# Tutor profile: Neil Gerson A.

## Questions

### Subject: Basic Chemistry

Balance the following chemical reaction: $$ Cu \; + \; HNO_3 \; \longrightarrow \; Cu(NO_3)_2 \; + \; NO \; + \; H_2O$$

$$ 3Cu \; + \; 8HNO_3 \; \longrightarrow \; 3Cu(NO_3)_2 \; + \; 2NO \; + \; 4H_2O$$

### Subject: C++ Programming

Write a C++ code that recursively computes the nth power of a.

function power (int a, int n) { // In this code, let the problem look like a^n = ? if (n == 1) { return a; } else { return a * power(a, n - 1); } }

### Subject: Algebra

How many distinct solutions (not necessarily real) exist for the equation? $(x-1)(x^2-1)(x^3-1)\ldots(x^10-1)=0$1$.

Recall that if we have two numbers or terms that when multiplied together results to 0, its either one of them is 0, or both of them are 0, i.e. if $AB = 0$, then $A = 0$ or $B = 0$. Applying this concept to the equation, we can let $A$ and $B$ (take note we can arbitrarily partition the equation in two parts) to be $A = (x - 1)(x^2 - 1)$ and the remaining parts of the equation to be $B$, giving us either $(x - 1)(x^2 - 1) = 0$ or $(x^3 - 1)(x^4 - 1) \ldots (x^10 - 1) = 0$. Consider $A$ to be 0, as stated in the concept above. Take note that it can again be divided into two terms, $(x - 1)$ and $(x^2 - 1)$. Applying the concept again, we have either $(x - 1) = 0$ or $(x^2 - 1) = 0$. Solving the first one, we have $x = 1$. The second one gives us $x = \pm 1$. Applying the same concept repetitively on $B$, we have two distinct solutions for $B$, which is $x = 1$ and $x = -1$. Thus, taking the solutions of $A$ and $B$, we have two distinct solutions for the equation.

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