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Simplifying Expressions and Combining Like Terms

34 questions

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Question 21Question →

Match each algebraic expression on the left with its fully simplified equivalent expression on the right.

Click a left item, then click its matching right item

Items

−−(a2−2ab)+a(a−3b)--(a^2 - 2ab) + a(a - 3b)
2(a2b−ab)−ab(2a−3)2(a^2b - ab) - ab(2a - 3)
a2(a−b)−a(a2−ab)a^2(a - b) - a(a^2 - ab)

Matches

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Answer

The expression −−(a2−2ab)+a(a−3b)--(a^2 - 2ab) + a(a - 3b) simplifies to −ab-ab; the expression 2(a2b−ab)−ab(2a−3)2(a^2b - ab) - ab(2a - 3) simplifies to abab; and the expression a2(a−b)−a(a2−ab)a^2(a - b) - a(a^2 - ab) simplifies to 00.
Each expression is correctly simplified by distributing the external factors across parenthetical terms and combining the resulting like terms.

Step-by-Step Solution

1
Simplify the first expression by distributing coefficients and combining like terms.
−−a2+2ab+a2−3b=−ab--a^2 + 2ab + a^2 - 3b = -ab
Distributing the negative sign yields −−a2+2ab--a^2 + 2ab and distributing aa yields a2−3aba^2 - 3ab. Adding them eliminates the a2a^2 terms, leaving −ab-ab.
2
Simplify the second expression by distributing coefficients and combining like terms.
2a2b−2ab−2a2b+3ab=ab2a^2b - 2ab - 2a^2b + 3ab = ab
Distributing 22 yields 2a2b−2ab2a^2b - 2ab and distributing −ab-ab yields −2a2b+3ab-2a^2b + 3ab. Adding them eliminates the 2a2b2a^2b terms, leaving abab.
3
Simplify the third expression by distributing coefficients and combining like terms.
a3−a2b−a3+a2b=0a^3 - a^2b - a^3 + a^2b = 0
Distributing a2a^2 yields a3−a2ba^3 - a^2b and distributing −a-a yields −a3+a2b-a^3 + a^2b. Adding them eliminates all terms, leaving 00.

Key Concept

Simplifying algebraic expressions containing multiple variables and higher powers by applying the distributive property and combining like terms.
Question 22Question →

Match each algebraic expression on the left with its fully simplified equivalent expression on the right for all real values of mm and nn.

Click a left item, then click its matching right item

Items

m(m2−3mn)−2n(m2−n2)−(m3−mn2)m(m^2 - 3mn) - 2n(m^2 - n^2) - (m^3 - mn^2)
(m−n)3−m(m2−3n2)+n3(m - n)^3 - m(m^2 - 3n^2) + n^3
3mn(m−n)−2m(n2−mn)−(m2n−5mn2)3mn(m - n) - 2m(n^2 - mn) - (m^2n - 5mn^2)
m2(2m−n)−n(m2−n2)−(m3+n3)m^2(2m - n) - n(m^2 - n^2) - (m^3 + n^3)

Matches

Show answer & explanation

Answer

The correct matches are: (1) m(m2−3mn)−2n(m2−n2)−(m3−mn2)m(m^2 - 3mn) - 2n(m^2 - n^2) - (m^3 - mn^2) matches with −5m2n+mn2+2n3-5m^2n + mn^2 + 2n^3; (2) (m−n)3−m(m2−3n2)+n3(m - n)^3 - m(m^2 - 3n^2) + n^3 matches with −3m2n+6mn2-3m^2n + 6mn^2; (3) 3mn(m−n)−2m(n2−mn)−(m2n−5mn2)3mn(m - n) - 2m(n^2 - mn) - (m^2n - 5mn^2) matches with 4m2n4m^2n; and (4) m2(2m−n)−n(m2−n2)−(m3+n3)m^2(2m - n) - n(m^2 - n^2) - (m^3 + n^3) matches with m3−2m2nm^3 - 2m^2n.
Each expression is simplified by systematically expanding parenthetical groups and collecting like terms with identical variable powers.

Step-by-Step Solution

1
Simplify the expression m(m2−3mn)−2n(m2−n2)−(m3−mn2)m(m^2 - 3mn) - 2n(m^2 - n^2) - (m^3 - mn^2)
−5m2n+mn2+2n3-5m^2n + mn^2 + 2n^3
Distribute each term across the parenthetical expressions: m3−3m2n−2m2n+2n3−m3+mn2m^3 - 3m^2n - 2m^2n + 2n^3 - m^3 + mn^2. Group the like terms: (m3−m3)+(−3m2n−2m2n)+mn2+2n3(m^3 - m^3) + (-3m^2n - 2m^2n) + mn^2 + 2n^3, which simplifies to −5m2n+mn2+2n3-5m^2n + mn^2 + 2n^3.
2
Simplify the expression (m−n)3−m(m2−3n2)+n3(m - n)^3 - m(m^2 - 3n^2) + n^3
−3m2n+6mn2-3m^2n + 6mn^2
Use the binomial expansion formula to expand (m−n)3=m3−3m2n+3mn2−n3(m - n)^3 = m^3 - 3m^2n + 3mn^2 - n^3. Distribute the −m-m term to obtain −m3+3mn2-m^3 + 3mn^2. Sum all the expressions and combine like terms: (m3−m3)−3m2n+(3mn2+3mn2)+(−n3+n3)=−3m2n+6mn2(m^3 - m^3) - 3m^2n + (3mn^2 + 3mn^2) + (-n^3 + n^3) = -3m^2n + 6mn^2.
3
Simplify the expression 3mn(m−n)−2m(n2−mn)−(m2n−5mn2)3mn(m - n) - 2m(n^2 - mn) - (m^2n - 5mn^2)
4m2n4m^2n
Expand the terms by distributing the outer coefficients to get 3m2n−3mn2−2mn2+2m2n−m2n+5mn23m^2n - 3mn^2 - 2mn^2 + 2m^2n - m^2n + 5mn^2. Grouping similar variables yields (3+2−1)m2n+(−3−2+5)mn2=4m2n+0=4m2n(3 + 2 - 1)m^2n + (-3 - 2 + 5)mn^2 = 4m^2n + 0 = 4m^2n.
4
Simplify the expression m2(2m−n)−n(m2−n2)−(m3+n3)m^2(2m - n) - n(m^2 - n^2) - (m^3 + n^3)
m3−2m2nm^3 - 2m^2n
Expand the terms by distributing the multiplication: 2m3−m2n−m2n+n3−m3−n32m^3 - m^2n - m^2n + n^3 - m^3 - n^3. Grouping like terms yields (2m3−m3)+(−m2n−m2n)+(n3−n3)=m3−2m2n(2m^3 - m^3) + (-m^2n - m^2n) + (n^3 - n^3) = m^3 - 2m^2n.

Key Concept

Simplifying multivariable expressions by distributing terms (including negative signs) and combining like terms.
Question 23Question →

For all real values of xx and yy, the expression 5x3y−2x2(3xy−y)−4x2y5x^3y - 2x^2(3xy - y) - 4x^2y can be simplified. Which of the following is equivalent to this simplified expression?

Show answer & explanation

Answer: −x3y−2x2y-x^3y - 2x^2y

Answer

−x3y−2x2y-x^3y - 2x^2y
Distributing −2x2-2x^2 across the parentheses (3xy−y)(3xy - y) yields −6x3y+2x2y-6x^3y + 2x^2y. Substituting this back gives 5x3y−6x3y+2x2y−4x2y5x^3y - 6x^3y + 2x^2y - 4x^2y. Combining the x3yx^3y terms (5x3y−6x3y=−x3y5x^3y - 6x^3y = -x^3y) and the x2yx^2y terms (2x2y−4x2y=−2x2y2x^2y - 4x^2y = -2x^2y) yields the simplified expression −x3y−2x2y-x^3y - 2x^2y.

Step-by-Step Solution

1
Distribute the term −2x2-2x^2 to both terms inside the parentheses (3xy−y)(3xy - y).
−6x3y+2x2y-6x^3y + 2x^2y
Multiply −2x2-2x^2 by 3xy3xy (yielding −6x3y-6x^3y) and multiply −2x2-2x^2 by −y-y (yielding +2x2y+2x^2y).
2
Substitute the expanded terms back into the original expression.
5x3y−6x3y+2x2y−4x2y5x^3y - 6x^3y + 2x^2y - 4x^2y
Replace −2x2(3xy−y)-2x^2(3xy - y) with the expanded terms −6x3y+2x2y-6x^3y + 2x^2y.
3
Combine like terms by grouping the x3yx^3y terms and the x2yx^2y terms.
−x3y−2x2y-x^3y - 2x^2y
Combine 5x3y−6x3y=−x3y5x^3y - 6x^3y = -x^3y and 2x2y−4x2y=−2x2y2x^2y - 4x^2y = -2x^2y.

Key Concept

Simplifying expressions by distributing terms and combining like terms.
Question 24Question →

Which of the following is equivalent to the expression 2a(3a−b)2−3b(2a2−ab)−a2(18a−15b)2a(3a - b)^2 - 3b(2a^2 - ab) - a^2(18a - 15b) for all real values of aa and bb?

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Answer: −3a2b+5ab2-3a^2b + 5ab^2

Answer

The simplified expression is −3a2b+5ab2-3a^2b + 5ab^2.
The correct expression is −3a2b+5ab2-3a^2b + 5ab^2. Expanding the three parts yields 18a3−12a2b+2ab218a^3 - 12a^2b + 2ab^2, −6a2b+3ab2-6a^2b + 3ab^2, and −18a3+15a2b-18a^3 + 15a^2b. Summing these parts together cancels out the cubic a3a^3 terms (18a3−18a3=018a^3 - 18a^3 = 0), combines the a2ba^2b terms (−12a2b−6a2b+15a2b=−3a2b-12a^2b - 6a^2b + 15a^2b = -3a^2b), and combines the ab2ab^2 terms (2ab2+3ab2=5ab22ab^2 + 3ab^2 = 5ab^2).

Step-by-Step Solution

1
Expand the first term 2a(3a−b)22a(3a - b)^2
18a3−12a2b+2ab218a^3 - 12a^2b + 2ab^2
First expand the squared binomial (3a−b)2=9a2−6ab+b2(3a - b)^2 = 9a^2 - 6ab + b^2, then distribute the 2a2a to each term.
2
Expand the second term −3b(2a2−ab)-3b(2a^2 - ab)
−6a2b+3ab2-6a^2b + 3ab^2
Distribute −3b-3b to both terms inside the parentheses, ensuring that the negative sign is applied to both terms.
3
Expand the third term −a2(18a−15b)-a^2(18a - 15b)
−18a3+15a2b-18a^3 + 15a^2b
Distribute −a2-a^2 to both terms inside the parentheses, changing the sign of both terms.
4
Combine the expanded expressions and group like terms
(18a3−18a3)+(−12a2b−6a2b+15a2b)+(2ab2+3ab2)(18a^3 - 18a^3) + (-12a^2b - 6a^2b + 15a^2b) + (2ab^2 + 3ab^2)
Grouping terms with identical variable parts allows them to be added or subtracted.
5
Perform the final simplification
−3a2b+5ab2-3a^2b + 5ab^2
The a3a^3 terms cancel out, the a2ba^2b terms combine to −3a2b-3a^2b, and the ab2ab^2 terms combine to 5ab25ab^2.

Key Concept

Simplifying algebraic expressions by expanding parentheses, applying exponent rules, and combining like terms.
Question 25Question →

For all real numbers mm and nn, the expression 4m(2m−3n)2−n(5m2−36mn)4m(2m - 3n)^2 - n(5m^2 - 36mn) can be written in the form am3+bm2n+cmn2am^3 + bm^2n + cmn^2, where aa, bb, and cc are real constants. What is the value of the coefficient bb?

Show answer & explanation

Answer: -53

Answer

The value of the coefficient bb is −53-53.
Expanding the entire expression yields 16m3−53m2n+72mn216m^3 - 53m^2n + 72mn^2. Comparing this to the template form am3+bm2n+cmn2am^3 + bm^2n + cmn^2 shows that b=−53b = -53.

Step-by-Step Solution

1
Expand the squared binomial (2m−3n)2(2m - 3n)^2
4m2−12mn+9n24m^2 - 12mn + 9n^2
Apply the identity (x−y)2=x2−2xy+y2(x - y)^2 = x^2 - 2xy + y^2 to expand the expression inside the parentheses.
2
Distribute 4m4m through the trinomial
16m3−48m2n+36mn216m^3 - 48m^2n + 36mn^2
Multiply each term of 4m2−12mn+9n24m^2 - 12mn + 9n^2 by 4m4m using properties of exponents.
3
Distribute −n-n across (5m2−36mn)(5m^2 - 36mn)
−5m2n+36mn2-5m^2n + 36mn^2
Multiply −n-n by both terms inside the parentheses, paying attention to sign rules.
4
Group and combine the like terms
16m3+(−48m2n−5m2n)+(36mn2+36mn2)16m^3 + (-48m^2n - 5m^2n) + (36mn^2 + 36mn^2)
Identify terms with the same variables and exponents to combine them.
5
Combine the coefficients of the like terms
16m3−53m2n+72mn216m^3 - 53m^2n + 72mn^2
Perform arithmetic on the coefficients: −48−5=−53-48 - 5 = -53 for the m2nm^2n terms and 36+36=7236 + 36 = 72 for the mn2mn^2 terms.
6
Identify the coefficient bb
−53-53
Compare the simplified polynomial to the form am3+bm2n+cmn2am^3 + bm^2n + cmn^2 to find the value of bb.

Key Concept

Simplifying algebraic expressions by expanding binomials, distributing variables and signs, and combining like terms.

Alternative Method

Instead of expanding the entire expression, focus only on terms that produce m2nm^2n: from the first part, 4m×(−12mn)=−48m2n4m \times (-12mn) = -48m^2n, and from the second part, −n×5m2=−5m2n-n \times 5m^2 = -5m^2n. Adding these gives −53m2n-53m^2n.
Estimated Time:1m 30s
Question 26Question →

If xx and yy are real numbers, the expression 3x(x−2y)2−x2(3x−10y)−y2(2x−y)3x(x - 2y)^2 - x^2(3x - 10y) - y^2(2x - y) can be simplified to which of the following?

Show answer & explanation

Answer: −2x2y+10xy2+y3-2x^2y + 10xy^2 + y^3

Answer

The correct expression is −2x2y+10xy2+y3-2x^2y + 10xy^2 + y^3.
The correct expression −2x2y+10xy2+y3-2x^2y + 10xy^2 + y^3 is obtained by correctly expanding the binomial (x−2y)2(x-2y)^2 into x2−4xy+4y2x^2 - 4xy + 4y^2, distributing the outer terms 3x3x, −x2-x^2, and −y2-y^2 to all terms inside their respective parentheses, and then combining the coefficients of the matching variable terms: the x3x^3 terms cancel out, the x2yx^2y terms combine to −2x2y-2x^2y, the xy2xy^2 terms combine to 10xy210xy^2, and the y3y^3 term remains.

Step-by-Step Solution

1
Expand the binomial expression (x−2y)2(x - 2y)^2.
x2−4xy+4y2x^2 - 4xy + 4y^2
Before distributing 3x3x, the squared binomial must be expanded using the identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2.
2
Multiply the expanded binomial by the coefficient 3x3x.
3x3−12x2y+12xy23x^3 - 12x^2y + 12xy^2
Distribute 3x3x to each of the three terms in the expanded expression: 3x(x2)=3x33x(x^2) = 3x^3, 3x(−4xy)=−12x2y3x(-4xy) = -12x^2y, and 3x(4y2)=12xy23x(4y^2) = 12xy^2.
3
Distribute −x2-x^2 to (3x−10y)(3x - 10y).
−3x3+10x2y-3x^3 + 10x^2y
Distribute −x2-x^2 to both terms, noting that multiplying two negative signs yields a positive term: −x2(3x)=−3x3-x^2(3x) = -3x^3 and −x2(−10y)=+10x2y-x^2(-10y) = +10x^2y.
4
Distribute −y2-y^2 to (2x−y)(2x - y).
−2xy2+y3-2xy^2 + y^3
Distribute −y2-y^2 to both terms, adding the exponents of yy where appropriate: −y2(2x)=−2xy2-y^2(2x) = -2xy^2 and −y2(−y1)=+y3-y^2(-y^1) = +y^3.
5
Combine all parts and group the like terms together.
−2x2y+10xy2+y3-2x^2y + 10xy^2 + y^3
Group and add the coefficients of identical variable terms: (3x3−3x3)+(−12x2y+10x2y)+(12xy2−2xy2)+y3=0x3−2x2y+10xy2+y3(3x^3 - 3x^3) + (-12x^2y + 10x^2y) + (12xy^2 - 2xy^2) + y^3 = 0x^3 - 2x^2y + 10xy^2 + y^3.

Key Concept

Simplifying polynomial expressions with multiple variables by expanding binomials, distributing coefficients (including negative signs), and combining like terms.
Question 27Question →

When the expression 2x(3x−y)2−3y2(x−2y)−x2(18x−15y)2x(3x - y)^2 - 3y^2(x - 2y) - x^2(18x - 15y) is simplified by combining like terms, what is the coefficient of x2yx^2y?

Show answer & explanation

Answer: 3

Answer

The coefficient of x2yx^2y is 33.
The correct coefficient of 33 is obtained by expanding the expression step-by-step. First, the square of the binomial (3x−y)2(3x - y)^2 is 9x2−6xy+y29x^2 - 6xy + y^2. Distributing 2x2x gives 18x3−12x2y+2xy218x^3 - 12x^2y + 2xy^2. Next, distributing −x2-x^2 to (18x−15y)(18x - 15y) gives −18x3+15x2y-18x^3 + 15x^2y. Combining the x2yx^2y terms gives −12x2y+15x2y=3x2y-12x^2y + 15x^2y = 3x^2y, meaning the coefficient is 33.

Step-by-Step Solution

1
Expand the squared binomial term (3x−y)2(3x - y)^2 using the algebraic identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2.
(3x−y)2=9x2−6xy+y2(3x - y)^2 = 9x^2 - 6xy + y^2
This is necessary to remove the parentheses before distributing the outer variable.
2
Multiply the term 2x2x by each term inside the expanded binomial expression: 2x(9x2−6xy+y2)2x(9x^2 - 6xy + y^2).
18x3−12x2y+2xy218x^3 - 12x^2y + 2xy^2
Applying the distributive property expands the first part of the expression.
3
Distribute −x2-x^2 to the terms inside the parentheses (18x−15y)(18x - 15y), paying close attention to the signs.
−18x3+15x2y-18x^3 + 15x^2y
This expands the third part of the expression and correctly distributes the negative sign.
4
Identify and combine the like terms of the form x2yx^2y from the expanded parts of the expression.
−12x2y+15x2y=3x2y-12x^2y + 15x^2y = 3x^2y
To find the coefficient of x2yx^2y, we only need to sum the coefficients of the terms that contain exactly x2yx^2y.

Key Concept

Simplifying Expressions and Combining Like Terms
Estimated Time:1m 30s
Question 28Question →

Match each unsimplified algebraic expression on the left with its equivalent simplified form on the right.

Click a left item, then click its matching right item

Items

3x(x−2y)−2x(x−3y)3x(x - 2y) - 2x(x - 3y)
2(x2−xy)−3(xy−y2)2(x^2 - xy) - 3(xy - y^2)
x2(x−2y)−x(x2−xy)x^2(x - 2y) - x(x^2 - xy)
(x+y)2−(x−y)2(x + y)^2 - (x - y)^2

Matches

Show answer & explanation

Answer

The correct pairings match 3x(x−2y)−2x(x−3y)3x(x - 2y) - 2x(x - 3y) to x2x^2; 2(x2−xy)−3(xy−y2)2(x^2 - xy) - 3(xy - y^2) to 2x2−5xy+3y22x^2 - 5xy + 3y^2; x2(x−2y)−x(x2−xy)x^2(x - 2y) - x(x^2 - xy) to −x2y-x^2y; and (x+y)2−(x−y)2(x + y)^2 - (x - y)^2 to 4xy4xy.
Each unsimplified expression is correctly matched to its simplified equivalent by expanding parenthetical terms (taking care to distribute negative signs) and combining like terms.

Step-by-Step Solution

1
Distribute and combine like terms for the first expression 3x(x−2y)−2x(x−3y)3x(x - 2y) - 2x(x - 3y).
The expression simplifies to x2x^2.
First distribute the coefficients to get 3x2−6xy−2x2+6xy3x^2 - 6xy - 2x^2 + 6xy. Then combine 3x2−2x2=x23x^2 - 2x^2 = x^2 and −6xy+6xy=0-6xy + 6xy = 0.
2
Distribute and combine like terms for the second expression 2(x2−xy)−3(xy−y2)2(x^2 - xy) - 3(xy - y^2).
The expression simplifies to 2x2−5xy+3y22x^2 - 5xy + 3y^2.
Distribute the coefficients to get 2x2−2xy−3xy+3y22x^2 - 2xy - 3xy + 3y^2. Note that distributing the negative sign of −3-3 to −y2-y^2 results in +3y2+3y^2. Then combine −2xy−3xy=−5xy-2xy - 3xy = -5xy.
3
Distribute and combine like terms for the third expression x2(x−2y)−x(x2−xy)x^2(x - 2y) - x(x^2 - xy).
The expression simplifies to −x2y-x^2y.
Distribute to get x3−2x2y−x3+x2yx^3 - 2x^2y - x^3 + x^2y. Note that distributing −x-x to −xy-xy gives +x2y+x^2y. The x3−x3x^3 - x^3 terms cancel, leaving −2x2y+x2y=−x2y-2x^2y + x^2y = -x^2y.
4
Expand and simplify the fourth expression (x+y)2−(x−y)2(x + y)^2 - (x - y)^2.
The expression simplifies to 4xy4xy.
Expand both squared binomials: (x2+2xy+y2)−(x2−2xy+y2)(x^2 + 2xy + y^2) - (x^2 - 2xy + y^2). Distribute the negative sign to get x2+2xy+y2−x2+2xy−y2x^2 + 2xy + y^2 - x^2 + 2xy - y^2. Combine terms to cancel x2x^2 and y2y^2, leaving 4xy4xy.

Key Concept

Simplifying Expressions and Combining Like Terms
Estimated Time:1m 30s
Question 29Question →

For all real numbers uu and vv, which of the following is equivalent to the expression 12(2u−3v)2−23u(3u−9v)−23v2\frac{1}{2}(2u - 3v)^2 - \frac{2}{3}u(3u - 9v) - \frac{2}{3}v^2?

Show answer & explanation

Answer: 236v2\frac{23}{6}v^2

Answer

The simplified equivalent expression is 236v2\frac{23}{6}v^2.
The correct answer is obtained by expanding (2u−3v)2(2u - 3v)^2 to 4u2−12uv+9v24u^2 - 12uv + 9v^2, multiplying it by 12\frac{1}{2} to get 2u2−6uv+92v22u^2 - 6uv + \frac{9}{2}v^2, distributing −23u-\frac{2}{3}u to get −2u2+6uv-2u^2 + 6uv, and combining the terms: (2u2−2u2)+(−6uv+6uv)+(92−23)v2=236v2(2u^2 - 2u^2) + (-6uv + 6uv) + (\frac{9}{2} - \frac{2}{3})v^2 = \frac{23}{6}v^2.

Step-by-Step Solution

1
Expand the squared binomial (2u−3v)2(2u - 3v)^2.
4u2−12uv+9v24u^2 - 12uv + 9v^2
Using the binomial square formula (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2 allows us to expand the expression before applying the outer coefficient.
2
Multiply the expanded binomial by the coefficient 12\frac{1}{2}.
2u2−6uv+92v22u^2 - 6uv + \frac{9}{2}v^2
Distributing the constant factor of 12\frac{1}{2} to each term of the expanded binomial.
3
Distribute the term −23u-\frac{2}{3}u across the parenthetical expression (3u−9v)(3u - 9v).
−2u2+6uv-2u^2 + 6uv
Multiplying each term inside the parentheses by −23u-\frac{2}{3}u, making sure to distribute the negative sign properly: −23u⋅3u=−2u2-\frac{2}{3}u \cdot 3u = -2u^2 and −23u⋅(−9v)=6uv-\frac{2}{3}u \cdot (-9v) = 6uv.
4
Combine all terms and group like terms.
2u2−6uv+92v2−2u2+6uv−23v22u^2 - 6uv + \frac{9}{2}v^2 - 2u^2 + 6uv - \frac{2}{3}v^2
Write the full expression with all distributed terms to identify and combine like terms.
5
Combine the u2u^2, uvuv, and v2v^2 terms.
236v2\frac{23}{6}v^2
Combining the coefficients: 2u2−2u2=02u^2 - 2u^2 = 0, −6uv+6uv=0-6uv + 6uv = 0, and 92v2−23v2=(276−46)v2=236v2\frac{9}{2}v^2 - \frac{2}{3}v^2 = (\frac{27}{6} - \frac{4}{6})v^2 = \frac{23}{6}v^2.

Key Concept

Simplifying expressions by expanding binomials, distributing negative signs, and combining like terms with fractional coefficients.

Alternative Method

Instead of algebraic expansion, you can substitute simple non-zero values for uu and vv (e.g., u=3u = 3 and v=2v = 2) into the original expression and evaluate it. Then, substitute the same values into the answer choices to find which one yields the same result.
Estimated Time:1m 30s
Question 30Question →

When the expression 2a(a2−3ab)−3b(a2+2b2)−(a3−5ab2)2a(a^2 - 3ab) - 3b(a^2 + 2b^2) - (a^3 - 5ab^2) is simplified by combining like terms, what is the coefficient of the a2ba^2b term?

Show answer & explanation

Answer: -9

Answer

The coefficient of the a2ba^2b term is −9-9.
Expanding the entire expression yields 2a3−6a2b−3a2b−6b3−a3+5ab22a^3 - 6a^2b - 3a^2b - 6b^3 - a^3 + 5ab^2. Combining the a2ba^2b terms gives (−6−3)a2b=−9a2b(-6 - 3)a^2b = -9a^2b. Therefore, the coefficient of the a2ba^2b term is −9-9.

Step-by-Step Solution

1
Distribute 2a2a to the terms inside the first set of parentheses: 2a(a2−3ab)2a(a^2 - 3ab)
2a3−6a2b2a^3 - 6a^2b
To expand the first part of the expression.
2
Distribute −3b-3b to the terms inside the second set of parentheses: −3b(a2+2b2)-3b(a^2 + 2b^2)
−3a2b−6b3-3a^2b - 6b^3
To expand the second part of the expression, ensuring the negative sign is distributed to all terms inside.
3
Distribute the negative sign to the terms inside the third set of parentheses: −(a3−5ab2)-(a^3 - 5ab^2)
−a3+5ab2-a^3 + 5ab^2
To expand the third part of the expression, reversing the sign of each term inside.
4
Identify and combine the like terms for the a2ba^2b variable combination: −6a2b−3a2b-6a^2b - 3a^2b
−9a2b-9a^2b
To simplify the expression by combining terms with the same variable components.

Key Concept

Simplifying expressions by distributing terms and combining like terms
Estimated Time:1m 30s
Question 31Question →

For all real numbers pp and qq, which of the following is equivalent to the expression 2p2(p−3q)−3q(p2−q2)−(2p3−5p2q)2p^2(p - 3q) - 3q(p^2 - q^2) - (2p^3 - 5p^2q)?

Show answer & explanation

Answer: −4p2q+3q3-4p^2q + 3q^3

Answer

−4p2q+3q3-4p^2q + 3q^3
The correct answer is obtained by distributing all factors and negative signs, then combining the coefficients of the like terms. This yields −4p2q+3q3-4p^2q + 3q^3.

Step-by-Step Solution

1
Distribute the term 2p22p^2 into the first parenthesis, −3q-3q into the second parenthesis, and the negative sign into the third parenthesis.
2p2(p−3q)=2p3−6p2q2p^2(p - 3q) = 2p^3 - 6p^2q
−3q(p2−q2)=−3p2q+3q3-3q(p^2 - q^2) = -3p^2q + 3q^3
−(2p3−5p2q)=−2p3+5p2q-(2p^3 - 5p^2q) = -2p^3 + 5p^2q
Expanding the terms removes the parentheses and prepares the expression for combining like terms.
2
Combine the expanded parts into a single expression and group the like terms together.
(2p3−2p3)+(−6p2q−3p2q+5p2q)+3q3(2p^3 - 2p^3) + (-6p^2q - 3p^2q + 5p^2q) + 3q^3
Grouping terms with identical variable parts makes combining coefficients straightforward.
3
Simplify the coefficients for each group of like terms.
0p3+(−6−3+5)p2q+3q3=−4p2q+3q30p^3 + (-6 - 3 + 5)p^2q + 3q^3 = -4p^2q + 3q^3
Simplifying the combined coefficients yields the final, simplified expression.

Key Concept

Simplifying Expressions and Combining Like Terms
Question 32Question →

Match each of the unsimplified algebraic expressions on the left with its equivalent simplified form on the right. (Assume all variables represent real numbers.)

Click a left item, then click its matching right item

Items

3r(2r−s)−2s(r−3s)3r(2r - s) - 2s(r - 3s)
(2r−s)2−2(r2−rs)(2r - s)^2 - 2(r^2 - rs)
12(4r2−6rs)−(r−s)(2r+3s)\frac{1}{2}(4r^2 - 6rs) - (r - s)(2r + 3s)
r2(6−s)−s(r2−5s)r^2(6 - s) - s(r^2 - 5s)

Matches

Show answer & explanation

Answer

The correct matches pair the first expression with 6r2−5rs+6s26r^2 - 5rs + 6s^2, the second expression with 2r2−2rs+s22r^2 - 2rs + s^2, the third expression with 3s2−4rs3s^2 - 4rs, and the fourth expression with 6r2−2r2s+5s26r^2 - 2r^2s + 5s^2.
Each expression is correctly simplified by performing distribution first (carefully tracking negative signs and binomial expansion rules) and then combining terms that share the exact same variable powers.

Step-by-Step Solution

1
Simplify the expression 3r(2r−s)−2s(r−3s)3r(2r - s) - 2s(r - 3s).
6r2−5rs+6s26r^2 - 5rs + 6s^2
Distribute 3r3r to get 6r2−3rs6r^2 - 3rs. Then distribute −2s-2s to get −2rs+6s2-2rs + 6s^2 (noting that a negative times a negative is positive). Combine the like terms −3rs-3rs and −2rs-2rs to get −5rs-5rs.
2
Simplify the expression (2r−s)2−2(r2−rs)(2r - s)^2 - 2(r^2 - rs).
2r2−2rs+s22r^2 - 2rs + s^2
Square the binomial (2r−s)2(2r - s)^2 using the formula (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2 to get 4r2−4rs+s24r^2 - 4rs + s^2. Distribute −2-2 to get −2r2+2rs-2r^2 + 2rs. Combine 4r2−2r24r^2 - 2r^2 to get 2r22r^2, and −4rs+2rs-4rs + 2rs to get −2rs-2rs.
3
Simplify the expression 12(4r2−6rs)−(r−s)(2r+3s)\frac{1}{2}(4r^2 - 6rs) - (r - s)(2r + 3s).
3s2−4rs3s^2 - 4rs
Distribute 12\frac{1}{2} to get 2r2−3rs2r^2 - 3rs. Expand the binomial product to get 2r2+3rs−2rs−3s2=2r2+rs−3s22r^2 + 3rs - 2rs - 3s^2 = 2r^2 + rs - 3s^2. Subtract this entire expression: (2r2−3rs)−(2r2+rs−3s2)=2r2−3rs−2r2−rs+3s2(2r^2 - 3rs) - (2r^2 + rs - 3s^2) = 2r^2 - 3rs - 2r^2 - rs + 3s^2. Combine like terms to get −4rs+3s2-4rs + 3s^2.
4
Simplify the expression r2(6−s)−s(r2−5s)r^2(6 - s) - s(r^2 - 5s).
6r2−2r2s+5s26r^2 - 2r^2s + 5s^2
Distribute r2r^2 to get 6r2−r2s6r^2 - r^2s. Distribute −s-s to get −sr2+5s2-sr^2 + 5s^2. Combine the like terms −r2s-r^2s and −sr2-sr^2 (since multiplication is commutative) to get −2r2s-2r^2s.

Key Concept

Simplifying expressions by applying the distributive property, expanding products of binomials, and combining like terms.
Question 33Question →

When the expression 3x2(2x−y)−2y(x2−3xy)−(4x3−xy2)3x^2(2x - y) - 2y(x^2 - 3xy) - (4x^3 - xy^2) is simplified to the form Ax3+Bx2y+Cxy2Ax^3 + Bx^2y + Cxy^2, where AA, BB, and CC are integers, what is the value of A+B+CA + B + C?

Show answer & explanation

Answer: 4

Answer

4
The correct sum of the coefficients is 4. By distributing all terms correctly: 3x2(2x−y)=6x3−3x2y3x^2(2x - y) = 6x^3 - 3x^2y, −2y(x2−3xy)=−2x2y+6xy2-2y(x^2 - 3xy) = -2x^2y + 6xy^2, and −(4x3−xy2)=−4x3+xy2-(4x^3 - xy^2) = -4x^3 + xy^2. Combining the terms yields 2x3−5x2y+7xy22x^3 - 5x^2y + 7xy^2, which gives A=2A = 2, B=−5B = -5, and C=7C = 7. Summing these values gives 2−5+7=42 - 5 + 7 = 4.

Step-by-Step Solution

1
Distribute 3x23x^2 to the first parenthetical expression: 3x2(2x−y)3x^2(2x - y)
6x3−3x2y6x^3 - 3x^2y
Applying the distributive property multiplies 3x23x^2 by both 2x2x and −y-y.
2
Distribute −2y-2y to the second parenthetical expression: −2y(x2−3xy)-2y(x^2 - 3xy)
−2x2y+6xy2-2x^2y + 6xy^2
Applying the distributive property multiplies −2y-2y by both x2x^2 and −3xy-3xy, noting that a negative times a negative is a positive.
3
Distribute the negative sign to the third parenthetical expression: −(4x3−xy2)-(4x^3 - xy^2)
−4x3+xy2-4x^3 + xy^2
Distributing the negative sign changes the signs of both terms inside the parenthesis.
4
Combine the expanded parts: (6x3−3x2y)+(−2x2y+6xy2)+(−4x3+xy2)(6x^3 - 3x^2y) + (-2x^2y + 6xy^2) + (-4x^3 + xy^2) and group like terms
(6x3−4x3)+(−3x2y−2x2y)+(6xy2+xy2)(6x^3 - 4x^3) + (-3x^2y - 2x^2y) + (6xy^2 + xy^2)
Grouping like terms together makes it easier to combine their coefficients.
5
Combine the coefficients of the like terms
2x3−5x2y+7xy22x^3 - 5x^2y + 7xy^2
Combining the coefficients gives A=2A = 2, B=−5B = -5, and C=7C = 7.
6
Calculate the sum A+B+CA + B + C
2+(−5)+7=42 + (-5) + 7 = 4
Adding the coefficients together yields the final numerical value.

Key Concept

Simplifying algebraic expressions by distributing coefficients (including negative signs) and combining like terms.
Question 34Question →

When the expression 4x(x−2y)−(2x−3y)2+5y(2x−y)4x(x - 2y) - (2x - 3y)^2 + 5y(2x - y) is simplified to the form Ax2+Bxy+Cy2Ax^2 + Bxy + Cy^2, where AA, BB, and CC are constants, what is the value of BB?

Show answer & explanation

Answer: 14

Answer

The value of the coefficient BB is 14.
Expanding the entire expression yields 4x2−8xy−4x2+12xy−9y2+10xy−5y24x^2 - 8xy - 4x^2 + 12xy - 9y^2 + 10xy - 5y^2. Grouping and combining the xyxy terms gives (−8+12+10)xy=14xy(-8 + 12 + 10)xy = 14xy. Therefore, the coefficient BB is 14.

Step-by-Step Solution

1
Expand the first term
4x2−8xy4x^2 - 8xy
Distribute 4x4x to both terms inside the parentheses: 4x(x)−4x(2y)=4x2−8xy4x(x) - 4x(2y) = 4x^2 - 8xy.
2
Expand the squared binomial and apply the negative sign
−4x2+12xy−9y2-4x^2 + 12xy - 9y^2
Use the binomial expansion formula (2x−3y)2=4x2−12xy+9y2(2x - 3y)^2 = 4x^2 - 12xy + 9y^2, then multiply each term by −1-1.
3
Expand the third term
10xy−5y210xy - 5y^2
Distribute 5y5y to both terms inside the parentheses: 5y(2x)−5y(y)=10xy−5y25y(2x) - 5y(y) = 10xy - 5y^2.
4
Combine the coefficients of the like terms
0x2+14xy−14y20x^2 + 14xy - 14y^2
Sum the coefficients for each corresponding variable group: (4−4)x2+(−8+12+10)xy+(−9−5)y2(4 - 4)x^2 + (-8 + 12 + 10)xy + (-9 - 5)y^2.

Key Concept

Simplifying Algebraic Expressions and Combining Like Terms
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