Tutor profile: Jiwoo C.

Let $$ f$$ be a function on the real numbers such that $$f(x-4) = 9x^3 + 1.$$ What is $$f(6)$$?

We let $$x-4 = 6$$ and find $$ x = 10$$. Then, $$ f(6) = f(10 - 4) = 9(10)^3 + 1 = 9001$$.

Prove that $$ x^2 \equiv y^2 \hspace{.02in} (\text{mod } p)$$ if and only if $$ x \equiv y \hspace{.02in} (\text{mod }p)$$ or $$ x \equiv -y \hspace{.02in} (\text{mod }p)$$ where $$p$$ is prime.

We recall that $$ a \equiv b \hspace{.02in} (\text{mod } n)$$ if and only if $$n \mid (a-b)$$. Then, $$ x^2 \equiv y^2 \hspace{.02in} (\text{mod } p)$$ if and only if $$ p \mid (x^2-y^2)$$. By factoring the difference of squares, we get $$ x^2 \equiv y^2 \hspace{.02in} (\text{mod } p)$$ if and only if $$ p \mid (x+y)(x-y)$$. Since $$p$$ is prime, either $$ p \mid (x+y)$$ or $$ p \mid (x-y) $$. Using the same identity that we started this proof with, $$ p \mid (x+y)$$ if and only if $$ x \equiv y \hspace{.02in} (\text{mod } p)$$, and $$ p \mid (x-y)$$ if and only if $$ x \equiv -y \hspace{.02in} (\text{mod } p)$$. Therefore, we conclude that for a prime $$p$$, $$ x^2 \equiv y^2 \hspace{.02in} (\text{mod } p)$$ if and only if $$ x \equiv y \hspace{.02in} (\text{mod }p)$$ or $$ x \equiv -y \hspace{.02in} (\text{mod }p)$$ .

Identify the degree of the polynomial $$ 21x^3 - 9x^2 + 63x - 27$$ and find its real zeroes.

The degree is the value of the highest powered variable in the polynomial, so the degree of the given polynomial is 3. To find the zeroes, we can graph the function or factor the polynomial. We demonstrate factoring the polynomial: Next, we notice that both the first and the third terms are divisible by 7 and that both the second and third terms are divisible by 3, so we consider factoring by grouping. 21 and 9 are both divisible by 3, and 63 and 27 are both divisible by 9, and factoring those out of the coefficients give us $( 3(7x^3 - 3x^2) + 9(7x - 3) $) We also factor out $$ x^2 $$ from the first two terms $( 3x^2(7x - 3) + 9(7x - 3) $) And by grouping we get $( (3x^2 + 9)(7x - 3)$) Now by letting each of those terms equal to zero $( 3x^2 + 9 = 0, \quad 7x-3 = 0 $) we find $$ x^2 = -3$$ and $$x = \frac{3}{7}$$ so the only real solution to the given polynomial is $$x = \frac{3}{7}$$.