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 know that (P(2) = 0), we can factor (P(x)) using ((x - 2)) as one of the factors. We will perform polynomial long division to factor (P(x)):

Divide:

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

Thus, we can state:

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

Vectors are perpendicular when their dot product is zero:

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

Evaluating the dot product:

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

Using the quadratic formula: $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-2)}}{2(2)} = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4}\n$ Thus, we find:
$a = 2 \text{ or } a = -\frac{1}{2}.$

The function (y = f(x)) is a downward facing quadratic which opens upwards between the roots. The graph will have vertical asymptotes where (f(x) = 0). Since (f(x)) crosses the x-axis at two points (root points), the graph of (y = \frac{1}{f(x)}) will approach zero at these points. The sketch should display:

• The location of the zeros of (f(x)) becoming vertical asymptotes.
• The general shape reflecting the behavior of (f(x)).