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 we found that $P(2) = 0$, we can conclude that $(x - 2)$ is a factor of $P(x)$. To factor $P(x)$ completely, we perform polynomial long division:

1. Divide $P(x)$ by $(x - 2)$:
   • Start with $x^3$ divided by $x$ gives $x^2$.
   • Multiply $(x - 2)$ by $x^2$, and subtract:
   $x^3 + 3x^2 - 13x + 6$ $-(x^3 - 2x^2)$ $$$$

ightarrow 5x^2 - 13x + 6$$

2. Now repeat with $5x^2$: $5x^2 / x = 5x$.
   • Multiply $(x - 2)$ by $5x$ and subtract:
   $5x^2 - 13x + 6$ $-(5x^2 - 10x)$ $$$$

ightarrow -3x + 6$$

3. Finally, divide $-3x + 6$ by $(x - 2)$:
   • This gives $-3$, and multiplying gives:
   $-3x + 6$ $-(-3x + 6)$ $$$$

ightarrow 0$$

We can now express the polynomial as:

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

For two vectors to be perpendicular, their dot product must equal zero. Let's set the dot product of the vectors equal to zero:

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

Calculating the dot product:

• $a(2a - 3) + (-1)(2) = 0$

Thus:

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

Now we can solve this quadratic equation using the quadratic formula:

$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = -3$, and $c = -2$. Solving the roots will give you the required values of $a$.

To sketch the graph of $y = \frac{1}{f(x)}$, we need to consider the following features of the graph $y = f(x)$:

• Locate the roots of $f(x)$, where $f(x) = 0$, since these will be vertical asymptotes for $y = \frac{1}{f(x)}$.
• Identify the behavior of $f(x)$ near these roots; if the graph approaches zero from above, $y = \frac{1}{f(x)}$ will approach infinity, and vice versa.
• Since $f(x)$ is a quadratic equation, it will have a parabolic shape opening upwards or downwards. Using this information, we ensure the sketch reflects the points where the curve is defined and where it becomes undefined due to vertical asymptotes.