Determine the exact critical value of the real parameter $k$ such that the degree of the simplified polynomial expression $E(x) = (2k^2 - 7k + 3)x^3 - (k - 3)x^2 + (4k + 1)x - 9$ strictly drops from $3$ to $1$ after combining all like terms.

Final answer

$k = 3$

Steps

Understand the degree conditions

To determine the value of $k$ that reduces the degree of the polynomial expression $E(x)$ from $3$ to $1$ , examine the conditions required for each power of $x$ :

The polynomial is given in standard form: $E(x) = (2k^2 - 7k + 3)x^3 - (k - 3)x^2 + (4k + 1)x - 9$

For the degree of $E(x)$ to strictly drop from $3$ to $1$ :

The coefficient of $x^3$ must equal $0$ (eliminating the degree- $3$ term).
The coefficient of $x^2$ must equal $0$ (ensuring the degree does not remain at $2$ ).
The coefficient of $x$ must be non-zero (ensuring the degree is exactly $1$ , not $0$ ).

Eliminate the $x^3$ term

Set the coefficient of $x^3$ to zero: $2k^2 - 7k + 3 = 0$

Factor the quadratic expression: $(2k - 1)(k - 3) = 0$

This yields two possible values: $k = \frac{1}{2} \quad \text{or} \quad k = 3$

Eliminate the $x^2$ term

Set the coefficient of $x^2$ to zero: $-(k - 3) = 0 \implies k = 3$

Common Pitfall: If $k = \frac{1}{2}$ is selected, the $x^3$ term vanishes, but the $x^2$ term becomes $-\left(\frac{1}{2} - 3\right)x^2 = \frac{5}{2}x^2 \neq 0$ . In that case, the degree of the polynomial drops to $2$ , not $1$ .

Verify the linear term

Both the cubic and quadratic terms vanish simultaneously only when: $k = 3$

Now verify that the degree- $1$ coefficient is non-zero at $k = 3$ : $4k + 1 = 4(3) + 1 = 13 \neq 0$

Substituting $k = 3$ into $E(x)$ : $E(x) = 0\cdot x^3 - 0\cdot x^2 + 13x - 9 = 13x - 9$

Since $13 \neq 0$ , the simplified expression is linear (degree $1$ ).

Thus, the exact critical value is $k = 3$ .

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student only requires the cubic term to vanish without eliminating the quadratic term, leading them to incorrectly accept both roots of 2k² - 7k + 3 = 0.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
2k^2 - 7k + 3 = 0 $\implies$ (2k - 1)(k - 3) = 0
Step 2
k = $\frac{1}{2}$ $\text{ or }$ k = 3
Correct
\lbrace 2k^2 - 7k + 3 = 0, -(k - 3) = 0 $\implies$ k = 3
You treated both roots as valid final answers. However, if $k = \frac{1}{2}$ , the polynomial becomes $\frac{5}{2}x^2 + 3x - 9$ , which has degree $2$ . To drop the degree strictly to $1$ , the coefficient of $x^2$ , $-(k - 3)$ , must also equal $0$ , which uniquely requires $k = 3$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly recognized that the coefficient of the highest-degree term $x^3$ must vanish.

You factored the quadratic expression $2k^2 - 7k + 3$ accurately into $(2k - 1)(k - 3)$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

incomplete degree condition

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You focused exclusively on eliminating the cubic term and assumed that making the leading coefficient zero would automatically make the polynomial linear, forgetting to enforce that intermediate terms with degree greater than $1$ must also vanish simultaneously.

REASONING PATH

Your ACEpath teacher’s suggested approach

Degree Reduction Condition: For a polynomial's degree to drop from $n$ to $m$ (where $m < n$ ), all coefficients for powers $x^p$ with $m < p \le n$ must vanish simultaneously, and the coefficient of $x^m$ must remain non-zero.
Always list and solve the system of equations for every power of $x$ that needs to disappear, then check the remaining leading coefficient.

CORRECTED PATH

ACEpath Teacher's Demonstration

Set the coefficient of $x^2$ equal to zero: $-(k - 3) = 0 \implies k = 3$ .
Find the common solution between the $x^3$ condition ( $k = \frac{1}{2}$ or $k = 3$ ) and the $x^2$ condition ( $k = 3$ ), which gives $k = 3$ .
Verify the linear term at $k = 3$ : $4k + 1 = 4(3) + 1 = 13 \neq 0$ .

Final answer

k = 3

NEXT STEPS

Final Takeaways from Your Teacher

When a problem specifies a target degree, write down an explicit condition for every power above that target degree.
Take the intersection (common values) of the parameters that satisfy all vanishing conditions simultaneously.
Always substitute your candidate values back into the expression to verify the resulting degree.

OverviewACEpath Teacher's Diagnosis

02 The student selects k = 1/2 by focusing only on the smaller root of the x³ coefficient while ignoring the requirement that the x² term must also vanish.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
2k^2 - 7k + 3 = 0 $\implies$ (2k - 1)(k - 3) = 0
Step 2
2k - 1 = 0 $\implies$ k = $\frac{1}{2}$
Correct
\lbrace 2k^2 - 7k + 3 = 0, -(k - 3) = 0, 4k + 1 $\neq$ 0 $\implies$ k = 3
To reduce the degree of $E(x)$ to strictly $1$ , the coefficient of $x^2$ must also vanish simultaneously. If $k = \frac{1}{2}$ , the $x^2$ term has coefficient $-\left(\frac{1}{2} - 3\right) = \frac{5}{2} \neq 0$ , leaving the degree at $2$ . Requiring $-(k - 3) = 0$ gives $k = 3$ , which eliminates both $x^3$ and $x^2$ while keeping the linear coefficient $4(3) + 1 = 13 \neq 0$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly recognized that the degree- $3$ term must be eliminated by setting its coefficient to zero.

You accurately factored the quadratic expression $2k^2 - 7k + 3$ into $(2k - 1)(k - 3)$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Overlooked intermediate degree condition

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You focused solely on eliminating the highest-degree term ( $x^3$ ) and did not consider that dropping the degree all the way down to $1$ requires the intermediate term ( $x^2$ ) to vanish at the same time.

REASONING PATH

Your ACEpath teacher’s suggested approach

Check all intermediate degrees: For a polynomial's degree to drop from $n$ to $m$ (where $m < n - 1$ ), the coefficients of all powers $x^p$ with $m < p \le n$ must simultaneously equal zero.
Set up a system of equations making every coefficient from degree $n$ down to degree $m+1$ zero, and find their common solution.
Verify non-zero target degree: The coefficient of the target degree $x^m$ must not equal zero.
Substitute your candidate value of the parameter into the degree- $m$ term to confirm it remains non-zero.

CORRECTED PATH

ACEpath Teacher's Demonstration

Set the degree- $2$ coefficient to zero: $-(k - 3) = 0 \implies k = 3$ .
Find the intersection of the conditions: $k \in \left\{\frac{1}{2}, 3\right\} \cap \{3\} \implies k = 3$ .
Verify that the linear coefficient is non-zero: $4(3) + 1 = 13 \neq 0$ .

Final answer

k = 3

NEXT STEPS

Final Takeaways from Your Teacher

Review the conditions needed when a polynomial skips one or more degrees.
Practice checking all intermediate degree coefficients whenever a problem asks for a degree drop greater than $1$ .

OverviewACEpath Teacher's Diagnosis

03 The student sets the x³ coefficient to zero and solves it with a sign error in factoring, finding k = -3 instead of k = 3.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
$2k^2 - 7k + 3 = 0 \quad \text{and} \quad k - 3 = 0$
Step 2
$k - 3 = 0 \text{ gives } k = 3 \implies 2(3)^2 - 7(3) + 3 = 18 - 21 + 3 = 0$
Step 3
$k = -3$
Correct
$k = 3$
After correctly establishing and verifying that $k = 3$ makes both coefficients zero, you wrote $k = -3$ as the final answer due to an accidental sign reversal.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly set up the simultaneous conditions requiring both the $x^3$ and $x^2$ coefficients to equal zero.

You correctly solved $k - 3 = 0$ to get $k = 3$ .

You verified that $k = 3$ satisfies the quadratic equation $2k^2 - 7k + 3 = 0$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Final sign reversal

You correctly derived and verified $k = 3$ in your work, but wrote $k = -3$ in your concluding line.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

A sign-inversion transcription slip occurred between solving the linear equation $k - 3 = 0 \implies k = 3$ and stating the final solution, likely conflating the negative sign inside the factor/equation $(k - 3 = 0)$ with the sign of the root itself.

REASONING PATH

Your ACEpath teacher’s suggested approach

Root vs. Subtraction Sign: Solving $k - c = 0$ yields $k = +c$ , not $-c$ .
Ensure that the sign of the solved root matches the positive value obtained from adding $c$ to both sides.
Verify Linear Coefficient: For the degree to be exactly 1, the linear coefficient must not vanish: $4k + 1 \neq 0$ .
Always check that the surviving target term has a non-zero coefficient at the critical value ( $4(3) + 1 = 13 \neq 0$ ).

CORRECTED PATH

ACEpath Teacher's Demonstration

The common solution satisfying both $2k^2 - 7k + 3 = 0$ and $-(k - 3) = 0$ is $k = 3$ .
Check the linear coefficient: $4k + 1 = 4(3) + 1 = 13 \neq 0$ , confirming the simplified expression has degree 1.

Final answer

$k = 3$

NEXT STEPS

Final Takeaways from Your Teacher

Double-check your final boxed answer directly against your working step above it to avoid transcription sign flips.
Always test the non-zero condition for the remaining leading coefficient to guarantee the required degree.

Lock it in

Question

If a polynomial $P(x) = a x^3 + b x^2 + c x + d$ must have degree strictly equal to $1$ , what conditions must the coefficients satisfy?

Answer

$a = 0$ , $b = 0$ , and $c \neq 0$ . Every term of degree higher than $1$ must vanish simultaneously while the linear term must remain non-zero.

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student only requires the cubic term to vanish without eliminating the quadratic term, leading them to incorrectly accept both roots of 2k² - 7k + 3 = 0.

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 selects k = 1/2 by focusing only on the smaller root of the x³ coefficient while ignoring the requirement that the x² term must also vanish.

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 sets the x³ coefficient to zero and solves it with a sign error in factoring, finding k = -3 instead of k = 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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