SAT Equivalent Expressions Question 213 | Free SAT Prep | Aniko

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Question 213·Hard·Equivalent Expressions

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$p(x)=x^{4}-5x^{3}+6x^{2}$

$q(x)=x^{3}-4x^{2}+4x$

For all values of $x$ for which both expressions are defined, which expression is equivalent to $\dfrac{p(x)}{q(x)}$ ?

For rational expressions involving polynomials, your fastest approach is almost always to factor the numerator and denominator completely and then cancel common factors. Start by pulling out any greatest common factor, then factor remaining quadratics (using sum–product, completing the square, or other familiar methods). Write the fraction using these factored forms, cancel only whole factors (never cancel terms within sums), and keep domain restrictions in mind by noting where the original denominator is zero. Finally, match your simplified expression to the answer choices instead of trying to manipulate each choice individually.

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Look for common factors

Before trying any long division, see if you can factor a greatest common factor out of $p(x)$ and $q(x)$ . What can you factor out of each expression immediately?

Factor the quadratics

After factoring out the greatest common factor from $p(x)$ and $q(x)$ , you will be left with quadratics. Can you factor each quadratic into two binomials?

Use the factored form to simplify the fraction

Once both numerator and denominator are written as products of factors, write $\dfrac{p(x)}{q(x)}$ using those products. Which factors appear in both the numerator and the denominator, and can therefore be canceled?

Desmos Guide

Enter the original rational expression

Type y1 = (x^4 - 5x^3 + 6x^2) / (x^3 - 4x^2 + 4x) into Desmos. This is the graph of $\dfrac{p(x)}{q(x)}$ .

Graph each answer choice for comparison

On new lines, enter each choice as y2, y3, y4, and y5, for example:

y2 = (x-3)/(x-2)
y3 = x(x-3)/(x-2)
y4 = x(x-3)/(x-2)^2
y5 = x(x-3)^2/(x-2) Zoom out if needed so you can see the overall shapes of the graphs.

Compare the graphs and domains

Look for the answer-choice graph that lies exactly on top of the graph of $y1$ everywhere they are both defined, except possibly at points where $y1$ has holes (where the original denominator is zero). The choice whose graph matches $y1$ in this way is the equivalent expression.

Step-by-step Explanation

Factor the numerator $p(x)$

Start with

p(x)=x^{4}-5x^{3}+6x^{2}.

First, factor out the greatest common factor $x^2$ :

p(x)=x^2(x^2 - 5x + 6).

Now factor the quadratic $x^2 - 5x + 6$ . You need two numbers that multiply to $6$ and add to $-5$ . Those numbers are $-2$ and $-3$ , so:

x^2 - 5x + 6 = (x-2)(x-3).

Thus the fully factored numerator is

p(x)=x^2(x-2)(x-3).

Factor the denominator $q(x)$

Now factor

q(x)=x^{3}-4x^{2}+4x.

First, factor out the greatest common factor $x$ :

q(x)=x(x^2 - 4x + 4).

Next, factor the quadratic $x^2 - 4x + 4$ . You need two numbers that multiply to $4$ and add to $-4$ . Those numbers are $-2$ and $-2$ , so:

x^2 - 4x + 4 = (x-2)^2.

Thus the fully factored denominator is

q(x)=x(x-2)^2.

Write the rational expression in factored form and cancel common factors

Now write $\dfrac{p(x)}{q(x)}$ using the factored forms:

\frac{p(x)}{q(x)} = \frac{x^2(x-2)(x-3)}{x(x-2)^2}.

Identify common factors in the numerator and denominator:

There is a common factor of $x$ .
There is also a common factor of $(x-2)$ .

Cancel one $x$ from $x^2$ over $x$ and cancel one $(x-2)$ from $(x-2)$ over $(x-2)^2$ . After canceling, the numerator has a single $x$ and a factor of $(x-3)$ , and the denominator has a single factor of $(x-2)$ .

(Remember that this simplification is valid only for $x$ -values where the original denominator $q(x)$ is not zero.)

Write the simplified expression and match the answer choice

After canceling common factors, the simplified expression is

\frac{p(x)}{q(x)} = \frac{x(x-3)}{x-2},

for all $x$ where $q(x) \neq 0$ (that is, $x \neq 0$ and $x \neq 2$ ).

This matches answer choice B) $\dfrac{x(x-3)}{x-2}$ .

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation