Consider the polynomial expression $E(x) = (ax^2 + 5x - 3)(2x^2 - bx + 4) - (6x^4 + 7x^3 - 14x^2 + 23x - 12)$. Determine the unique values of the real coefficients $a$ and $b$ such that the degree of $E(x)$ is strictly less than $3$ (that is, the coefficients of both $x^4$ and $x^3$ vanish identically).

Consider the polynomial expression $E(x) = (ax^2 + 5x - 3)(2x^2 - bx + 4) - (6x^4 + 7x^3 - 14x^2 + 23x - 12)$ . Determine the unique values of the real coefficients $a$ and $b$ such that the degree of $E(x)$ is strictly less than $3$ (that is, the coefficients of both $x^4$ and $x^3$ vanish identically).

Final answer

$a = 3$ , $b = 1$

Steps

Identify the coefficient of $x^4$

To make the degree of the polynomial $E(x)$ strictly less than $3$ , the coefficients of both $x^4$ and $x^3$ in the expanded expression must equal zero.

The term containing $x^4$ in the product $(ax^2 + 5x - 3)(2x^2 - bx + 4)$ arises solely from multiplying the $x^2$ terms of each factor: $(ax^2)(2x^2) = 2a x^4$

Subtracting the $x^4$ term from the second polynomial, $6x^4$ , gives the total $x^4$ term in $E(x)$ : $(2a - 6)x^4$

For this coefficient to vanish: $2a - 6 = 0 \implies 2a = 6 \implies a = 3$

Identify the coefficient of $x^3$

The terms containing $x^3$ in the product come from combining:

The $x^2$ term of the first factor and the $x$ term of the second factor: $(ax^2)(-bx) = -ab x^3$
The $x$ term of the first factor and the $x^2$ term of the second factor: $(5x)(2x^2) = 10 x^3$

Adding these gives the total $x^3$ term from the product: $(-ab + 10)x^3$

Subtracting the $x^3$ term from the second polynomial, $7x^3$ , gives the total $x^3$ term in $E(x)$ : $(-ab + 10 - 7)x^3 = (3 - ab)x^3$

For this coefficient to vanish: $3 - ab = 0 \implies ab = 3$

Solve for $b$

Substitute $a = 3$ into the equation $ab = 3$ : $3b = 3 \implies b = 1$

Verification of the degree

Substitute $a = 3$ and $b = 1$ back to check the remaining terms:

The $x^2$ term in the product is $(3x^2)(4) + (5x)(-x) + (-3)(2x^2) = (12 - 5 - 6)x^2 = 1x^2$ . Subtracting $-14x^2$ yields $1x^2 - (-14x^2) = 15x^2 \neq 0$ .
The resulting polynomial is $E(x) = 15x^2$ , which has degree $2$ .

Since $2 < 3$ , the condition is satisfied with $a = 3$ and $b = 1$ .

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to distribute the minus sign to the subtracted polynomial when setting up the x³ coefficient equation, resulting in b = 17/3.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
2a - 6 = 0 $\implies$ a = 3
Step 2
-ab + 10 + 7 = 0
Correct
-ab + 10 - 7 = 0
The second polynomial is preceded by a subtraction sign: $-(6x^4 + 7x^3 - \dots)$ . Distributing the negative sign gives $-7x^3$ , so the net coefficient equation should be $-ab + 10 - 7 = 0$ , not $-ab + 10 + 7 = 0$ .
Step 3
-3b + 17 = 0 $\implies$ b = $\frac{17}{3}$
Correct
-3b + 3 = 0 $\implies$ b = 1
Your algebraic solving technique was mathematically sound from your previous equation, but substituting $a = 3$ into the incorrect equation propagated the error.
Step 4
a = 3, b = $\frac{17}{3}$
Correct
a = 3, b = 1
Because of the sign error in setting up the $x^3$ coefficient equation, your final value for $b$ is incorrect.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified that the degree condition requires the coefficients of both $x^4$ and $x^3$ to vanish.

You correctly calculated the $x^4$ coefficient and solved for $a = 3$ .

You correctly identified the two cross-product terms that contribute to the $x^3$ degree from the product of the two polynomials.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Sign distribution error when subtracting polynomials

You forgot to distribute the leading minus sign to the $+7x^3$ term inside the subtracted polynomial, writing $+7$ instead of $-7$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating a subtracted polynomial $-(P(x))$ as if terms retain their original positive signs inside individual degree equations rather than distributing the negative sign to every term.

REASONING PATH

Your ACEpath teacher’s suggested approach

Distribute the negative sign completely: $-(A + B) = -A - B$
Always explicitly distribute the leading minus sign across every term of the second polynomial before collecting like terms.
Isolate degree coefficients systematically: [x^k] (P(x) - Q(x)) = [x^k]P(x) - [x^k]Q(x)
Write the subtraction explicitly as $(\text{product term}) - (\text{subtracted term})$ to avoid dropping the minus sign.

CORRECTED PATH

ACEpath Teacher's Demonstration

Set up the coefficient of $x^3$ : $-ab + 10 - 7 = 0 \implies 3 - ab = 0$ .
Substitute $a = 3$ into the equation: $3 - 3b = 0$ .
Solve for $b$ : $3b = 3 \implies b = 1$ .

Final answer

a = 3, b = 1

NEXT STEPS

Final Takeaways from Your Teacher

Practice distributing negative signs across grouped polynomial subtractions.
Verify your final coefficients by testing the remaining terms to ensure the degree condition is strictly satisfied.

OverviewACEpath Teacher's Diagnosis

02 The student neglects the negative sign on the -bx term when multiplying ax² by -bx, obtaining ab + 3 = 0 and b = -1.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
2a - 6 = 0 $\implies$ a = 3
Step 2
ab + 10 - 7 = 0 $\implies$ ab + 3 = 0
Correct
-ab + 10 - 7 = 0 $\implies$ 3 - ab = 0
When multiplying the terms that produce $x^3$ , the product of $ax^2$ and $-bx$ is $-ab x^3$ , not $+ab x^3$ . Combining with $(5x)(2x^2) = 10x^3$ and subtracting $7x^3$ yields $-ab + 10 - 7 = 0$ .
Step 3
3b + 3 = 0 $\implies$ b = -1
Correct
3(1 - b) = 0 $\implies$ 3 - 3b = 0 $\implies$ b = 1
Your algebraic solving of $3b + 3 = 0$ was executed correctly, but the equation itself was inherited from the sign error in the previous step.
Step 4
a = 3,\ b = -1
Correct
a = 3,\ b = 1
The final value for $b$ is incorrect due to the propagated sign error.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified the condition that the degree being strictly less than $3$ requires the coefficients of $x^4$ and $x^3$ to vanish.

You accurately extracted the $x^4$ coefficient and solved for $a = 3$ .

You correctly identified all pairs of terms whose product produces an $x^3$ term.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Sign error in polynomial multiplication

You treated the linear term in the second factor as $+bx$ rather than $-bx$ when expanding $(ax^2)(-bx)$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Overlooking the subtraction sign attached to a term inside parentheses when selecting individual terms to multiply.

REASONING PATH

Your ACEpath teacher’s suggested approach

Include the Operator with the Term: Every term in a polynomial carries the sign immediately preceding it: in $(2x^2 - bx + 4)$ , the linear term is $(-bx)$ .
Always circle or write parentheses around the full signed term, such as $(ax^2) \cdot (-bx) = -ab x^3$ , before combining like terms.

CORRECTED PATH

ACEpath Teacher's Demonstration

Identify the signed terms that contribute to $x^3$ : $(ax^2)(-bx) = -ab x^3$ and $(5x)(2x^2) = 10x^3$ .
Subtract the $7x^3$ term from the second polynomial: $(-ab + 10 - 7)x^3 = (3 - ab)x^3$ .
Set the coefficient to zero: $3 - ab = 0 \implies ab = 3$ .
Substitute $a = 3$ : $3b = 3 \implies b = 1$ .

Final answer

a = 3,\ b = 1

NEXT STEPS

Final Takeaways from Your Teacher

Review the expansion of polynomial products containing negative coefficients.
Practice tracking signs explicitly by writing terms in parentheses before multiplying.
Complete the targeted flashcards and practice problems below to reinforce sign accuracy.

OverviewACEpath Teacher's Diagnosis

03 The student omits the contribution of the cross-term 5x · 2x² when collecting the x³ terms, leading to b = -7/3.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
2a - 6 = 0 $\implies$ a = 3
Step 2
-ab - 7 = 0 $\implies$ -3b - 7 = 0 $\implies$ b = - $\frac{7}{3}$
Correct
(-ab + 10 - 7) = 0 $\implies$ 3 - 3b = 0 $\implies$ b = 1
You accounted for the cross-term $(ax^2)(-bx) = -ab x^3$ and the subtracted term $-7x^3$ , but you overlooked the second cross-term from the product: $(5x)(2x^2) = 10x^3$ . The complete $x^3$ coefficient is $-ab + 10 - 7 = 3 - ab$ , which gives $b = 1$ when $a = 3$ .
Step 3
a = 3, b = - $\frac{7}{3}$
Correct
a = 3, b = 1
This final answer carries over the value of $b$ computed from the incomplete $x^3$ coefficient in Step 2.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly set up and solved the equation for the $x^4$ coefficient to find $a = 3$ .

You recognized that eliminating higher-degree terms requires their combined coefficients to equal zero.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Omitted cross-term in polynomial multiplication

You omitted the cross-term $(5x)(2x^2) = 10x^3$ when gathering all degree-3 terms from the product.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

When multiplying two multi-term polynomials, a target degree $k$ can be formed by several exponent pairs $(i, j)$ such that $i + j = k$ . You only took the pair $(2, 1)$ and missed $(1, 2)$ before subtracting the second polynomial's $x^3$ term.

REASONING PATH

Your ACEpath teacher’s suggested approach

Exponent Sum Check: To collect terms of degree $k$ in $(P)(Q)$ , find all pairs $(i, j)$ of term degrees such that $i + j = k$ .
For degree 3, check both $(x^2 \cdot x^1)$ and $(x^1 \cdot x^2)$ before proceeding to simplify.

CORRECTED PATH

ACEpath Teacher's Demonstration

List all pairs of terms in $(ax^2 + 5x - 3)(2x^2 - bx + 4)$ that multiply to give $x^3$ : $(ax^2)(-bx) = -ab x^3$ and $(5x)(2x^2) = 10x^3$ .
Add the product cross-terms: $(-ab + 10)x^3$ .
Subtract the $x^3$ term from the second polynomial: $(-ab + 10 - 7)x^3 = (3 - ab)x^3$ .
Set the net coefficient to zero: $3 - ab = 0$ .
Substitute $a = 3$ : $3 - 3b = 0 \implies b = 1$ .

Final answer

a = 3, b = 1

NEXT STEPS

Final Takeaways from Your Teacher

Practice explicitly writing out all exponent pairs $(i, j)$ whose sum matches the target power before multiplying.
Check every bracketed term systematically using the distributive property to ensure no cross-product is missed.

Lock it in

Question

When expanding $-(c x^3 + d x^2)$ , what are the coefficients of $x^3$ and $x^2$ ?

Answer

$-c$ and $-d$ . The negative sign outside parentheses distributes to every term inside: $-(A + B) = -A - B$ .

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student forgets to distribute the minus sign to the subtracted polynomial when setting up the x³ coefficient equation, resulting in b = 17/3.

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 neglects the negative sign on the -bx term when multiplying ax² by -bx, obtaining ab + 3 = 0 and b = -1.

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 omits the contribution of the cross-term 5x · 2x² when collecting the x³ terms, leading to b = -7/3.

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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