In the expansion of the expression $(4x^2 + k y^3)^2$, the middle term simplifies to $-56x^2 y^3$. Determine the value of the constant coefficient $k$, and state the resulting final term of the expanded trinomial.

In the expansion of the expression $(4x^2 + k y^3)^2$ , the middle term simplifies to $-56x^2 y^3$ . Determine the value of the constant coefficient $k$ , and state the resulting final term of the expanded trinomial.

Final answer

k = -7; $\text{ final term: }$ 49y^6

Steps

To expand the square of a binomial, recall the special product formula: $(a + b)^2 = a^2 + 2ab + b^2$

Step 1: Identify the terms of the binomial

In the expression $(4x^2 + ky^3)^2$ , set:

First term: $a = 4x^2$
Second term: $b = ky^3$

Step 2: Set up and solve for the coefficient $k$

The middle term in the expansion corresponds to $2ab$ : $\text{Middle term} = 2(4x^2)(ky^3) = 8k x^2 y^3$

We are given that this middle term simplifies to $-56x^2 y^3$ . Equating the two expressions: $8k x^2 y^3 = -56x^2 y^3$

Dividing both sides by the variable part $8x^2 y^3$ : $k = \frac{-56}{8} = -7$

Step 3: Determine the final term of the trinomial

The final term of the expansion is $b^2$ : $\text{Final term} = (ky^3)^2$

Substitute $k = -7$ : $(-7y^3)^2 = (-7)^2 \cdot (y^3)^2 = 49y^6$

Common Pitfall: Remember to apply the power of a power rule, $(y^m)^n = y^{m \cdot n}$ , so $(y^3)^2 = y^{3 \cdot 2} = y^6$ , not $y^{3^2} = y^9$ . Additionally, squaring a negative coefficient yields a positive result: $(-7)^2 = +49$ .

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets the factor of 2 in the middle term of the binomial expansion, equating the middle term directly to 4k.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$\text{Middle term}$ = (4x^2)(ky^3) = 4k x^2 y^3 $\implies$ 4k = -56 $\implies$ k = -14
Correct
$\text{Middle term}$ = 2(4x^2)(ky^3) = 8k x^2 y^3 $\implies$ 8k = -56 $\implies$ k = -7
When expanding a squared binomial $(a + b)^2 = a^2 + 2ab + b^2$ , the middle term is $2ab$ , not simply $ab$ . Multiplying $2(4x^2)(ky^3)$ gives $8k x^2 y^3$ , which leads to $8k = -56$ and $k = -7$ .
Step 2
$\text{Final term}$ = (ky^3)^2 = (-14y^3)^2 $\implies$ $\text{Final term}$ = 196y^6
Correct
$\text{Final term}$ = (ky^3)^2 = (-7y^3)^2 = 49y^6
Your squaring process and exponent rule application were completely sound, but using the incorrect coefficient $k = -14$ propagated to an incorrect final term.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified $a = 4x^2$ and $b = ky^3$ as the component terms of the binomial.

You correctly applied the power of a power rule and squared both the negative coefficient and variable part to compute the final term.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Omitted factor of 2 in binomial square expansion

You set the middle term equal to the simple product of the two terms $ab$ rather than doubling it to $2ab$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Incomplete recall of the perfect square trinomial pattern $(a + b)^2 = a^2 + 2ab + b^2$ , where the cross-term must be doubled because $(a+b)(a+b)$ produces two identical cross-products ( $ab + ba$ ).

REASONING PATH

Your ACEpath teacher’s suggested approach

Middle Term Rule: For any squared binomial $(a + b)^2$ , the middle term is always $2ab$ .
Write out the template $(\dots)^2 + 2(\dots)(\dots) + (\dots)^2$ before substituting terms.

CORRECTED PATH

ACEpath Teacher's Demonstration

Set the middle term using the formula $2ab$ : $2(4x^2)(ky^3) = 8kx^2y^3$ .
Equate with the given value: $8k = -56$ , which gives $k = -7$ .
Substitute $k = -7$ into $b^2$ : $(-7y^3)^2 = 49y^6$ .

Final answer

k = -7; $\text{ final term: }$ 49y^6

NEXT STEPS

Final Takeaways from Your Teacher

Review the special product identity $(a + b)^2 = a^2 + 2ab + b^2$ using geometric area models to reinforce why the factor of $2$ appears.
Practice expanding binomials with non-unit coefficients by explicitly writing $2 \cdot a \cdot b$ as an unsimplified step.

OverviewACEpath Teacher's Diagnosis

02 The student correctly finds k = -7 but forgets to square the constant k when computing the final term.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
2(4x^2)(ky^3) = 8kx^2y^3 $\implies$ 8k = -56 $\implies$ k = -7
Step 2
$\text{Final term}$ = k(y^3)^2 $\implies$ $\text{Final term}$ = -7y^6
Correct
$\text{Final term}$ = (ky^3)^2 = k^2(y^3)^2 = (-7)^2 y^6 = 49y^6
When expanding $(a + b)^2 = a^2 + 2ab + b^2$ , the final term is $b^2 = (ky^3)^2$ . The entire term must be squared, including its coefficient $k$ , which yields $(-7)^2 y^6 = 49y^6$ rather than $-7y^6$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly applied the middle-term structure $2ab$ for a squared binomial.

You accurately equated coefficients and solved for $k = -7$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Power of a Product Omission

You squared the variable factor $(y^3)^2$ to get $y^6$ , but left the coefficient $k$ unsquared.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the square of an entire term $(c \cdot v)^2$ with multiplying the coefficient by the squared variable $c \cdot (v)^2$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Square the Entire Second Term: In $(a + b)^2$ , the last term is $b^2 = (ky^n)^2 = k^2 y^{2n}$ .
Always wrap the complete second term in parentheses before applying the exponent $2$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Identify the second term of the binomial as $b = ky^3 = -7y^3$ .
Apply the exponent to the entire term: $b^2 = (-7y^3)^2$ .
Distribute the power to both the coefficient and the variable factor: $(-7)^2 \cdot (y^3)^2 = 49y^6$ .

Final answer

k = -7; $\text{ final term: }$ 49y^6

NEXT STEPS

Final Takeaways from Your Teacher

Always write full parentheses around monomial terms before squaring them, such as $(c y^n)^2 = c^2 y^{2n}$ .
Double-check that squaring any real-coefficient term always results in a non-negative coefficient.

OverviewACEpath Teacher's Diagnosis

03 The student squares the term (-7 y³)² but adds exponents instead of multiplying them, writing y⁵ instead of y⁶.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
2(4)(k) = 8k $\implies$ 8k = -56 $\implies$ k = -7
Step 2
$\text{Final term}$ = (-7y^3)^2 $\implies$ (-7)^2 = 49
Step 3
(y^3)^2 = y^{3+2} = y^5
Correct
(y^3)^2 = y^{3 $\cdot$ 2} = y^6
When raising a power to another power, you must multiply the exponents via $(y^m)^n = y^{m \cdot n}$ , rather than adding them.
Step 4
$\text{Final term} = 49y^5$
Correct
$\text{Final term} = 49y^6$
This final expression is affected by the exponent addition mistake made in the previous step.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly used the middle term pattern $2ab$ to set up an equation and solve for $k = -7$ .

You correctly set up the final term as $(-7y^3)^2$ and evaluated $(-7)^2 = 49$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Exponent rule confusion: Power of a Power vs. Product of Powers

You wrote $(y^3)^2 = y^{3+2} = y^5$ instead of multiplying the exponents to get $y^{3 \cdot 2} = y^6$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the power of a power rule $(a^m)^n = a^{m \cdot n}$ with the product rule $a^m \cdot a^n = a^{m+n}$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Power of a Power Rule: When raising a base with an exponent to an outside power, $(a^m)^n = a^{m \cdot n}$ .
Multiply the inner exponent by the outer exponent.
Product of Powers Rule: When multiplying two like bases, $a^m \cdot a^n = a^{m+n}$ .
Only add exponents when multiplying separate exponential terms with the same base.

CORRECTED PATH

ACEpath Teacher's Demonstration

Apply the power of a power rule: $(y^3)^2 = y^{3 \cdot 2} = y^6$ .
Combine the squared coefficient with the variable factor: $49 \cdot y^6 = 49y^6$ .

Final answer

k = -7; $\text{ final term: }$ 49y^6

NEXT STEPS

Final Takeaways from Your Teacher

Review the distinction between $(x^a)^b = x^{ab}$ and $x^a \cdot x^b = x^{a+b}$ .
Complete the targeted flashcards and practice exercises below to build automatic recall for exponent rules.

Lock it in

Question

What is the algebraic identity for expanding $(a + b)^2$ ?

Answer

$a^2 + 2ab + b^2$ . Always remember to double the product of the terms for the middle term.

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student forgets the factor of 2 in the middle term of the binomial expansion, equating the middle term directly to 4k.

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 correctly finds k = -7 but forgets to square the constant k when computing the final term.

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 squares the term (-7 y³)² but adds exponents instead of multiplying them, writing y⁵ instead of y⁶.

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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