2. The expression is

$$
X=\left(a+a^{-1}\right)^{-1}\left(a^{2}+3 a+2\right)\left(a^{2}-3 a+2\right)\left(a^{2}-4\right)^{-1}
$$

a) Simplify the expression $X$ and write it in the form of a product.

(8 points)

b) Calculate the value of the expression $X$ for $a=-\frac{1}{3}$.

## 19th Mathematics Knowledge Competition for Students of Secondary Technical and Vocational Schools  Selection Competition, March 21, 2019

## Problems for 3rd Year

| N1 | N2 |
| :--- | :--- |
|  |  |
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2. The first factor is transformed into $\left(a+a^{-1}\right)^{-1}=\frac{a}{a^{2}+1}$.

In the product $\left(a^{2}+3 a+2\right)\left(a^{2}-3 a+2\right)$, each term in the first parenthesis can be multiplied by each term in the second parenthesis. A faster way is to transform this into the product of the sum and difference of the same terms, resulting in the difference of squares: $\left(a^{2}+3 a+2\right)\left(a^{2}-3 a+2\right)=\left(\left(a^{2}+2\right)+3 a\right)\left(\left(a^{2}+2\right)-3 a\right)=\left(a^{2}+2\right)^{2}-(3 a)^{2}=a^{4}+4 a^{2}+4-9 a^{2}=a^{4}-5 a^{2}+4$.

Using Vieta's formulas, we factorize $a^{4}-5 a^{2}+4=\left(a^{2}-1\right)\left(a^{2}-4\right)$.

We also transform the last factor $\left(a^{2}-4\right)^{-1}=\frac{1}{a^{2}-4}$. Multiplying everything together, we get $X=\frac{a}{a^{2}+1} \cdot\left(a^{2}-1\right)\left(a^{2}-4\right) \cdot \frac{1}{a^{2}-4}=\frac{a}{a^{2}+1} \cdot\left(a^{2}-1\right)=\frac{a(a-1)(a+1)}{a^{2}+1}$. Substituting $a=-\frac{1}{3}$ into the simplified expression $X$, we get $\frac{4}{15}$.

Using $a^{-1}=\frac{1}{a}$ .1 point

The simplified first factor, written as a fraction $\frac{a}{a^{2}+1}$ 1 point

Multiplying the second and third factors: $\left(a^{2}+3 a+2\right)\left(a^{2}-3 a+2\right)=a^{4}-5 a^{2}+4 \ldots \ldots \ldots \ldots 1^{*}+1$ point Using Vieta's formulas $a^{4}-5 a^{2}+4=\left(a^{2}-1\right)\left(a^{2}-4\right)$, which can be factored to $(a+1)(a+2)(a-1)(a-2)$ 1 point

Transforming the last factor $\left(a^{2}-4\right)^{-1}=\frac{1}{a^{2}-4}$, which can be written as $\frac{1}{(a+2)(a-2)} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ Multiplying the obtained factors and simplifying.....................................................................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_18400104441c5ee1e47ag-14.jpg?height=83&width=1727&top_left_y=1055&top_left_x=176)

![](https://cdn.mathpix.com/cropped/2024_06_07_18400104441c5ee1e47ag-14.jpg?height=60&width=1727&top_left_y=1124&top_left_x=176)

![](https://cdn.mathpix.com/cropped/2024_06_07_18400104441c5ee1e47ag-14.jpg?height=60&width=1742&top_left_y=1175&top_left_x=174)

## 19th Mathematics Knowledge Competition for Students of Secondary Technical and Vocational Schools

## Preliminary Competition, March 21, 2019

## Solutions and Scoring Guide

A contestant who arrives at the solution by any correct method (even if not anticipated by the scoring guide) receives all possible points.

A correct method is one that:

- sensibly takes into account the problem statement,
- leads to the solution of the problem,
- is mathematically correct and complete.

If an intermediate or final result can be recognized, guessed, read from a diagram, or calculated mentally, the contestant generally receives all the points provided for. If, however, the solution is guessed (and cannot be arrived at without calculation), even if it is correct, it is scored with 0 points.

A contestant who only partially solves the problem, with correct steps in the solution but no clear path to the final solution, cannot receive more than half of the possible points.

The asterisk '*' next to points means that the contestant can earn the point or points for a correct procedure, even if the calculation is incorrect.

## Third Year