Tutor profile: Jacob W.

Find every rational zero of the function given below: $$ f(x) = 2x^4-7x^3+3x^2+8x-4 $$

The Rational Zeros Theorem gives us a way to find every possible rational zero of a given polynomial. Every possible rational zero has the form: $$ \text{possible rational zero} = \pm\frac{\text{factor of constant term}}{\text{factor of leading coefficient}} $$ First, we list out all the possible rational zeros using the form given to us above: $$ \pm\frac{1}{1}, \pm\frac{2}{1}, \pm\frac{4}{1}, \pm\frac{1}{2}, \pm\frac{2}{2}, \pm\frac{4}{2}, $$ These results leave us with four possible rational zeros (some of the results are equivalent, so they will not appear twice): $$ \pm1, \pm2, \pm4, \pm\frac{1}{2}$$ Next, we use synthetic division to see if these possible rational zeros are real zeros of our given polynomial. If the remainder is $$0$$, then the zero divides evenly and is a real zero of the polynomial. If the remainder is anything else, then it is not a real zero. $$ \begin{matrix} 1 & | & 2 & -7 & 3 & 8 & -4\\ & | & & 2 & -5 & -2 & 6\\ & | & - & - & - & - & -\\ & | & 2 & -5 & -2 & 6 & 2 \end{matrix} $$ $$1$$ is NOT a real zero. $$ \begin{matrix} -1 & | & 2 & -7 & 3 & 8 & -4\\ & | & & -2 & 9 & -12 & 4\\ & | & - & - & - & - & -\\ & | & 2 & -9 & 12 & -4 & 0 \end{matrix} $$ $$-1$$ IS a real zero. $$ \begin{matrix} 2 & | & 2 & -7 & 3 & 8 & -4\\ & | & & 4 & -6 & -6 & 4\\ & | & - & - & - & - & -\\ & | & 2 & -3 & -3 & 2 & 0 \end{matrix} $$ $$2$$ IS a real zero. $$ \begin{matrix} -2 & | & 2 & -7 & 3 & 8 & -4\\ & | & & -4 & 22 & -50 & 84\\ & | & - & - & - & - & -\\ & | & 2 & -11 & 25 & -42 & 80 \end{matrix} $$ $$-2$$ is NOT a real zero. $$ \begin{matrix} 4 & | & 2 & -7 & 3 & 8 & -4\\ & | & & 8 & 4 & 28 & 144\\ & | & - & - & - & - & -\\ & | & 2 & 1 & 7 & 36 & 140 \end{matrix} $$ $$4$$ is NOT a real zero. $$ \begin{matrix} -4 & | & 2 & -7 & 3 & 8 & -4\\ & | & & -8 & 60 & -252 & 976\\ & | & - & - & - & - & -\\ & | & 2 & -15 & 63 & -244 & 972 \end{matrix} $$ $$-4$$ is NOT a real zero. $$ \begin{matrix} \frac{1}{2} & | & 2 & -7 & 3 & 8 & -4\\ & | & & 1 & -3 & 0 & 4\\ & | & - & - & - & - & -\\ & | & 2 & -6 & 0 & 8 & 0 \end{matrix} $$ $$\frac{1}{2}$$ IS a real zero. $$ \begin{matrix} -\frac{1}{2} & | & 2 & -7 & 3 & 8 & -4\\ & | & & -1 & 4 & -\frac{7}{2} & -\frac{9}{4}\\ & | & - & - & - & - & -\\ & | & 2 & -8 & 7 & \frac{9}{2} & -\frac{25}{4} \end{matrix} $$ $$-\frac{1}{2}$$ is NOT a real zero. We see that the rational zeros of the given function are: $$ -1, 2, \frac{1}{2} $$

How would you print out a random integer on the interval [42, 169]?

#include <iostream> #include <ctime> #include <cstdlib> int main() { /* Guarantees that the given sequence of random numbers is not the same every time the program is executed. */ std::srand(std::time(NULL)); /* std::srand() gives us a random integer from the entire interval of available integers. Using the modulus operator, we can prevent the number given to us by std::rand() to not go past a certain number. Remember that modulus gives us the remainder of an expression, so if we give the number above what our upper bound is, we will receive all integers below and including our upper bound. However, in order get the lower bound of 42, we must add 42 because std::rand() gives us all integers below a certain upper bound. To adjust for adding 42 to get a desired lower bound, we must adjust the upper bound of our modulus operand so that we get the correct upper bound. To clarify, if we did std::rand() % 170, we would get [0, 169]. If we added 42 after the expression std::rand() % 170 is executed, the bound would change to [42, 211]. This is why we have std::srand() % 128 instead of std::srand() % 170. We got 128 from the difference of 170 - 42. */ std::cout << (std::rand() % 128) + 42 << std::endl; return 0; }

Simplify the given expression: $$ \frac{\sqrt{52x}}{\sqrt{117x^3}} $$

First, we can apply the quotient property of radical expressions to rewrite the given expression so that we can work with the fraction itself: $$ \sqrt{\frac{52x}{117x^3}} $$ Next, we can simplify the expression in the root by factoring out the numerator and denominator as such: $$ \sqrt{\frac{13 \cdot 4 \cdot x}{13 \cdot 9 \cdot x^3}} $$ Then, we can cancel common factors, which gives us: $$ \sqrt{\frac{13 \cdot 4 \cdot x}{13 \cdot 9 \cdot x^3}} = \sqrt{\frac{4}{9x^2}} $$ Finally, we can apply the roots back to the numerator and denominator separately and get our final answer by solving the expressions in the numerator and denominator: $$ \frac{\sqrt{4}}{\sqrt{9x^2}} = \frac{2}{3x} $$