: $(3x^2 - (-2x^2)) = 5x^2$
4. Bring down next term, repeat: $P(x) = (x - 2)(x^2 + 5x - 3)$

Therefore, we can factor it as $P(x) = (x - 2)(x^2 + 5x - 3)$.

Vectors are perpendicular if their dot product equals 0. Setting up the equation:

$$\begin{pmatrix} a \ -1 \ \end{pmatrix} \cdot \begin{pmatrix} 2a - 3 \ 2 \ \end{pmatrix} = 0$$

Calculating the dot product:

$$(a)(2a - 3) + (-1)(2) = 0$$

Expanding this gives:

$$2a^2 - 3a - 2 = 0$$

Using the quadratic formula:

$$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-2)}}{2(2)} = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4}.$$

So, $a = 2$ or $a = -\frac{1}{2}$.

We begin by identifying the characteristics of $f(x)$. It has a local minimum at $(1, y)$ and crosses the x-axis at $(2,0)$, which suggests that it approaches 0 as $x$ tends to infinity or negative infinity.

The reciprocal graph $y = \frac{1}{f(x)}$ will have vertical asymptotes where $f(x) = 0$, which occurs at $x = 2$. At the local minimum, $f(x)$ is at its lowest, thus the graph of $\frac{1}{f(x)}$ will be at a maximum. Key features include:

• Vertical asymptote at $x = 2$
• Horizontal asymptotes approaching $y = 0$ as $x \to \pm \infty$.

The sketch will reflect these features.