In a pre-calculus analysis of rational functions, a student considers the difference quotient parameter $h$ approaching zero. Consider the expanded product $f(x + h) = 2(x + h)^3 - 5(x + h)$. (a) Use the binomial expansion and the distributive property to expand $f(x + h)$ completely. (b) Form the difference $f(x + h) - f(x)$, distribute the negative sign, factor out $h$, and evaluate the asymptotic limit $\lim_{h \to 0} \frac{f(x + h)

In a pre-calculus analysis of rational functions, a student considers the difference quotient parameter $h$ approaching zero. Consider the expanded product $f(x + h) = 2(x + h)^3 - 5(x + h)$ . (a) Use the binomial expansion and the distributive property to expand $f(x + h)$ completely. (b) Form the difference $f(x + h) - f(x)$ , distribute the negative sign, factor out $h$ , and evaluate the asymptotic limit $\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ .

Final answer

(a) $f(x + h) = 2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h$ ; (b) $\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = 6x^2 - 5$

Steps

Part (a): Expand $f(x + h)$ completely

Apply the binomial expansion to $(x + h)^3$ : Using the binomial formula $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ : $(x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$
Distribute the constants:
- Distribute $2$ across the expanded cubic expression: $2(x + h)^3 = 2(x^3 + 3x^2h + 3xh^2 + h^3) = 2x^3 + 6x^2h + 6xh^2 + 2h^3$
- Distribute $-5$ across $(x + h)$ : $-5(x + h) = -5x - 5h$
Combine the terms: $f(x + h) = 2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h$

Part (b): Form the difference quotient and evaluate the limit

Identify $f(x)$ and subtract it from $f(x + h)$ : From the definition, $f(x) = 2x^3 - 5x$ . $f(x + h) - f(x) = (2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h) - (2x^3 - 5x)$
Distribute the negative sign and cancel opposite terms: $f(x + h) - f(x) = 2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h - 2x^3 + 5x$

Common Pitfall: Be careful to distribute the negative sign across both terms of $f(x)$ , turning $-(-5x)$ into $+5x$ .

The non- $h$ terms cancel out ( $2x^3 - 2x^3 = 0$ and $-5x + 5x = 0$ ): $f(x + h) - f(x) = 6x^2h + 6xh^2 + 2h^3 - 5h$
Factor out $h$ and simplify the fraction: Factor the common parameter $h$ from the numerator: $f(x + h) - f(x) = h(6x^2 + 6xh + 2h^2 - 5)$

Substitute this into the difference quotient for $h \neq 0$ : $\frac{f(x + h) - f(x)}{h} = \frac{h(6x^2 + 6xh + 2h^2 - 5)}{h} = 6x^2 + 6xh + 2h^2 - 5$
Evaluate the limit as $h \to 0$ : Using direct substitution: $\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \to 0} (6x^2 + 6xh + 2h^2 - 5)$ $\lim_{h \to 0} (6x^2 + 6xh + 2h^2 - 5) = 6x^2 + 6x(0) + 2(0)^2 - 5 = 6x^2 - 5$

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student misremembers the binomial expansion of (x + h)³ as simply x³ + h³, ignoring the intermediate terms.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$(x + h)^3 = x^3 + h^3$
Correct
$(x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$
Powers do not distribute over addition. Expanding $(x + h)^3 = (x + h)(x^2 + 2xh + h^2)$ produces the two middle cross terms $3x^2h$ and $3xh^2$ .
Step 2
$f(x + h) = 2(x^3 + h^3) - 5(x + h) \implies f(x + h) = 2x^3 + 2h^3 - 5x - 5h$
Correct
$f(x + h) = 2(x^3 + 3x^2h + 3xh^2 + h^3) - 5(x + h) = 2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h$
You distributed $2$ and $-5$ correctly across your expression, but the expression was missing the cubic cross terms from Step 1.
Step 3
$f(x + h) - f(x) = (2x^3 + 2h^3 - 5x - 5h) - (2x^3 - 5x) \implies f(x + h) - f(x) = 2h^3 - 5h$
Correct
$f(x + h) - f(x) = (2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h) - (2x^3 - 5x) = 6x^2h + 6xh^2 + 2h^3 - 5h$
You subtracted $f(x)$ and distributed the negative sign accurately, but the resulting difference reflects the missing terms.
Step 4
$h(2h^2 - 5) \implies \frac{f(x + h) - f(x)}{h} = 2h^2 - 5$
Correct
$\frac{f(x + h) - f(x)}{h} = \frac{h(6x^2 + 6xh + 2h^2 - 5)}{h} = 6x^2 + 6xh + 2h^2 - 5$
You correctly factored out $h$ and cancelled it with the denominator.
Step 5
$\lim_{h \to 0} (2h^2 - 5) = -5$
Correct
$\lim_{h \to 0} (6x^2 + 6xh + 2h^2 - 5) = 6x^2 - 5$
You evaluated the limit correctly for your simplified difference quotient, but the correct limit retains $6x^2$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly set up the difference quotient definition and substituted $f(x + h)$ and $f(x)$ .

You distributed negative signs and constant coefficients across parentheses without sign errors.

You properly factored out $h$ from all remaining terms to cancel the denominator.

You evaluated the algebraic limit as $h \to 0$ using direct substitution.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Freshman's Dream in Binomial Expansion

none

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated exponentiation as distributive over addition, writing $(x + h)^3$ as $x^3 + h^3$ rather than using the full binomial expansion $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Binomial Expansion of a Cube: For any real terms $a$ and $b$ , $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ . Exponents never distribute directly across addition.
Whenever expanding $(x + h)^3$ , write out all four terms using coefficients from Pascal's triangle ( $1, 3, 3, 1$ ) before distributing outside constants.

CORRECTED PATH

ACEpath Teacher's Demonstration

Expand $(x + h)^3$ fully: $(x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$ .
Substitute into $f(x + h)$ and distribute: $f(x + h) = 2(x^3 + 3x^2h + 3xh^2 + h^3) - 5(x + h) = 2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h$ .
Subtract $f(x)$ : $f(x + h) - f(x) = (2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h) - (2x^3 - 5x) = 6x^2h + 6xh^2 + 2h^3 - 5h$ .
Factor out $h$ and cancel: $\frac{f(x + h) - f(x)}{h} = 6x^2 + 6xh + 2h^2 - 5$ .
Take the limit: $\lim_{h \to 0} (6x^2 + 6xh + 2h^2 - 5) = 6x^2 - 5$ .

Final answer

(a) $f(x + h) = 2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h$ ; (b) $\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = 6x^2 - 5$

NEXT STEPS

Final Takeaways from Your Teacher

Memorize the binomial coefficients for $n = 3$ , which are $1, 3, 3, 1$ .
Practice expanding cubes of binomials of the form $(x + c)^3$ and $(x + h)^3$ until including cross terms becomes second nature.
When finding the derivative of a polynomial using the difference quotient, verify that terms without $h$ completely cancel out, leaving linear $x$ -terms in the final derivative.

OverviewACEpath Teacher's Diagnosis

02 The student fails to distribute the negative sign to the second term of f(x) when computing the difference f(x + h) - f(x).

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
(x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 $\implies$ f(x + h) = 2(x^3 + 3x^2h + 3xh^2 + h^3) - 5(x + h) $\implies$ f(x + h) = 2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h
Step 2
f(x + h) - f(x) = 2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h - 2x^3 - 5x
Correct
f(x + h) - f(x) = (2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h) - (2x^3 - 5x) = 6x^2h + 6xh^2 + 2h^3 - 5h
When subtracting $f(x)$ , you need parentheses around $f(x)$ : $-(2x^3 - 5x) = -2x^3 + 5x$ . Omitting parentheses caused $-5x$ to remain negative instead of flipping to $+5x$ .
Step 3
f(x + h) - f(x) = 6x^2h + 6xh^2 + 2h^3 - 10x - 5h
Correct
f(x + h) - f(x) = 6x^2h + 6xh^2 + 2h^3 - 5h
This step combined like terms based on your prior sign error ( $-5x - 5x = -10x$ ). Had the sign been distributed correctly, the non- $h$ terms $-5x$ and $+5x$ would have cancelled to $0$ .
Step 4
$\frac{f(x + h) - f(x)}{h}$ = 6x^2 + 6xh + 2h^2 - $\frac{10x}{h}$ - 5 $\implies$ $\text{The limit does not exist due to }$ - $\frac{10x}{h}$ $\text{ as }$ h $\to$ 0
Correct
$\lim_{h \to 0} \frac{h(6x^2 + 6xh + 2h^2 - 5)}{h} = \lim_{h \to 0} (6x^2 + 6xh + 2h^2 - 5) = 6x^2 - 5$
Because of the leftover $-10x$ term, $h$ could not be completely factored out of the numerator, leading you to conclude that the limit does not exist.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly used the binomial expansion formula to expand $(x + h)^3$ .

You accurately applied the distributive property to find the expanded expression for $f(x + h)$ .

You recognized that dividing by $h$ with a remaining non- $h$ term prevents evaluating the limit by direct substitution.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Sign distribution error

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Failing to enclose a multi-term polynomial in parentheses when subtracting it, which causes the negative sign to be applied only to the first term rather than distributed to all terms.

REASONING PATH

Your ACEpath teacher’s suggested approach

Parenthesize Subtracted Polynomials: $-(A - B) = -A + B$
Always wrap $f(x)$ in parentheses, writing $- (f(x))$ , and change the sign of every term inside before combining like terms.

CORRECTED PATH

ACEpath Teacher's Demonstration

Write out the subtraction with explicit grouping: $f(x + h) - f(x) = (2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h) - (2x^3 - 5x)$ .
Distribute the negative sign across both terms: $- (2x^3 - 5x) = -2x^3 + 5x$ .
Cancel opposite terms: $2x^3 - 2x^3 = 0$ and $-5x + 5x = 0$ , leaving $f(x + h) - f(x) = 6x^2h + 6xh^2 + 2h^3 - 5h$ .
Factor out $h$ from the numerator: $h(6x^2 + 6xh + 2h^2 - 5)$ .
Divide by $h$ and evaluate the limit as $h \to 0$ : $\lim_{h \to 0} (6x^2 + 6xh + 2h^2 - 5) = 6x^2 - 5$ .

Final answer

(a) $f(x + h) = 2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h$ ; (b) $\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = 6x^2 - 5$

NEXT STEPS

Final Takeaways from Your Teacher

Always enclose multi-term expressions in parentheses when substituting after a minus sign.
Remember the self-check rule for polynomial difference quotients: all terms without a factor of $h$ must cancel completely before dividing by $h$ .

OverviewACEpath Teacher's Diagnosis

03 The student forgets to distribute the leading coefficient 2 to all terms in the expanded binomial.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3
Step 2
f(x+h) = 2x^3 + 3x^2h + 3xh^2 + h^3 - 5x - 5h
Correct
f(x+h) = 2(x^3 + 3x^2h + 3xh^2 + h^3) - 5(x+h) = 2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h
When multiplying $2$ across $(x + h)^3$ , you distributed $2$ to $x^3$ but left $3x^2h$ , $3xh^2$ , and $h^3$ without multiplying them by $2$ .
Step 3
f(x+h)-f(x) = (2x^3+3x^2h+3xh^2+h^3-5x-5h)-(2x^3-5x) $\implies$ f(x+h)-f(x) = 3x^2h+3xh^2+h^3-5h
Correct
f(x+h)-f(x) = (2x^3+6x^2h+6xh^2+2h^3-5x-5h)-(2x^3-5x) = 6x^2h+6xh^2+2h^3-5h
You distributed the negative sign and canceled opposite terms properly, but the expression carries the incorrect coefficients from Step 2.
Step 4
h(3x^2+3xh+h^2-5) $\implies$ $\frac{f(x+h)-f(x)}{h}$ = 3x^2+3xh+h^2-5
Correct
$\frac{f(x+h)-f(x)}{h}$ = $\frac{h(6x^2+6xh+2h^2-5)}{h}$ = 6x^2+6xh+2h^2-5
Your factoring of $h$ and division by the denominator were executed correctly based on your previous numerator.
Step 5
$\lim_{h $\to$ 0}$ (3x^2+3xh+h^2-5) = 3x^2-5_
Correct
$\lim_{h $\to$ 0}$ (6x^2+6xh+2h^2-5) = 6x^2-5_
You evaluated the limit as $h \to 0$ accurately using direct substitution, yielding $3x^2 - 5$ instead of $6x^2 - 5$ due to the earlier distribution slip.

WHAT YOU DID WELL

Compliments from Your Teacher

You recalled and expanded the cubic binomial $(x + h)^3$ without error.

You handled the negative distribution across $-5(x + h)$ and $-(2x^3 - 5x)$ perfectly.

You correctly factored the common parameter $h$ to cancel the indeterminate quotient.

You evaluated the resulting limit cleanly via direct substitution.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Incomplete distribution across polynomial terms

You wrote $2x^3$ for the first term of $2(x+h)^3$ , but kept the remaining terms as $+ 3x^2h + 3xh^2 + h^3$ instead of multiplying each by $2$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Multiplying a multi-term expanded polynomial by an outer constant without keeping grouping parentheses around the polynomial often leads to multiplying only the leading term.

REASONING PATH

Your ACEpath teacher’s suggested approach

Keep Parentheses Until Distributed: For any expression $c \cdot P$ , replace $P$ with its full expansion inside parentheses before dropping grouping symbols: $c(a_1 + a_2 + \dots + a_n) = ca_1 + ca_2 + \dots + ca_n$ .
Write $2(x^3 + 3x^2h + 3xh^2 + h^3)$ explicitly as a middle step before expanding each term individually.

CORRECTED PATH

ACEpath Teacher's Demonstration

Distribute $2$ across all four terms of the cubic expansion: $2(x^3 + 3x^2h + 3xh^2 + h^3) = 2x^3 + 6x^2h + 6xh^2 + 2h^3$ .
Combine with $-5(x+h)$ to get $f(x+h) = 2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h$ .
Subtract $f(x)$ : $f(x+h) - f(x) = (2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h) - (2x^3 - 5x) = 6x^2h + 6xh^2 + 2h^3 - 5h$ .
Factor $h$ and divide: $\frac{h(6x^2 + 6xh + 2h^2 - 5)}{h} = 6x^2 + 6xh + 2h^2 - 5$ .
Take the limit: $\lim_{h \to 0} (6x^2 + 6xh + 2h^2 - 5) = 6x^2 - 5$ .

Final answer

(a) $f(x + h) = 2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h$ ; (b) $\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = 6x^2 - 5$

NEXT STEPS

Final Takeaways from Your Teacher

Always write an intermediate step with explicit grouping brackets, such as $2[\dots]$ , before expanding.
Perform a term-count check: an expression with 4 terms inside multiplied by a scalar must produce 4 scaled terms.
Work through the flashcards below to reinforce complete distribution across multi-term expressions.

Lock it in

Question

What is the complete algebraic expansion of $(a + b)^3$ ?

Answer

$a^3 + 3a^2b + 3ab^2 + b^3$ . The coefficients come from row 3 of Pascal's triangle ( $1, 3, 3, 1$ ).

Resources

Answer and steps

Final answer

Steps

Part (a): Expand f(x+h)f(x + h)f(x+h) completely

Part (b): Form the difference quotient and evaluate the limit

Typical errors

01 The student misremembers the binomial expansion of (x + h)³ as simply x³ + h³, ignoring the intermediate terms.

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 fails to distribute the negative sign to the second term of f(x) when computing the difference f(x + h) - f(x).

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 forgets to distribute the leading coefficient 2 to all terms in the expanded binomial.

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

Similar question

Part (a): Expand $f(x + h)$ completely