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Expand and simplify the algebraic expression using special product patterns: (3x−5y)2−(3x+5y)(3x−5y)(3x - 5y)^2 - (3x + 5y)(3x - 5y).

Answer and steps

Final answer

50y^2 - 30xy

Steps

01

Expand the first part using the square of a binomial

To expand and simplify the expression (3x−5y)2−(3x+5y)(3x−5y)(3x - 5y)^2 - (3x + 5y)(3x - 5y), we can apply two fundamental special product formulas where a=3xa = 3x and b=5yb = 5y.

Recall the square of a difference identity: (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2

Substitute a=3xa = 3x and b=5yb = 5y: (3x−5y)2=(3x)2−2(3x)(5y)+(5y)2(3x - 5y)^2 = (3x)^2 - 2(3x)(5y) + (5y)^2 (3x−5y)2=9x2−30xy+25y2(3x - 5y)^2 = 9x^2 - 30xy + 25y^2

02

Expand the second part using the difference of squares

Recall the difference of squares identity: (a+b)(a−b)=a2−b2(a + b)(a - b) = a^2 - b^2

Substitute a=3xa = 3x and b=5yb = 5y: (3x+5y)(3x−5y)=(3x)2−(5y)2=9x2−25y2(3x + 5y)(3x - 5y) = (3x)^2 - (5y)^2 = 9x^2 - 25y^2

03

Subtract the second expression from the first

Substitute the expansions back into the original expression: (3x−5y)2−(3x+5y)(3x−5y)=(9x2−30xy+25y2)−(9x2−25y2)(3x - 5y)^2 - (3x + 5y)(3x - 5y) = (9x^2 - 30xy + 25y^2) - (9x^2 - 25y^2)

Common Pitfall: Keep parentheses around the second expression. The minus sign in front must be distributed to both terms inside: −(9x2−25y2)=−9x2+25y2-(9x^2 - 25y^2) = -9x^2 + 25y^2

Distribute the negative sign: =9x2−30xy+25y2−9x2+25y2= 9x^2 - 30xy + 25y^2 - 9x^2 + 25y^2

04

Combine like terms

Group like terms together:

  • x2x^2-terms: 9x2−9x2=09x^2 - 9x^2 = 0
  • xyxy-terms: −30xy-30xy
  • y2y^2-terms: 25y2+25y2=50y225y^2 + 25y^2 = 50y^2

=50y2−30xy= 50y^2 - 30xy

Alternative Method (Factoring Common Terms)

Notice that (3x−5y)(3x - 5y) is a common factor in both parts: (3x−5y)2−(3x+5y)(3x−5y)=(3x−5y)[(3x−5y)−(3x+5y)](3x - 5y)^2 - (3x + 5y)(3x - 5y) = (3x - 5y)\big[(3x - 5y) - (3x + 5y)\big] =(3x−5y)(3x−5y−3x−5y)= (3x - 5y)(3x - 5y - 3x - 5y) =(3x−5y)(−10y)= (3x - 5y)(-10y) =−30xy+50y2=50y2−30xy= -30xy + 50y^2 = 50y^2 - 30xy

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to distribute the negative sign to the second term of the expanded difference of squares, subtracting 25y² instead of adding it.

Sample Student Work

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

    (3x - 5y)^2 - (3x + 5y)(3x - 5y) = (9x^2 - 30xy + 25y^2) - (9x^2 - 25y^2)

  2. Step 2

    = 9x^2 - 30xy + 25y^2 - 9x^2 - 25y^2

    Correct

    = 9x^2 - 30xy + 25y^2 - 9x^2 + 25y^2

    When subtracting a binomial, the minus sign applies to both terms inside: −(9x2−25y2)=−9x2−(−25y2)=−9x2+25y2-(9x^2 - 25y^2) = -9x^2 - (-25y^2) = -9x^2 + 25y^2.

  3. Step 3

    = -30xy

    Correct

    = 50y^2 - 30xy

    You correctly combined the like terms from your previous line, but because +25y2+25y^2 was written as −25y2-25y^2, the y2y^2 terms cancelled out incorrectly.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly used the identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2 to expand (3x−5y)2(3x - 5y)^2.

You correctly used the difference of squares identity (a+b)(a−b)=a2−b2(a + b)(a - b) = a^2 - b^2 to expand (3x+5y)(3x−5y)(3x + 5y)(3x - 5y).

You remembered to place parentheses around the second product before subtracting.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Sign distribution error across parentheses

You distributed the negative sign to the first term 9x29x^2 but forgot to flip the sign of the second term −25y2-25y^2.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Treating the leading minus sign as simply replacing the first term's sign rather than distributing −1-1 to every term inside the parentheses.

REASONING PATH
Your ACEpath teacher’s suggested approach
Distribute negative signs completely

  • −(A−B)=−A+B-(A - B) = -A + B

  • Always multiply each term inside the bracket by −1-1, changing subtraction to addition for negative terms.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Distribute the negative sign across both terms of the second group: −(9x2−25y2)=−9x2+25y2-(9x^2 - 25y^2) = -9x^2 + 25y^2.

  2. Rewrite the full expression: 9x2−30xy+25y2−9x2+25y29x^2 - 30xy + 25y^2 - 9x^2 + 25y^2.

  3. Combine like terms: (9x2−9x2)−30xy+(25y2+25y2)=50y2−30xy(9x^2 - 9x^2) - 30xy + (25y^2 + 25y^2) = 50y^2 - 30xy.

Final answer

50y^2 - 30xy

NEXT STEPS
Final Takeaways from Your Teacher

  • Whenever you remove parentheses preceded by a minus sign, circle the minus sign and draw arrows to every term inside.

  • Practice distributing −1-1 across binomials containing negative terms, such as −(u2−v2)=−u2+v2-(u^2 - v^2) = -u^2 + v^2.

OverviewACEpath Teacher's Diagnosis

02 The student misapplies the square of a binomial as the sum of squares, omitting the middle cross term -30xy.

Sample Student Work

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

    (3x - 5y)^2 - (3x + 5y)(3x - 5y) = (9x^2 + 25y^2) - (9x^2 - 25y^2)

    Correct

    (9x^2 - 30xy + 25y^2) - (9x^2 - 25y^2)

    You expanded the difference of squares (3x+5y)(3x−5y)=9x2−25y2(3x + 5y)(3x - 5y) = 9x^2 - 25y^2 correctly, but you expanded (3x−5y)2(3x - 5y)^2 as 9x2+25y29x^2 + 25y^2 instead of a2−2ab+b2=9x2−30xy+25y2a^2 - 2ab + b^2 = 9x^2 - 30xy + 25y^2, omitting the middle term −2(3x)(5y)=−30xy-2(3x)(5y) = -30xy.

  2. Step 2

    = 9x^2 + 25y^2 - 9x^2 + 25y^2

    Correct

    = 9x^2 - 30xy + 25y^2 - 9x^2 + 25y^2

    You correctly distributed the subtraction across the second set of parentheses, but this step carries forward the missing −30xy-30xy term from Step 1.

  3. Step 3

    = 50y^2

    Correct

    = 50y^2 - 30xy

    Your arithmetic in combining like terms was accurate based on your previous lines, but the final expression is missing −30xy-30xy.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly recognized and applied the difference of squares pattern (a+b)(a−b)=a2−b2(a + b)(a - b) = a^2 - b^2.

You properly kept parentheses around the subtracted expression.

You accurately distributed the negative sign to both terms inside the second set of parentheses.

You correctly combined like terms in the final algebraic step.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Freshman's Dream / Missing Middle Term

You squared the individual terms 3x3x and 5y5y directly without including the middle cross term −2ab-2ab.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing the square of a binomial (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2 with distributing an exponent over multiplication, leading to the omission of the middle term.

REASONING PATH
Your ACEpath teacher’s suggested approach
Square of a Binomial Formula

  • (a±b)2=a2±2ab+b2(a \pm b)^2 = a^2 \pm 2ab + b^2

  • Always write out the three terms: square the first term, add or subtract twice the product of both terms, and square the last term.

Two Terms vs. Three Terms Check

  • Multiplying conjugates (a+b)(a−b)(a + b)(a - b) produces 2 terms (a2−b2a^2 - b^2), while squaring a binomial (a±b)2(a \pm b)^2 produces 3 terms.

  • Count the expanded terms before combining: (a±b)2(a \pm b)^2 must yield a trinomial.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Expand both parts using their respective formulas: (3x−5y)2=(3x)2−2(3x)(5y)+(5y)2=9x2−30xy+25y2(3x - 5y)^2 = (3x)^2 - 2(3x)(5y) + (5y)^2 = 9x^2 - 30xy + 25y^2, and (3x+5y)(3x−5y)=(3x)2−(5y)2=9x2−25y2(3x + 5y)(3x - 5y) = (3x)^2 - (5y)^2 = 9x^2 - 25y^2.

  2. Substitute the expansions back into the difference: (9x2−30xy+25y2)−(9x2−25y2)(9x^2 - 30xy + 25y^2) - (9x^2 - 25y^2).

  3. Distribute the negative sign: 9x2−30xy+25y2−9x2+25y29x^2 - 30xy + 25y^2 - 9x^2 + 25y^2.

  4. Combine like terms: (9x2−9x2)−30xy+(25y2+25y2)=50y2−30xy(9x^2 - 9x^2) - 30xy + (25y^2 + 25y^2) = 50y^2 - 30xy.

Final answer

50y^2 - 30xy

NEXT STEPS
Final Takeaways from Your Teacher

  • Review the expansion of (a−b)2(a - b)^2 using the box method or FOIL to see why the middle term −2ab-2ab always appears.

  • Practice identifying whether a product produces a binomial (difference of squares) or a trinomial (perfect square trinomial).

  • Re-solve this expression from the start, making sure (3x−5y)2(3x - 5y)^2 expands into three terms.

OverviewACEpath Teacher's Diagnosis

03 The student squares the variable but fails to square the numerical coefficient 3, writing 3x² instead of 9x².

Sample Student Work

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

    (3x^2 - 30xy + 25y^2) - (3x^2 - 25y^2)

    Correct

    (9x^2 - 30xy + 25y^2) - (9x^2 - 25y^2)

    When expanding (3x−5y)2(3x - 5y)^2 and (3x+5y)(3x−5y)(3x + 5y)(3x - 5y), the first term squared is (3x)2=32x2=9x2(3x)^2 = 3^2 x^2 = 9x^2. You squared the variable xx but did not square the coefficient 33.

  2. Step 2

    3x^2 - 30xy + 25y^2 - 3x^2 + 25y^2

    Correct

    9x^2 - 30xy + 25y^2 - 9x^2 + 25y^2

    You correctly distributed the minus sign across the second group of parentheses, though it carried forward the un-squared 3x23x^2 terms.

  3. Step 3

    -30xy + 50y^2

    Correct

    50y^2 - 30xy

    You combined like terms correctly. Although your final expression matches the correct simplified answer because the x2x^2 terms canceled each other out (3x2−3x2=03x^2 - 3x^2 = 0), the intermediate expansion was algebraically invalid.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly identified and applied the square of a difference identity to get −30xy-30xy and +25y2+25y^2.

You correctly identified the difference of squares pattern for the second factor pair.

You accurately distributed the subtraction sign over the second polynomial.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Coefficient Not Squared in Power of a Product

You wrote 3x23x^2 instead of (3x)2=9x2(3x)^2 = 9x^2 in both binomial expansions.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Treating (ax)n(ax)^n as axna x^n by omitting the exponent on the numerical coefficient, rather than applying the power of a product rule (ax)n=anxn(ax)^n = a^n x^n.

REASONING PATH
Your ACEpath teacher’s suggested approach
Power of a Product Rule

  • For any product raised to a power, distribute the exponent to every factor inside: (ab)n=anbn(ab)^n = a^n b^n.

  • Always write parentheses around composite terms before applying powers, e.g., (3x)2=(3)2(x)2=9x2(3x)^2 = (3)^2 (x)^2 = 9x^2.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Apply the power of a product rule to the first term: (3x)2=32x2=9x2(3x)^2 = 3^2 x^2 = 9x^2.

  2. Write the complete expansions: (9x2−30xy+25y2)−(9x2−25y2)(9x^2 - 30xy + 25y^2) - (9x^2 - 25y^2).

  3. Distribute the negative sign: 9x2−30xy+25y2−9x2+25y29x^2 - 30xy + 25y^2 - 9x^2 + 25y^2.

  4. Combine like terms: (9x2−9x2)−30xy+(25y2+25y2)=50y2−30xy(9x^2 - 9x^2) - 30xy + (25y^2 + 25y^2) = 50y^2 - 30xy.

Final answer

50y^2 - 30xy

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice squaring monomial terms containing both coefficients and variables using parentheses.

  • Double-check that every factor inside parentheses receives the exponent before combining like terms.

Lock it in

Question

What does −(A−B)-(A - B) simplify to when parentheses are removed?

Answer

−A+B-A + B. Distributing the minus sign flips the sign of every term inside: −1×(−B)=+B-1 \times (-B) = +B.

Similar question

01

Expand and simplify: (2x−3y)2−(2x+3y)(2x−3y)(2x - 3y)^2 - (2x + 3y)(2x - 3y)

Answer

18y2−12xy18y^2 - 12xy Expanding gives (4x2−12xy+9y2)−(4x2−9y2)(4x^2 - 12xy + 9y^2) - (4x^2 - 9y^2). Distributing the negative gives 4x2−12xy+9y2−4x2+9y2=18y2−12xy4x^2 - 12xy + 9y^2 - 4x^2 + 9y^2 = 18y^2 - 12xy.

02

Expand and simplify: 2(x−4)2−(x+4)(x−4)2(x - 4)^2 - (x + 4)(x - 4)

Answer

x2−16x+48x^2 - 16x + 48 2(x2−8x+16)−(x2−16)=(2x2−16x+32)−x2+16=x2−16x+482(x^2 - 8x + 16) - (x^2 - 16) = (2x^2 - 16x + 32) - x^2 + 16 = x^2 - 16x + 48.

03

Expand and simplify: (a−2b)2−(a+2b)2(a - 2b)^2 - (a + 2b)^2

Answer

−8ab-8ab (a2−4ab+4b2)−(a2+4ab+4b2)=a2−4ab+4b2−a2−4ab−4b2=−8ab(a^2 - 4ab + 4b^2) - (a^2 + 4ab + 4b^2) = a^2 - 4ab + 4b^2 - a^2 - 4ab - 4b^2 = -8ab.

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