Rewrite the algebraic fraction $\frac{2x^2 + 5x - 12}{2x^2 - 7x + 6}$ in its lowest equivalent reduced form $\frac{f(x)}{g(x)}$, and state all real boundary values of $x$ for which this equivalence fails to hold due to domain restrictions.

Rewrite the algebraic fraction $\frac{2x^2 + 5x - 12}{2x^2 - 7x + 6}$ in its lowest equivalent reduced form $\frac{f(x)}{g(x)}$ , and state all real boundary values of $x$ for which this equivalence fails to hold due to domain restrictions.

Final answer

$\frac{x + 4}{x - 2}$ for $x \neq \frac{3}{2}$ and $x \neq 2$

Steps

Factor the numerator

To simplify the rational expression and find its domain restrictions, follow these steps:

The numerator is a quadratic expression: $2x^2 + 5x - 12$ . To factor by grouping, find two numbers whose product is $2 \cdot (-12) = -24$ and whose sum is $+5$ . These numbers are $8$ and $-3$ : $2x^2 + 8x - 3x - 12 = 2x(x + 4) - 3(x + 4) = (2x - 3)(x + 4)$

Factor the denominator

The denominator is: $2x^2 - 7x + 6$ . Find two numbers whose product is $2 \cdot 6 = 12$ and whose sum is $-7$ . These numbers are $-4$ and $-3$ : $2x^2 - 4x - 3x + 6 = 2x(x - 2) - 3(x - 2) = (2x - 3)(x - 2)$

Determine the domain restrictions

A rational expression is undefined wherever its denominator equals zero. Setting the original denominator equal to zero gives: $(2x - 3)(x - 2) = 0$ $2x - 3 = 0 \implies x = \frac{3}{2}$ $x - 2 = 0 \implies x = 2$

Therefore, the original expression is only defined for $x \neq \frac{3}{2}$ and $x \neq 2$ .

Cancel common factors to write the reduced form

Rewrite the original fraction in terms of its factored polynomials: $\frac{2x^2 + 5x - 12}{2x^2 - 7x + 6} = \frac{(2x - 3)(x + 4)}{(2x - 3)(x - 2)}$

Canceling the non-zero common factor $(2x - 3)$ gives: $\frac{x + 4}{x - 2}$

Pitfall Note: While $\frac{x + 4}{x - 2}$ is algebraically defined at $x = \frac{3}{2}$ , the original expression is not. For the two expressions to be strictly equivalent, the restrictions of the original expression must be preserved: $x \neq \frac{3}{2}$ and $x \neq 2$ .

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student only identifies the removable discontinuity as the boundary value where equivalence fails, forgetting that the vertical asymptote is also excluded from the original domain.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
$\frac{2x^2 + 5x - 12}{2x^2 - 7x + 6}$ = $\frac{(2x - 3)(x + 4)}{(2x - 3)(x - 2)}$ = $\frac{x + 4}{x - 2}$
Step 2
2x - 3 = 0 $\implies$ x = $\frac{3}{2}$
Correct
(2x - 3)(x - 2) = 0 $\implies$ x = $\frac{3}{2}$ $\text{ and }$ x = 2
You only set the cancelled factor $2x - 3$ to zero. The original rational expression is undefined whenever its entire original denominator equals zero, which requires excluding both $x = \frac{3}{2}$ and $x = 2$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly factored both quadratic polynomials into linear factors.

You simplified the fraction to its lowest terms by cancelling the common factor.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Incomplete domain restrictions

You solved only for the root of the cancelled factor instead of checking all roots of the original denominator.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the hole (removable discontinuity) with the full set of domain restrictions of the original expression.

REASONING PATH

Your ACEpath teacher’s suggested approach

Find Original Domain First: Domain restrictions must be identified from the unreduced denominator before cancelling any common factors.
Set every factor of the original denominator to zero: $(2x - 3)(x - 2) = 0$ yields $x = \frac{3}{2}$ and $x = 2$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Set the entire factored denominator of the original expression equal to zero: $(2x - 3)(x - 2) = 0$ .
Solve for all excluded values: $2x - 3 = 0 \implies x = \frac{3}{2}$ , and $x - 2 = 0 \implies x = 2$ .
State the reduced form alongside all domain restrictions from the original expression.

Final answer

$\frac{x + 4}{x - 2}$ $\text{ for }$ x $\neq$ $\frac{3}{2}$ $\text{ and }$ $x \neq 2$

NEXT STEPS

Final Takeaways from Your Teacher

Always write down the domain restrictions from the original factored denominator before crossing out any common factors.
Remember that an algebraic equivalence between rational expressions is valid only on the intersection of their domains.

OverviewACEpath Teacher's Diagnosis

02 The student makes a sign error when factoring the trinomial in the denominator, leading to an incorrect cancellation and wrong boundary values.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
2x^2 + 5x - 12 = (2x - 3)(x + 4)
Step 2
2x^2 - 7x + 6 = (2x + 3)(x - 2)
Correct
2x^2 - 7x + 6 = (2x - 3)(x - 2)
Expanding $(2x + 3)(x - 2)$ yields $2x^2 - x - 6$ , not $2x^2 - 7x + 6$ . Because the middle term is $-7x$ and the constant term is $+6$ , both binomial factors must have negative constant terms: $-4x - 3x = -7x$ and $(-3)(-2) = +6$ .
Step 3
= $\frac{(2x - 3)(x + 4)}{(2x + 3)(x - 2)}$
Correct
$\frac{(2x - 3)(x + 4)}{(2x - 3)(x - 2)}$ = $\frac{x + 4}{x - 2}$
Due to the sign error in the denominator, you were unable to identify and cancel the shared factor $(2x - 3)$ .
Step 4
(2x + 3)(x - 2) = 0 $\implies$ x = - $\frac{3}{2}$ ,; x = 2
Correct
(2x - 3)(x - 2) = 0 $\implies$ x = $\frac{3}{2}$ ,; x = 2
Setting your factored denominator equal to zero correctly followed your previous step, but inherited the sign error from Step 2.

WHAT YOU DID WELL

Compliments from Your Teacher

You factored the numerator $2x^2 + 5x - 12$ into $(2x - 3)(x + 4)$ accurately.

You set the denominator equal to zero to identify the restricted boundary values.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Factoring quadratic with signs error

You wrote $(2x + 3)$ instead of $(2x - 3)$ when factoring $2x^2 - 7x + 6$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Overlooking the signs of the product and sum when splitting the middle term of $ax^2 + bx + c$ : when $ac > 0$ and $b < 0$ , both factors must be negative.

REASONING PATH

Your ACEpath teacher’s suggested approach

Sign check for quadratic trinomials: For $ax^2 + bx + c$ , if $ac > 0$ and $b < 0$ , both split factors must have negative signs.
Always multiply your factors back out using FOIL to verify the middle and constant terms before proceeding.

CORRECTED PATH

ACEpath Teacher's Demonstration

Factor the denominator correctly: $2x^2 - 7x + 6 = 2x^2 - 4x - 3x + 6 = 2x(x - 2) - 3(x - 2) = (2x - 3)(x - 2)$ .
Substitute both factored forms into the fraction: $\frac{(2x - 3)(x + 4)}{(2x - 3)(x - 2)}$ .
Cancel the common factor $(2x - 3)$ for all $x \neq \frac{3}{2}$ to obtain $\frac{x + 4}{x - 2}$ .
Identify all values where the original denominator equals zero: $(2x - 3)(x - 2) = 0 \implies x = \frac{3}{2}$ and $x = 2$ .

Final answer

$\frac{x + 4}{x - 2}$$\text{ for }$ x $\neq$ $\frac{3}{2} $\text{ and }$ x \neq 2$$

NEXT STEPS

Final Takeaways from Your Teacher

Practice factoring quadratics with negative middle terms and positive constant terms.
Perform a quick mental FOIL expansion on any factored quadratic before canceling factors.
Remember that domain restrictions must always come from the original unreduced denominator.

OverviewACEpath Teacher's Diagnosis

03 The student improperly cancels terms across the numerator and denominator before factoring.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$\frac{2x^2 + 5x - 12}{2x^2 - 7x + 6}$ = $\frac{5x - 12}{-7x + 6}$
Correct
$\frac{2x^2 + 5x - 12}{2x^2 - 7x + 6}$ = $\frac{(2x - 3)(x + 4)}{(2x - 3)(x - 2)}$ = $\frac{x + 4}{x - 2}$
You canceled the additive term $2x^2$ from both the numerator and denominator. Fractions can only be simplified by dividing out common multiplicative factors, never individual terms separated by addition or subtraction.
Step 2
-7x + 6 = 0 $\implies$ x = $\frac{6}{7}$
Correct
(2x - 3)(x - 2) = 0 $\implies$ x = $\frac{3}{2}$ , x = 2
You correctly set your resulting denominator to zero to find the restricted value, but because the denominator was simplified incorrectly in the previous step, this yielded an incorrect boundary value.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified that finding domain restrictions involves setting the denominator equal to zero and solving for $x$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Canceling terms instead of factors

Canceling the leading $2x^2$ terms across the fraction line as if they were factors.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated terms separated by addition/subtraction as common factors that could be canceled directly, bypassing the need to factor the quadratics first.

REASONING PATH

Your ACEpath teacher’s suggested approach

Factor before canceling: Only identical multiplicative factors can cancel across a fraction: $\frac{a \cdot b}{a \cdot c} = \frac{b}{c}$ , whereas $\frac{a + b}{a + c} \neq \frac{b}{c}$ .
Always fully factor both the numerator and denominator into products of linear or irreducible terms before crossing anything out.
Check original domain restrictions: Domain restrictions must be identified from the original, unsimplified denominator.
Set every factor of the original denominator to zero before canceling any common factors.

CORRECTED PATH

ACEpath Teacher's Demonstration

Factor the numerator: $2x^2 + 5x - 12 = (2x - 3)(x + 4)$ .
Factor the denominator: $2x^2 - 7x + 6 = (2x - 3)(x - 2)$ .
Find all domain restrictions from the denominator factors: $(2x - 3)(x - 2) = 0 \implies x = \frac{3}{2}$ or $x = 2$ .
Divide out the non-zero common factor $(2x - 3)$ .

Final answer

$\frac{x + 4}{x - 2}$ $\text{ for }$ x $\neq$ $\frac{3}{2}$ $\text{ and }$ $x \neq 2$

NEXT STEPS

Final Takeaways from Your Teacher

Review the fundamental rule of rational fractions: only factors (multiplied quantities) cancel, never terms (added/subtracted quantities).
Practice factoring quadratics with leading coefficients greater than 1 using the ac-method or grouping.
Remember to find domain restrictions from the factored original denominator before canceling identical factors.

Lock it in

Question

When simplifying a rational expression, from which expression should you determine the domain restrictions?

Answer

From the original (unreduced) denominator before cancelling any common factors.

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student only identifies the removable discontinuity as the boundary value where equivalence fails, forgetting that the vertical asymptote is also excluded from the original domain.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student makes a sign error when factoring the trinomial in the denominator, leading to an incorrect cancellation and wrong boundary values.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student improperly cancels terms across the numerator and denominator before factoring.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

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