SAT Equivalent Expressions Question 162 | Free SAT Prep | Aniko

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Question 162·Hard·Equivalent Expressions

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$p(t)=t^3-4t^2-t+4$

$q(t)=t^2-3t+2$

Which of the following expressions is equivalent to $\dfrac{p(t)}{q(t)}$ , for $t>4$ ?

For rational expressions built from polynomials on the SAT, first factor the numerator and denominator completely using basic techniques (trinomial factoring, grouping, and special products like difference of squares). Then rewrite the fraction as a product of factors, cancel only the factors that appear in both numerator and denominator, and pay attention to any domain restrictions given (like $t>4$ ) to be sure the canceled factors are never zero. Finally, match your simplified result to the answer choices rather than trying to perform long polynomial division.

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Start with the denominator

Try factoring $q(t)=t^2-3t+2$ . Look for two numbers that multiply to $2$ and add to $-3$ .

Factor the cubic by grouping

Rewrite $p(t)=t^3-4t^2-t+4$ as two groups, $(t^3-4t^2)+(-t+4)$ , and factor each group. Do you see a common factor after grouping?

Use common factors to simplify

Once both $p(t)$ and $q(t)$ are written as products of factors, write $\dfrac{p(t)}{q(t)}$ as a fraction of those products and look for a factor that appears in both numerator and denominator. Remember $t>4$ when deciding what can be canceled.

Desmos Guide

Define the given functions

In Desmos, enter p(t)=t^3-4t^2-t+4 and q(t)=t^2-3t+2. Then define r(t)=p(t)/q(t) to represent the original rational expression.

Enter each answer choice as its own function

Type the four choices as functions, for example: A(t)=((t+1)(t-2))/(t-4), B(t)=(t-4)/((t+1)(t-2)), C(t)=((t+1)(t-4))/(t-2), and D(t)=(t+1)/(t-2).

Compare values for $t>4$

Create a table for r(t) and each of A(t), B(t), C(t), and D(t) with $t$ -values greater than 4 (for example, $t=5,6,7$ ). The correct option is the one whose values match r(t) for all those $t$ -values.

Step-by-step Explanation

Factor the denominator $q(t)$

We are given

q(t)=t^2-3t+2.

To factor this quadratic, look for two numbers that multiply to $2$ and add to $-3$ .

The numbers $-1$ and $-2$ work: $(-1)(-2)=2$ and $-1+(-2)=-3$ .

q(t)=t^2-3t+2=(t-1)(t-2).

Factor the numerator $p(t)$ by grouping

Now factor

p(t)=t^3-4t^2-t+4.

Group the terms in pairs:

(t^3-4t^2)+(-t+4).

Factor each group:

From $t^3-4t^2$ , factor out $t^2$ : $t^3-4t^2=t^2(t-4)$ .
From $-t+4$ , factor out $-1$ : $-t+4=-1(t-4)$ .

p(t)=t^2(t-4)-1(t-4)=(t^2-1)(t-4).

Now factor $t^2-1$ as a difference of squares:

t^2-1=(t-1)(t+1),

p(t)=(t-1)(t+1)(t-4).

Simplify the rational expression and match a choice

Write $\dfrac{p(t)}{q(t)}$ using the factored forms:

\frac{p(t)}{q(t)} =\frac{(t-1)(t+1)(t-4)}{(t-1)(t-2)}.

There is a common factor $(t-1)$ in the numerator and denominator. Normally we must be careful not to cancel a factor that could be zero, but the problem states $t>4$ , so $t$ cannot be $1$ . That means $(t-1)\neq 0$ , and we can safely cancel it:

\frac{p(t)}{q(t)}=\frac{(t+1)(t-4)}{t-2}.

This simplified expression matches answer choice C) $\dfrac{(t+1)(t-4)}{t-2}$ .

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Step-by-step Explanation

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Step-by-step Explanation