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 $P(2) = 0$, this means $(x - 2)$ is a factor of $P(x)$. We can perform polynomial long division of $P(x)$ by $(x - 2)$:

$$P(x) = (x - 2)(x^2 + 5x - 3).$$

Thus, we can express $P(x)$ as:

$$P(x) = A(x)B(x) = (x - 2)(x^2 + 5x - 3),$$

where $B(x) = x^2 + 5x - 3$.

Two vectors are perpendicular if their dot product is zero. We calculate the dot product:

$$\begin{pmatrix} a \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 2a - 3 \\ 2 \end{pmatrix} = a(2a - 3) + (-1)(2) = 0.$$

Expanding and simplifying,

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

We can use the quadratic formula to find $a$:

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

Thus, we have two solutions:

$$a = 2 \quad \text{or} \quad a = -\frac{1}{2}.$$

To sketch the graph of $y = \frac{1}{f(x)}$, we analyze the properties of the original function $f(x)$. The points where $f(x)$ equals zero will become vertical asymptotes in the graph of $y = \frac{1}{f(x)}$.

Additionally, for intervals where $f(x)$ is positive, $\frac{1}{f(x)}$ will be positive, creating sections above the x-axis. Conversely, where $f(x)$ is negative, $\frac{1}{f(x)}$ will be negative, appearing below the x-axis.

Important features to include are:

• The location of vertical asymptotes where $f(x) = 0$ (which occur at the roots of $f$).
• The behavior of $\frac{1}{f(x)}$ as $x$ approaches these roots, tending to $\pm \infty$.