Question 111·Hard·Equivalent Expressions

Rewrite $x^2 - 1$ in factored form. How does $\dfrac{x^2 - 1}{(x-1)^2}$ simplify once you factor and cancel common terms?

Once you have the numerator as one fraction and the denominator as one fraction, remember that $\dfrac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b} \times \frac{d}{c}$. After you rewrite it as a product, which factor appears in both the numerator and the denominator so you can cancel it?

After simplifying, your result should be a single fraction whose numerator is $2x^2 - x + 3$. Check carefully which choice has the correct denominator based on the factors that remain after cancellation.

In one line, type the original expression as a function, for example f(x) = ( (2x/(x-1)) - (3/(x+1)) ) / ( (x^2 - 1)/(x-1)^2 ). Desmos will automatically handle the restrictions at $x=\pm1$ by leaving gaps or vertical asymptotes there.

On new lines, enter each option as g1(x) = (2x^2 - x + 3)/(x-1)^2, g2(x) = (2x^2 - x + 3)/(x^2 - 1), g3(x) = (2x^2 - x + 3)*(x+1)^2, and g4(x) = (2x^2 - x + 3)/(x+1)^2.

Look at the graphs and see which $g_i(x)$ lies exactly on top of $f(x)$ everywhere they are both defined (away from $x=\pm1$). You can also use a table in Desmos to compare numerical values at a few safe points like $x=0,2,-2$; the equivalent expression will match $f(x)$ at every tested value.

The numerator of the complex fraction is

Use a common denominator of $(x-1)(x+1)$:

So the numerator becomes

Now expand and simplify the numerator expression:

So the entire numerator of the complex fraction is

The denominator is

Factor the difference of squares $x^2 - 1$:

So the denominator becomes

using the fact that $x\neq 1$, so we can cancel one $(x-1)$ factor.

Now the original expression is

Dividing by a fraction is the same as multiplying by its reciprocal:

Cancel the common factor $(x-1)$ (allowed because $x\neq 1$):

So the expression is equivalent to $\dfrac{2x^2 - x + 3}{(x+1)^2}$, which matches choice D.