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In a continuous-flow chemical reactor, the mass flow rate of reactant A entering the vessel is modeled by Rin(t)=4t2−3t+5R_{\text{in}}(t) = 4t^2 - 3t + 5 (in g/min\text{g/min}), while the rate of consumption inside the tank is given by Rcons(t)=(t−2)(2t+1)R_{\text{cons}}(t) = (t - 2)(2t + 1) (in g/min\text{g/min}). Additionally, a cleaning purge introduces a dilution loss polynomial L(t)=−(t2−4t−3)L(t) = -(t^2 - 4t - 3) (in g/min\text{g/min}). Determine the net accumulation rate polynomial A(t)=Rin(t)−Rcons(t)−L(t)A(t) = R_{\text{in}}(t) - R_{\text{cons}}(t) - L(t) in standard form, and calculate the net rate of mass accumulation at t=5 mint = 5\text{ min}.

Answer and steps

Final answer

A(t)=3t2−4t+4A(t) = 3t^2 - 4t + 4; at t=5 mint = 5\text{ min}, A(5)=59 g/minA(5) = 59\text{ g/min}

Steps

01

Expand the consumption rate polynomial Rcons(t)R_{\text{cons}}(t)

To find the net accumulation rate polynomial A(t)A(t) and evaluate it at t=5 mint = 5\text{ min}, follow these algebraic steps:

Use the distributive property (FOIL method) to multiply the binomial factors: Rcons(t)=(t−2)(2t+1)=2t2+t−4t−2=2t2−3t−2R_{\text{cons}}(t) = (t - 2)(2t + 1) = 2t^2 + t - 4t - 2 = 2t^2 - 3t - 2

02

Set up the net accumulation expression

Substitute each component into the given formula A(t)=Rin(t)−Rcons(t)−L(t)A(t) = R_{\text{in}}(t) - R_{\text{cons}}(t) - L(t): A(t)=(4t2−3t+5)−(2t2−3t−2)−[−(t2−4t−3)]A(t) = (4t^2 - 3t + 5) - (2t^2 - 3t - 2) - [-(t^2 - 4t - 3)]

Notice that subtracting L(t)L(t) involves two negative signs, which simplify to addition: −L(t)=−[−(t2−4t−3)]=+(t2−4t−3)- L(t) = -[-(t^2 - 4t - 3)] = +(t^2 - 4t - 3)

Now, write out the terms by distributing the negative sign across Rcons(t)R_{\text{cons}}(t): A(t)=(4t2−3t+5)−2t2+3t+2+(t2−4t−3)A(t) = (4t^2 - 3t + 5) - 2t^2 + 3t + 2 + (t^2 - 4t - 3)

Pitfall: Be careful when subtracting polynomials. Ensure that the negative sign is distributed to every single term inside the parentheses (e.g., −(2t2−3t−2)=−2t2+3t+2-(2t^2 - 3t - 2) = -2t^2 + 3t + 2).

03

Combine like terms to write A(t)A(t) in standard form

Group terms by degree:

  • t2t^2-terms: (4−2+1)t2=3t2(4 - 2 + 1)t^2 = 3t^2
  • tt-terms: (−3+3−4)t=−4t(-3 + 3 - 4)t = -4t
  • Constant terms: 5+2−3=45 + 2 - 3 = 4

Combining these yields the polynomial in standard form: A(t)=3t2−4t+4A(t) = 3t^2 - 4t + 4

04

Evaluate the net accumulation rate at t=5 mint = 5\text{ min}

Substitute t=5t = 5 into the standard form of A(t)A(t): A(5)=3(5)2−4(5)+4A(5) = 3(5)^2 - 4(5) + 4 A(5)=3(25)−20+4=75−20+4=59 g/minA(5) = 3(25) - 20 + 4 = 75 - 20 + 4 = 59\text{ g/min}

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets that subtracting the negative polynomial L(t) results in addition, effectively distributing the negative sign only to the leading term or dropping the double negative entirely.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Partially correct
  1. Step 1

    Rcons(t)=2t2−3t−2R_{\text{cons}}(t) = 2t^2 - 3t - 2

  2. Step 2

    A(t)=(4t2−3t+5)−(2t2−3t−2)−(t2−4t−3)A(t) = (4t^2 - 3t + 5) - (2t^2 - 3t - 2) - (t^2 - 4t - 3)

    Correct

    A(t)=(4t2−3t+5)−(2t2−3t−2)−[−(t2−4t−3)]A(t) = (4t^2 - 3t + 5) - (2t^2 - 3t - 2) - [-(t^2 - 4t - 3)]

    The problem defines L(t)=−(t2−4t−3)L(t) = -(t^2 - 4t - 3). Subtracting L(t)L(t) requires writing −L(t)=−[−(t2−4t−3)]=+(t2−4t−3)-L(t) = -[-(t^2 - 4t - 3)] = +(t^2 - 4t - 3). By writing −(t2−4t−3)-(t^2 - 4t - 3), you dropped the original negative sign from L(t)L(t).

  3. Step 3

    A(t)=4t2−3t+5−2t2+3t+2−t2+4t+3  ⟹  A(t)=t2+4t+10A(t) = 4t^2 - 3t + 5 - 2t^2 + 3t + 2 - t^2 + 4t + 3 \implies A(t) = t^2 + 4t + 10

    Correct

    A(t)=4t2−3t+5−2t2+3t+2+t2−4t−3  ⟹  A(t)=3t2−4t+4A(t) = 4t^2 - 3t + 5 - 2t^2 + 3t + 2 + t^2 - 4t - 3 \implies A(t) = 3t^2 - 4t + 4

    Your distribution of negative signs and grouping of like terms followed your Step 2 line accurately, but the resulting polynomial was affected by the missing negative sign in L(t)L(t).

  4. Step 4

    A(5)=(5)2+4(5)+10  ⟹  A(5)=25+20+10=55 g/minA(5) = (5)^2 + 4(5) + 10 \implies A(5) = 25 + 20 + 10 = 55\text{ g/min}

    Correct

    A(5)=3(5)2−4(5)+4  ⟹  A(5)=75−20+4=59 g/minA(5) = 3(5)^2 - 4(5) + 4 \implies A(5) = 75 - 20 + 4 = 59\text{ g/min}

    You substituted t=5t = 5 and computed the arithmetic correctly according to your polynomial, but the final numerical value is affected by the earlier setup error.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly expanded the product (t−2)(2t+1)(t - 2)(2t + 1) using the distributive property.

You accurately distributed the subtraction sign across the terms of (2t2−3t−2)(2t^2 - 3t - 2).

You combined like terms correctly according to the expression you wrote.

You substituted t=5t = 5 into your quadratic polynomial and computed the value accurately.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Dropped double negative in polynomial substitution

You substituted (t2−4t−3)(t^2 - 4t - 3) instead of −(t2−4t−3)-(t^2 - 4t - 3) for L(t)L(t), effectively losing the outer negative sign defined in L(t)L(t).

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Overlooking that subtracting a quantity that already contains a leading negative sign creates a double negative: −L(t)=−[−(P(t))]=+P(t)-L(t) = -[-(P(t))] = +P(t).

REASONING PATH
Your ACEpath teacher’s suggested approach
Enclose substituted expressions in brackets

  • When substituting an expression with a leading negative sign into a subtraction, write −[−(...)]-[-(...)].

  • Always replace the variable with the exact full expression inside brackets before simplifying signs.

Simplify double negatives first

  • -[-P(t)] = +P(t)

  • Resolve adjacent negative signs into addition before distributing or combining terms.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Substitute L(t)=−(t2−4t−3)L(t) = -(t^2 - 4t - 3) completely: A(t)=(4t2−3t+5)−(2t2−3t−2)−[−(t2−4t−3)]A(t) = (4t^2 - 3t + 5) - (2t^2 - 3t - 2) - [-(t^2 - 4t - 3)].

  2. Simplify the double negative −[−(… )]-[-(\dots)] to addition: A(t)=(4t2−3t+5)−(2t2−3t−2)+(t2−4t−3)A(t) = (4t^2 - 3t + 5) - (2t^2 - 3t - 2) + (t^2 - 4t - 3).

  3. Distribute the subtraction across Rcons(t)R_{\text{cons}}(t): A(t)=4t2−3t+5−2t2+3t+2+t2−4t−3A(t) = 4t^2 - 3t + 5 - 2t^2 + 3t + 2 + t^2 - 4t - 3.

  4. Combine like terms: (4−2+1)t2+(−3+3−4)t+(5+2−3)=3t2−4t+4(4 - 2 + 1)t^2 + (-3 + 3 - 4)t + (5 + 2 - 3) = 3t^2 - 4t + 4.

  5. Evaluate at t=5t = 5: A(5)=3(5)2−4(5)+4=75−20+4=59 g/minA(5) = 3(5)^2 - 4(5) + 4 = 75 - 20 + 4 = 59\text{ g/min}.

Final answer

A(t)=3t2−4t+4A(t) = 3t^2 - 4t + 4; at t=5 mint = 5\text{ min}, A(5)=59 g/minA(5) = 59\text{ g/min}

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice substituting negative polynomials into subtraction operations using nested brackets.

  • Review the three targeted flashcards on double negatives and polynomial subtraction.

  • Solve the three follow-up transfer exercises below.

OverviewACEpath Teacher's Diagnosis

02 The student makes a sign error when expanding (t - 2)(2t + 1) by adding the linear terms incorrectly as -5t instead of -3t.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Partially correct
  1. Step 1

    Rcons(t)=2t2−5t−2R_{\text{cons}}(t) = 2t^2 - 5t - 2

    Correct

    Rcons(t)=(t−2)(2t+1)=2t2+t−4t−2=2t2−3t−2R_{\text{cons}}(t) = (t - 2)(2t + 1) = 2t^2 + t - 4t - 2 = 2t^2 - 3t - 2

    When multiplying (t−2)(2t+1)(t - 2)(2t + 1), the linear terms from the outer and inner products are +1t+1t and −4t-4t. Adding them gives t−4t=−3tt - 4t = -3t, not −5t-5t.

  2. Step 2

    L(t) = -t^2 + 4t + 3

  3. Step 3

    A(t) = (4t^2 - 3t + 5) - (2t^2 - 5t - 2) - (-t^2 + 4t + 3)   ⟹  \implies A(t) = 4t^2 - 3t + 5 - 2t^2 + 5t + 2 + t^2 - 4t - 3   ⟹  \implies A(t) = 3t^2 - 2t + 4

    Correct

    A(t) = (4t^2 - 3t + 5) - (2t^2 - 3t - 2) - (-t^2 + 4t + 3) = 3t^2 - 4t + 4

    Your subtraction and collection of like terms were algebraically consistent with your earlier lines, but the linear coefficient inherited the −5t-5t mistake from Step 1.

  4. Step 4

    A(5) = 3(5)^2 - 2(5) + 4   ⟹  \implies A(5) = 75 - 10 + 4 = 69 g/min\text{ g/min}

    Correct

    A(5) = 3(5)^2 - 4(5) + 4 = 75 - 20 + 4 = 59 g/min\text{ g/min}

    You evaluated your polynomial expression at t=5t = 5 accurately, but the final value differs from 59 g/min59\text{ g/min} because of the earlier propagated linear term.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly simplified L(t)L(t) by distributing the negative sign to every term inside the parentheses.

You correctly set up the overall rate formula A(t)=Rin(t)−Rcons(t)−L(t)A(t) = R_{\text{in}}(t) - R_{\text{cons}}(t) - L(t).

You distributed negative signs across all terms during subtraction without dropping any signs.

You substituted t=5t = 5 and computed the arithmetic accurately according to the order of operations.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Linear Term Sign Addition in Binomial Multiplication

When expanding (t−2)(2t+1)(t - 2)(2t + 1), you combined +1t+1t and −4t-4t as −5t-5t instead of −3t-3t.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You likely treated both the outer product +1t+1t and inner product −4t-4t as negative terms (adding −1t−4t=−5t-1t - 4t = -5t) rather than tracking the positive sign of (t)(+1)=+t(t)(+1) = +t.

REASONING PATH
Your ACEpath teacher’s suggested approach
FOIL Term-by-Term Check

  • When expanding (at+b)(ct+d)(at + b)(ct + d), write the outer product a⋅da \cdot d and inner product b⋅cb \cdot c with their explicit signs before combining.

  • Always write out +1t−4t+1t - 4t explicitly as an intermediate line before combining into −3t-3t.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Expand Rcons(t)=(t−2)(2t+1)=2t2+t−4t−2=2t2−3t−2R_{\text{cons}}(t) = (t - 2)(2t + 1) = 2t^2 + t - 4t - 2 = 2t^2 - 3t - 2.

  2. Substitute into A(t)A(t): A(t)=(4t2−3t+5)−(2t2−3t−2)−(−t2+4t+3)A(t) = (4t^2 - 3t + 5) - (2t^2 - 3t - 2) - (-t^2 + 4t + 3).

  3. Distribute signs: A(t)=4t2−3t+5−2t2+3t+2+t2−4t−3A(t) = 4t^2 - 3t + 5 - 2t^2 + 3t + 2 + t^2 - 4t - 3.

  4. Combine like terms: (4−2+1)t2+(−3+3−4)t+(5+2−3)=3t2−4t+4(4 - 2 + 1)t^2 + (-3 + 3 - 4)t + (5 + 2 - 3) = 3t^2 - 4t + 4.

  5. Evaluate at t=5t = 5: A(5)=3(5)2−4(5)+4=75−20+4=59 g/minA(5) = 3(5)^2 - 4(5) + 4 = 75 - 20 + 4 = 59\text{ g/min}.

Final answer

A(t) = 3t^2 - 4t + 4;  A(5) = 59 g/min\text{ g/min}

NEXT STEPS
Final Takeaways from Your Teacher

  • Write the intermediate step showing both the outer product and inner product explicitly with their signs when expanding binomials.

  • Verify the sign of each product pair individually before combining like degree terms.

OverviewACEpath Teacher's Diagnosis

03 The student treats the cleaning purge loss as an added term in the net accumulation formula rather than subtracting it, computing A(t) = R_in(t) - R_cons(t) + L(t).

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    Rcons(t)=2t2−3t−2  ⟹  L(t)=−t2+4t+3R_{\text{cons}}(t) = 2t^2 - 3t - 2 \implies L(t) = -t^2 + 4t + 3

  2. Step 2

    A(t) = (4t^2 - 3t + 5) - (2t^2 - 3t - 2) + (-t^2 + 4t + 3)

    Correct

    A(t) = (4t^2 - 3t + 5) - (2t^2 - 3t - 2) - (-t^2 + 4t + 3)

    The formula requires subtracting L(t)L(t), which means writing −L(t)- L(t) rather than +L(t)+ L(t).

  3. Step 3

    A(t) = 4t^2 - 3t + 5 - 2t^2 + 3t + 2 - t^2 + 4t + 3   ⟹  \implies A(t) = t^2 + 4t + 10

    Correct

    A(t) = 4t^2 - 3t + 5 - 2t^2 + 3t + 2 + t^2 - 4t - 3   ⟹  \implies A(t) = 3t^2 - 4t + 4

    You distributed signs and combined like terms accurately according to your setup, but the result was affected by the sign error in Step 2.

  4. Step 4

    A(5) = (5)^2 + 4(5) + 10   ⟹  \implies A(5) = 25 + 20 + 10 = 55 g/min\text{ g/min}

    Correct

    A(5) = 3(5)^2 - 4(5) + 4   ⟹  \implies A(5) = 75 - 20 + 4 = 59 g/min\text{ g/min}

    Your substitution and arithmetic were completely correct based on your polynomial, but the final value was affected by the earlier sign error.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly expanded the binomial product (t−2)(2t+1)(t - 2)(2t + 1) to get 2t2−3t−22t^2 - 3t - 2.

You accurately distributed the negative sign into L(t)=−(t2−4t−3)L(t) = -(t^2 - 4t - 3) to find −t2+4t+3-t^2 + 4t + 3.

You correctly evaluated your polynomial expression at t=5t = 5 with proper units.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Sign / Operation Error in Formula Substitution

You wrote +L(t)+ L(t) instead of −L(t)- L(t) when substituting the expanded loss polynomial into A(t)A(t).

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You likely noticed the leading negative sign in the expanded form L(t)=−t2+4t+3L(t) = -t^2 + 4t + 3 and treated the subtraction as already accounted for, inadvertently replacing the required minus sign with an addition sign.

REASONING PATH
Your ACEpath teacher’s suggested approach
Literal Formula Substitution

  • Always preserve the outer operation of the formula before considering the internal signs of the substituted expression: f(x)−g(x)f(x) - g(x) must be set up as f(x)−[g(x)]f(x) - [g(x)].

  • Wrap the substituted polynomial completely in brackets with the minus sign outside: −[−(t2−4t−3)]=+(t2−4t−3)- [-(t^2 - 4t - 3)] = +(t^2 - 4t - 3).

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Substitute L(t)L(t) into the formula with its required minus sign: A(t)=(4t2−3t+5)−(2t2−3t−2)−(−t2+4t+3)A(t) = (4t^2 - 3t + 5) - (2t^2 - 3t - 2) - (-t^2 + 4t + 3).

  2. Distribute the minus signs across both subtracted expressions: A(t)=4t2−3t+5−2t2+3t+2+t2−4t−3A(t) = 4t^2 - 3t + 5 - 2t^2 + 3t + 2 + t^2 - 4t - 3.

  3. Group and combine like terms: A(t)=(4−2+1)t2+(−3+3−4)t+(5+2−3)=3t2−4t+4A(t) = (4 - 2 + 1)t^2 + (-3 + 3 - 4)t + (5 + 2 - 3) = 3t^2 - 4t + 4.

  4. Evaluate at t=5t = 5: A(5)=3(5)2−4(5)+4=75−20+4=59 g/minA(5) = 3(5)^2 - 4(5) + 4 = 75 - 20 + 4 = 59\text{ g/min}.

Final answer

A(t)=3t2−4t+4A(t) = 3t^2 - 4t + 4; at t=5 mint = 5\text{ min}, A(5)=59 g/minA(5) = 59\text{ g/min}

NEXT STEPS
Final Takeaways from Your Teacher

  • Review the double-negative rule when subtracting negative expressions.

  • Practice using protective brackets whenever replacing a variable with an entire polynomial.

  • Complete the targeted flashcards and practice problems below.

Lock it in

Question

When a formula requires subtracting P(x)P(x), and P(x)=−(3x2−2x+1)P(x) = -(3x^2 - 2x + 1), how should −P(x)-P(x) be rewritten?

Answer

+[3x2−2x+1]+[3x^2 - 2x + 1]. Subtracting a negative polynomial creates a double negative: −[−(...)]=+(...)-[-(...)] = +(...).

Similar question

01

Given P(x)=2x2+5x−1P(x) = 2x^2 + 5x - 1 and Q(x)=−(x2−3x+4)Q(x) = -(x^2 - 3x + 4), determine the polynomial D(x)=P(x)−Q(x)D(x) = P(x) - Q(x) in standard form.

Answer

D(x)=3x2+2x+3D(x) = 3x^2 + 2x + 3. Key steps: D(x)=(2x2+5x−1)−[−(x2−3x+4)]=(2x2+5x−1)+(x2−3x+4)=(2+1)x2+(5−3)x+(−1+4)=3x2+2x+3D(x) = (2x^2 + 5x - 1) - [-(x^2 - 3x + 4)] = (2x^2 + 5x - 1) + (x^2 - 3x + 4) = (2 + 1)x^2 + (5 - 3)x + (-1 + 4) = 3x^2 + 2x + 3.

02

Let f(x)=3x2−2x+4f(x) = 3x^2 - 2x + 4, g(x)=(x−3)(x+2)g(x) = (x - 3)(x + 2), and h(x)=−(2x2−5x+1)h(x) = -(2x^2 - 5x + 1). Express M(x)=f(x)−g(x)−h(x)M(x) = f(x) - g(x) - h(x) in standard form.

Answer

M(x)=4x2−6x+11M(x) = 4x^2 - 6x + 11. Key steps: Expand g(x)=x2−x−6g(x) = x^2 - x - 6. Substitute: M(x)=(3x2−2x+4)−(x2−x−6)−[−(2x2−5x+1)]=3x2−2x+4−x2+x+6+2x2−5x+1=(3−1+2)x2+(−2+1−5)x+(4+6+1)=4x2−6x+11M(x) = (3x^2 - 2x + 4) - (x^2 - x - 6) - [-(2x^2 - 5x + 1)] = 3x^2 - 2x + 4 - x^2 + x + 6 + 2x^2 - 5x + 1 = (3 - 1 + 2)x^2 + (-2 + 1 - 5)x + (4 + 6 + 1) = 4x^2 - 6x + 11.

03

A reservoir's water volume change rate is modeled by V(t)=I(t)−O(t)−W(t)V(t) = I(t) - O(t) - W(t), where inflow is I(t)=5t2+2tI(t) = 5t^2 + 2t, outflow is O(t)=(2t−1)(t+3)O(t) = (2t - 1)(t + 3), and waste runoff is W(t)=−(t2−7t+4)W(t) = -(t^2 - 7t + 4). Find V(t)V(t) in standard form, and evaluate the net volume change rate at t=3t = 3.

Answer

V(t)=4t2+4t−1V(t) = 4t^2 + 4t - 1, and V(3)=47V(3) = 47. Key steps: Expand O(t)=2t2+5t−3O(t) = 2t^2 + 5t - 3. Notice −W(t)=−[−(t2−7t+4)]=+(t2−7t+4)-W(t) = -[-(t^2 - 7t + 4)] = +(t^2 - 7t + 4). Combine terms: V(t)=(5t2+2t)−(2t2+5t−3)+(t2−7t+4)=(5−2+1)t2+(2−5+7)t+(3−4)=4t2+4t−1V(t) = (5t^2 + 2t) - (2t^2 + 5t - 3) + (t^2 - 7t + 4) = (5 - 2 + 1)t^2 + (2 - 5 + 7)t + (3 - 4) = 4t^2 + 4t - 1. For t=3t = 3: V(3)=4(3)2+4(3)−1=36+12−1=47V(3) = 4(3)^2 + 4(3) - 1 = 36 + 12 - 1 = 47.

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