Question 270·Hard·Advanced Math

Equivalent expressions must agree at every allowed input, so one disagreement is enough to rule a choice out. Use $z=3$, the first whole number satisfying $z>2$. A match at $3$ is a useful test, though it doesn’t by itself prove a formula works for all $z>2$.

Type $p(z)=z^4-5z^2+4$ and $q(z)=z^2-3z+2$ for the given $P$ and $Q$. Then type $p(3)/q(3)$. Desmos prints $20$ for the quotient, so an equivalent choice must also give $20$ at $z=3$.

Substitute $3$ into each choice and type the four expressions below. Desmos prints $6$, $20$, $10$, and $16$, in order. Only $(z+1)(z+2)$ matches the original fraction at this allowed input; each other choice is ruled out. The factor route below shows why the match holds for every $z>2$. So the equivalent expression is $(z+1)(z+2)$. Choice B.

To factor $Q(z)$, find two numbers that multiply to $2$ and add to $-3$: they’re $-1$ and $-2$. Write the denominator as a product:

Its zeros are $1$ and $2$, and $z>2$ excludes both.

Treat $z^2$ as one quantity: $z^4$ is $(z^2)^2$. The numbers $-1$ and $-4$ add to $-5$ and multiply to $4$, so factor this quadratic in $z^2$:

A difference of squares has the form $a^2-b^2=(a-b)(a+b)$. Both $z^2-1$ and $z^2-4$ fit that pattern, so split both factors:

Replace the top and bottom of the rational expression, a fraction of polynomials, with their products:

Because $z>2$ makes $z-1$ and $z-2$ nonzero, divide the top and bottom by those whole common factors:

Canceling pieces of a sum instead would change the fraction. So $(z+1)(z+2)$ is equivalent to the original expression for every allowed $z$. Choice B.