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Let P(u)=12u2−23u+34P(u) = \frac{1}{2}u^2 - \frac{2}{3}u + \frac{3}{4} and Q(u)=35u−52Q(u) = \frac{3}{5}u - \frac{5}{2}. Compute the product P(u)⋅Q(u)P(u) \cdot Q(u), subtracting 110u3\frac{1}{10}u^3, and express all fractional coefficients in reduced rational form in standard order.

Answer and steps

Final answer

15u3−3320u2+canopy12760u−158\frac{1}{5}u^3 - \frac{33}{20}u^2 + canopy\frac{127}{60}u - \frac{15}{8}

Steps

01

Expand the product P(u)⋅Q(u)P(u) \cdot Q(u)

To find the required polynomial expression, expand the product P(u)⋅Q(u)P(u) \cdot Q(u) using the distributive property, subtract 110u3\frac{1}{10}u^3, and combine like terms.

Multiply each term of P(u)=12u2−23u+34P(u) = \frac{1}{2}u^2 - \frac{2}{3}u + \frac{3}{4} by each term of Q(u)=35u−52Q(u) = \frac{3}{5}u - \frac{5}{2}:

P(u)⋅Q(u)=(12u2−23u+34)(35u−52)P(u) \cdot Q(u) = \left(\frac{1}{2}u^2 - \frac{2}{3}u + \frac{3}{4}\right)\left(\frac{3}{5}u - \frac{5}{2}\right)

Distributing across terms:

  1. Cubic term (u3u^3): (12u2)(35u)=310u3\left(\frac{1}{2}u^2\right)\left(\frac{3}{5}u\right) = \frac{3}{10}u^3
  2. Quadratic terms (u2u^2): (12u2)(−52)+(−23u)(35u)=−54u2−25u2\left(\frac{1}{2}u^2\right)\left(-\frac{5}{2}\right) + \left(-\frac{2}{3}u\right)\left(\frac{3}{5}u\right) = -\frac{5}{4}u^2 - \frac{2}{5}u^2 Find a common denominator of 2020: −2520u2−820u2=−3320u2-\frac{25}{20}u^2 - \frac{8}{20}u^2 = -\frac{33}{20}u^2
  3. Linear terms (uu): (−23u)(−52)+(34)(35u)=53u+920u\left(-\frac{2}{3}u\right)\left(-\frac{5}{2}\right) + \left(\frac{3}{4}\right)\left(\frac{3}{5}u\right) = \frac{5}{3}u + \frac{9}{20}u Find a common denominator of 6060: 10060u+2760u=12760u\frac{100}{60}u + \frac{27}{60}u = \frac{127}{60}u
  4. Constant term: (34)(−52)=−158\left(\frac{3}{4}\right)\left(-\frac{5}{2}\right) = -\frac{15}{8}

Putting these together gives: P(u)⋅Q(u)=310u3−3320u2+12760u−158P(u) \cdot Q(u) = \frac{3}{10}u^3 - \frac{33}{20}u^2 + \frac{127}{60}u - \frac{15}{8}

02

Subtract 110u3\frac{1}{10}u^3 and simplify

Subtract 110u3\frac{1}{10}u^3 from the cubic term of the product:

(310u3−110u3)−3320u2+12760u−158=210u3−3320u2+12760u−158\left(\frac{3}{10}u^3 - \frac{1}{10}u^3\right) - \frac{33}{20}u^2 + \frac{127}{60}u - \frac{15}{8} = \frac{2}{10}u^3 - \frac{33}{20}u^2 + \frac{127}{60}u - \frac{15}{8}

Reduce the coefficient 210\frac{2}{10} to lowest terms:

210=15\frac{2}{10} = \frac{1}{5}

Thus, the polynomial in standard form is: 15u3−3320u2+12760u−158\frac{1}{5}u^3 - \frac{33}{20}u^2 + \frac{127}{60}u - \frac{15}{8}

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to distribute the negative sign when multiplying -2/3 u by -5/2, resulting in -5/3 u instead of +5/3 u.

Sample Student Work

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

    P(u)⋅Q(u)=(12u2−23u+34)(35u−52)  ⟹  =310u3−54u2−25u2−53u+920u−158P(u) \cdot Q(u) = \left(\frac{1}{2}u^2 - \frac{2}{3}u + \frac{3}{4}\right)\left(\frac{3}{5}u - \frac{5}{2}\right) \implies = \frac{3}{10}u^3 - \frac{5}{4}u^2 - \frac{2}{5}u^2 - \frac{5}{3}u + \frac{9}{20}u - \frac{15}{8}

    Correct

    P(u) ⋅\cdot Q(u) = 310\frac{3}{10}u^3 - 54\frac{5}{4}u^2 - 25\frac{2}{5}u^2 + 53\frac{5}{3}u + 920\frac{9}{20}u - 158\frac{15}{8}

    When multiplying the linear term −23u-\frac{2}{3}u by the constant term −52-\frac{5}{2}, the product of two negative numbers is positive: (−23)⋅(−52)=+106=+53\left(-\frac{2}{3}\right) \cdot \left(-\frac{5}{2}\right) = +\frac{10}{6} = +\frac{5}{3}. You wrote −53u-\frac{5}{3}u instead.

  2. Step 2

    = 310\frac{3}{10}u^3 - 3320\frac{33}{20}u^2 - 7360\frac{73}{60}u - 158\frac{15}{8}

    Correct

    = 310\frac{3}{10}u^3 - 3320\frac{33}{20}u^2 + 12760\frac{127}{60}u - 158\frac{15}{8}

    Your fraction addition −53+920=−10060+2760=−7360-\frac{5}{3} + \frac{9}{20} = -\frac{100}{60} + \frac{27}{60} = -\frac{73}{60} was algebraically consistent with your previous line, but it carries forward the earlier sign error. With the correct positive term, it should be +10060+2760=+12760+\frac{100}{60} + \frac{27}{60} = +\frac{127}{60}.

  3. Step 3

    Subtract \text{Subtract } 110\frac{1}{10}u^3:   ⟹  \implies = (310−110)\left(\frac{3}{10} - \frac{1}{10}\right)u^3 - 3320\frac{33}{20}u^2 - 7360\frac{73}{60}u - 158\frac{15}{8}   ⟹  \implies = 15\frac{1}{5}u^3 - 3320\frac{33}{20}u^2 - 7360\frac{73}{60}u - 158\frac{15}{8}

    Correct

    15\frac{1}{5}u^3 - 3320\frac{33}{20}u^2 + 12760\frac{127}{60}u - 158\frac{15}{8}

    You subtracted 110u3\frac{1}{10}u^3 and simplified 210\frac{2}{10} to 15\frac{1}{5} completely correctly, but the overall polynomial retains the incorrect linear term −7360u-\frac{73}{60}u.

WHAT YOU DID WELL
Compliments from Your Teacher

You systematically applied the distributive property to multiply each term of the trinomial by each term of the binomial.

You found correct common denominators and accurately combined the quadratic coefficients.

You followed the instruction to subtract 110u3\frac{1}{10}u^3 and simplified 210u3\frac{2}{10}u^3 to 15u3\frac{1}{5}u^3 in lowest terms.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Sign error when multiplying negative terms

You dropped the negative sign cancellation when computing (−23u)⋅(−52)\left(-\frac{2}{3}u\right) \cdot \left(-\frac{5}{2}\right), writing −53u-\frac{5}{3}u instead of +53u+\frac{5}{3}u.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Inattention to sign pairing during multi-term polynomial distribution, where terms with leading negative signs are treated as single negative factors rather than realizing that (negative)×(negative)=positive(\text{negative}) \times (\text{negative}) = \text{positive}.

REASONING PATH
Your ACEpath teacher’s suggested approach
Two-Sign Multiplication Rule

  • For any real numbers a,b>0a, b > 0, (−a)⋅(−b)=+ab(-a) \cdot (-b) = +ab.

  • Circle or highlight the sign attached to each factor before carrying out the numerical multiplication.

Term-by-Term Sign Audit

  • Each product generated in polynomial distribution must have an explicit sign determined prior to combining like terms.

  • Write out the product with explicit parentheses, such as (−23u)(−52)\left(-\frac{2}{3}u\right)\left(-\frac{5}{2}\right), before writing the combined sign in the expanded line.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Re-evaluate the linear term product: (−23u)(−52)=+106u=+53u\left(-\frac{2}{3}u\right)\left(-\frac{5}{2}\right) = +\frac{10}{6}u = +\frac{5}{3}u.

  2. Combine the linear terms using the common denominator 6060: +53u+920u=10060u+2760u=+12760u+\frac{5}{3}u + \frac{9}{20}u = \frac{100}{60}u + \frac{27}{60}u = +\frac{127}{60}u.

  3. Assemble the full polynomial: 15u3−3320u2+12760u−158\frac{1}{5}u^3 - \frac{33}{20}u^2 + \frac{127}{60}u - \frac{15}{8}.

Final answer

15\frac{1}{5}u^3 - 3320\frac{33}{20}u^2 + 12760\frac{127}{60}u - 158\frac{15}{8}

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice distributing binomials with multiple negative signs to build automatic sign tracking.

  • Always perform a quick sign check on every individual distributed term before adding or subtracting fractions.

OverviewACEpath Teacher's Diagnosis

02 The student computes the product P(u) · Q(u) correctly but completely overlooks the instruction to subtract 1/10 u³.

Sample Student Work

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

    P(u)⋅Q(u)=(12u2−23u+34)(35u−52)  ⟹  =310u3−54u2−25u2+53u+920u−158P(u) \cdot Q(u) = \left(\frac{1}{2}u^2 - \frac{2}{3}u + \frac{3}{4}\right)\left(\frac{3}{5}u - \frac{5}{2}\right) \implies = \frac{3}{10}u^3 - \frac{5}{4}u^2 - \frac{2}{5}u^2 + \frac{5}{3}u + \frac{9}{20}u - \frac{15}{8}

  2. Step 2

    = 310\frac{3}{10}u^3 - 3320\frac{33}{20}u^2 + 12760\frac{127}{60}u - 158\frac{15}{8}

  3. Step 3

    310\frac{3}{10}u^3 - 3320\frac{33}{20}u^2 + 12760\frac{127}{60}u - 158\frac{15}{8}

    Correct

    P(u)⋅Q(u)−110u3=(310−110)u3−3320u2+12760u−158=15u3−3320u2+12760u−158P(u) \cdot Q(u) - \frac{1}{10}u^3 = \left(\frac{3}{10} - \frac{1}{10}\right)u^3 - \frac{33}{20}u^2 + \frac{127}{60}u - \frac{15}{8} = \frac{1}{5}u^3 - \frac{33}{20}u^2 + \frac{127}{60}u - \frac{15}{8}

    You presented the product P(u)⋅Q(u)P(u) \cdot Q(u) as your final response, omitting the secondary instruction to subtract 110u3\frac{1}{10}u^3 from the result.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly applied the distributive property to expand all six terms of the polynomial product.

You handled positive and negative fraction multiplications without any sign errors.

You accurately found common denominators to combine the quadratic and linear terms.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Omitted Sub-Operation

You finished the primary calculation (P(u)⋅Q(u)P(u) \cdot Q(u)) and boxed it as the solution, forgetting to apply the additional step of subtracting 110u3\frac{1}{10}u^3.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

In multi-clause algebraic word problems, completing a heavy algebraic routine often creates closure fatigue, leading you to treat the intermediate polynomial product as the finished task rather than rereading the prompt to verify all required operations.

REASONING PATH
Your ACEpath teacher’s suggested approach
Prompt Annotation

  • Every distinct mathematical verb or operation clause in a prompt must be accounted for in your solution plan.

  • Underline or list each requested action before solving: (1) Compute P(u)⋅Q(u)P(u) \cdot Q(u), (2) Subtract 110u3\frac{1}{10}u^3, (3) Simplify to lowest terms.

Final Verification

  • Never box a final answer without matching the output variable or expression to the prompt's target quantity.

  • Check whether your result is P(u)⋅Q(u)P(u) \cdot Q(u) or P(u)⋅Q(u)−110u3P(u) \cdot Q(u) - \frac{1}{10}u^3 before finalizing.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Subtract 110u3\frac{1}{10}u^3 from the cubic term: (310−110)u3=210u3\left(\frac{3}{10} - \frac{1}{10}\right)u^3 = \frac{2}{10}u^3.

  2. Reduce the resulting fraction to lowest terms: 210=15\frac{2}{10} = \frac{1}{5}.

  3. Reassemble the polynomial with the unchanged lower-degree terms.

Final answer

15\frac{1}{5}u^3 - 3320\frac{33}{20}u^2 + 12760\frac{127}{60}u - 158\frac{15}{8}

NEXT STEPS
Final Takeaways from Your Teacher

  • Annotate multi-part problem statements by numbering each required calculation in the margin before writing algebraic lines.

  • Practice an end-of-problem check routine where you re-read the exact question sentence before declaring your final boxed answer.

OverviewACEpath Teacher's Diagnosis

03 The student miscalculates the common denominator addition for the u² coefficients by subtracting numerators directly without finding a common denominator.

Sample Student Work

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

    P(u)⋅Q(u)=(12u2−23u+34)(35u−52)  ⟹  =310u3−54u2−25u2+53u+920u−158P(u) \cdot Q(u) = \left(\frac{1}{2}u^2 - \frac{2}{3}u + \frac{3}{4}\right)\left(\frac{3}{5}u - \frac{5}{2}\right) \implies = \frac{3}{10}u^3 - \frac{5}{4}u^2 - \frac{2}{5}u^2 + \frac{5}{3}u + \frac{9}{20}u - \frac{15}{8}

  2. Step 2

    −54−25=−(5+2)4+5=−79-\frac{5}{4} - \frac{2}{5} = -\frac{(5+2)}{4+5} = -\frac{7}{9}

    Correct

    −54−25=−2520−820=−3320-\frac{5}{4} - \frac{2}{5} = -\frac{25}{20} - \frac{8}{20} = -\frac{33}{20}

    Fractions cannot be combined by adding their numerators and denominators together. You need a common denominator of 2020, which yields −2520−820=−3320-\frac{25}{20} - \frac{8}{20} = -\frac{33}{20}.

  3. Step 3

    =310u3−79u2+12760u−158= \frac{3}{10}u^3 - \frac{7}{9}u^2 + \frac{127}{60}u - \frac{15}{8}

    Correct

    =310u3−3320u2+12760u−158= \frac{3}{10}u^3 - \frac{33}{20}u^2 + \frac{127}{60}u - \frac{15}{8}

    You combined the linear terms 53u+920u=12760u\frac{5}{3}u + \frac{9}{20}u = \frac{127}{60}u correctly, but the quadratic coefficient carried over the calculation error from the previous step.

  4. Step 4

    Subtract 110u3:  ⟹  =15u3−79u2+12760u−158\text{Subtract } \frac{1}{10}u^3: \implies = \frac{1}{5}u^3 - \frac{7}{9}u^2 + \frac{127}{60}u - \frac{15}{8}

    Correct

    15u3−3320u2+12760u−158\frac{1}{5}u^3 - \frac{33}{20}u^2 + \frac{127}{60}u - \frac{15}{8}

    You correctly calculated 310u3−110u3=210u3=15u3\frac{3}{10}u^3 - \frac{1}{10}u^3 = \frac{2}{10}u^3 = \frac{1}{5}u^3, but the overall expression retains the incorrect quadratic coefficient.

WHAT YOU DID WELL
Compliments from Your Teacher

You expanded all six terms of the polynomial product completely and accurately.

You found the correct common denominator of 6060 to combine the linear terms into 12760u\frac{127}{60}u.

You correctly subtracted 110u3\frac{1}{10}u^3 from the leading cubic term and simplified 210\frac{2}{10} to 15\frac{1}{5}.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Adding Across Denominators in Fraction Subtraction

You accidentally added the numerators and denominators directly when combining −54−25-\frac{5}{4} - \frac{2}{5} instead of converting both fractions to the common denominator of 2020.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Lapse in fraction addition/subtraction procedure: treating fraction addition like multiplication by operating across numerators and denominators (−a+cb+d-\frac{a+c}{b+d}) instead of finding a common denominator lcm(b,d)\text{lcm}(b, d).

REASONING PATH
Your ACEpath teacher’s suggested approach
Common Denominator Rule

  • To add or subtract ab±cd\frac{a}{b} \pm \frac{c}{d}, never compute a±cb±d\frac{a \pm c}{b \pm d}. Convert both fractions to a shared denominator: ad±bcbd\frac{ad \pm bc}{bd}.

  • Identify lcm(4,5)=20\text{lcm}(4, 5) = 20, rewrite each fraction as an equivalent fraction with denominator 2020, and then add the numerators.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Convert −54-\frac{5}{4} and −25-\frac{2}{5} to equivalent fractions with denominator 2020: −5⋅54⋅5−2⋅45⋅4=−2520−820-\frac{5 \cdot 5}{4 \cdot 5} - \frac{2 \cdot 4}{5 \cdot 4} = -\frac{25}{20} - \frac{8}{20}

  2. Combine the numerators: −25+820=−3320-\frac{25 + 8}{20} = -\frac{33}{20}

  3. Replace the coefficient of u2u^2 in the final polynomial.

Final answer

15u3−3320u2+12760u−158\frac{1}{5}u^3 - \frac{33}{20}u^2 + \frac{127}{60}u - \frac{15}{8}

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice combining negative fractions with unlike denominators using common multiples.

  • Always pause before combining fractional coefficients to double-check that the denominators match before operating on numerators.

Lock it in

Question

What is the sign of the product when multiplying two negative terms, such as (−ax)⋅(−b)(-ax) \cdot (-b)?

Answer

Positive: (−ax)⋅(−b)=+abx(-ax) \cdot (-b) = +abx. A negative times a negative always yields a positive product.

Similar question

01

Expand and simplify the polynomial product: A(x)=(−34x+2)(−23x−56)A(x) = \left(-\frac{3}{4}x + 2\right)\left(-\frac{2}{3}x - \frac{5}{6}\right).

Answer

12x2−1724x−53\frac{1}{2}x^2 - \frac{17}{24}x - \frac{5}{3}

Key steps:

  1. Quadratic term: (−34x)(−23x)=+612x2=12x2\left(-\frac{3}{4}x\right)\left(-\frac{2}{3}x\right) = +\frac{6}{12}x^2 = \frac{1}{2}x^2
  2. Linear terms: (−34x)(−56)+(2)(−23x)=58x−43x=15−3224x=−1724x\left(-\frac{3}{4}x\right)\left(-\frac{5}{6}\right) + (2)\left(-\frac{2}{3}x\right) = \frac{5}{8}x - \frac{4}{3}x = \frac{15 - 32}{24}x = -\frac{17}{24}x
  3. Constant term: (2)(−56)=−53(2)\left(-\frac{5}{6}\right) = -\frac{5}{3}
02

Given f(y)=13y2−12yf(y) = \frac{1}{3}y^2 - \frac{1}{2}y and g(y)=−45y−23g(y) = -\frac{4}{5}y - \frac{2}{3}, compute f(y)⋅g(y)+415y3f(y) \cdot g(y) + \frac{4}{15}y^3 and write the result in standard form.

Answer

845y2+13y\frac{8}{45}y^2 + \frac{1}{3}y

Key steps:

  1. Expand f(y)⋅g(y)f(y) \cdot g(y):
    • Cubic: (13y2)(−45y)=−415y3\left(\frac{1}{3}y^2\right)\left(-\frac{4}{5}y\right) = -\frac{4}{15}y^3
    • Quadratic: (13y2)(−23)+(−12y)(−45y)=−29y2+25y2=−10+1845y2=845y2\left(\frac{1}{3}y^2\right)\left(-\frac{2}{3}\right) + \left(-\frac{1}{2}y\right)\left(-\frac{4}{5}y\right) = -\frac{2}{9}y^2 + \frac{2}{5}y^2 = \frac{-10 + 18}{45}y^2 = \frac{8}{45}y^2
    • Linear: (−12y)(−23)=+13y\left(-\frac{1}{2}y\right)\left(-\frac{2}{3}\right) = +\frac{1}{3}y
  2. Add 415y3\frac{4}{15}y^3: −415y3+415y3=0-\frac{4}{15}y^3 + \frac{4}{15}y^3 = 0
  3. Result: 845y2+13y\frac{8}{45}y^2 + \frac{1}{3}y
03

Let P(t)=23t2−34t+16P(t) = \frac{2}{3}t^2 - \frac{3}{4}t + \frac{1}{6} and Q(t)=−12t+45Q(t) = -\frac{1}{2}t + \frac{4}{5}. Compute P(t)⋅Q(t)−(−13t3)P(t) \cdot Q(t) - \left(-\frac{1}{3}t^3\right) and express all coefficients as reduced fractions in standard order.

Answer

109120t2−4160t+215\frac{109}{120}t^2 - \frac{41}{60}t + \frac{2}{15}

Key steps:

  1. Distribute terms:
    • t3t^3: (23t2)(−12t)=−13t3\left(\frac{2}{3}t^2\right)\left(-\frac{1}{2}t\right) = -\frac{1}{3}t^3
    • t2t^2: (23t2)(45)+(−34t)(−12t)=815t2+38t2=64+45120t2=109120t2\left(\frac{2}{3}t^2\right)\left(\frac{4}{5}\right) + \left(-\frac{3}{4}t\right)\left(-\frac{1}{2}t\right) = \frac{8}{15}t^2 + \frac{3}{8}t^2 = \frac{64 + 45}{120}t^2 = \frac{109}{120}t^2
    • tt: (−34t)(45)+(16)(−12t)=−35t−112t=−36−560t=−4160t\left(-\frac{3}{4}t\right)\left(\frac{4}{5}\right) + \left(\frac{1}{6}\right)\left(-\frac{1}{2}t\right) = -\frac{3}{5}t - \frac{1}{12}t = \frac{-36 - 5}{60}t = -\frac{41}{60}t
    • Constant: (16)(45)=+215\left(\frac{1}{6}\right)\left(\frac{4}{5}\right) = +\frac{2}{15}
  2. Subtract −13t3-\frac{1}{3}t^3: −13t3−(−13t3)=0-\frac{1}{3}t^3 - \left(-\frac{1}{3}t^3\right) = 0
  3. Combine to get 109120t2−4160t+215\frac{109}{120}t^2 - \frac{41}{60}t + \frac{2}{15}
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