Algebra 2 Quiz: Applying The Binomial Theorem
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Applying The Binomial TheoremQuestion 1 of 20

What is the 4th term (counting the first term as term 1) in the expansion of (x+y)6(x+y)^6 using Pascal's Triangle coefficients?

6x3y36x^3y^3
20x3y320x^3y^3
15x4y215x^4y^2
15x3y315x^3y^3
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Algebra 2 Quiz

Algebra 2 Quiz: Applying The Binomial Theorem

Practice Applying The Binomial Theorem in Algebra 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Applying The Binomial Theorem, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

What is the 4th term (counting the first term as term 1) in the expansion of (x+y)6(x+y)^6 using Pascal's Triangle coefficients?

  1. 6x3y36x^3y^3
  2. 20x3y320x^3y^3 (correct answer)
  3. 15x4y215x^4y^2
  4. 15x3y315x^3y^3
Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x+y)n(x + y)^n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! The Binomial Theorem says (x+y)n(x + y)^n expands to a sum of n+1n+1 terms where coefficients come from row nn of Pascal's Triangle, which saves enormous time compared to multiplying out by hand! For example, (x+y)4(x + y)^4 uses row 4 of Pascal's Triangle (1,4,6,4,11, 4, 6, 4, 1) to give: 1⋅x4+4⋅x3y+6⋅x2y2+4⋅xy3+1⋅y41 \cdot x^4 + 4 \cdot x^3 y + 6 \cdot x^2 y^2 + 4 \cdot x y^3 + 1 \cdot y^4. Each term has exponents summing to 4, and coefficients from the triangle make this work perfectly! For (x+y)6(x + y)^6, row 6 is 1,6,15,20,15,6,11,6,15,20,15,6,1, so the 4th term (starting from term 1 as x6x^6) is 20x3y320 x^3 y^3. Choice B correctly identifies the 4th term as 20x3y320x^3 y^3 using row 6. A mistake like 15x3y315x^3 y^3 (choice A) could be from picking the wrong position—count carefully: 1st: x6x^6, 2nd: 6x5y6x^5 y, 3rd: 15x4y215x^4 y^2, 4th: 20x3y320x^3 y^3. The expansion recipe using Pascal's Triangle: (1) Identify nn (the exponent on the binomial), (2) Write or construct row nn of Pascal's Triangle—you'll have n+1n+1 numbers, (3) Create n+1n+1 terms: first has coefficient 1 and is xnx^n, last has coefficient 1 and is yny^n, middle terms use Pascal's row with decreasing x-powers and increasing y-powers, (4) Write it out: [1[1st coefficient]⋅xn+[] \cdot x^n + [2nd coefficient]⋅xn−1⋅y+[] \cdot x^{n-1} \cdot y + [3rd coefficient]⋅xn−2⋅y2+…] \cdot x^{n-2} \cdot y^2 + \dots The pattern is systematic and reliable!

Question 2

Use Pascal's Triangle (row 3: 1,3,3,11,3,3,1) to expand (2x+1)3.(2x+1)^3.

  1. 8x3+12x2+6x+18x^3+12x^2+6x+1 (correct answer)
  2. 8x3+6x2+12x+18x^3+6x^2+12x+1
  3. 6x3+12x2+6x+16x^3+12x^2+6x+1
  4. 8x3+12x2+6x8x^3+12x^2+6x
Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x+y)n(x + y)^n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! For (2x+1)3(2x + 1)^3, we use row 3 of Pascal's Triangle (1, 3, 3, 1) but must be careful with the coefficient 2 on x. The expansion is: 1⋅(2x)3+3⋅(2x)2⋅1+3⋅(2x)⋅12+1⋅13=8x3+3⋅4x2+3⋅2x+1=8x3+12x2+6x+1.1 \cdot (2x)^3 + 3 \cdot (2x)^2 \cdot 1 + 3 \cdot (2x) \cdot 1^2 + 1 \cdot 1^3 = 8x^3 + 3 \cdot 4x^2 + 3 \cdot 2x + 1 = 8x^3 + 12x^2 + 6x + 1. Choice A correctly shows this expansion with proper handling of the 2: (2x)3=8x3(2x)^3 = 8x^3, (2x)2=4x2(2x)^2 = 4x^2, and 2x2x stays as 2x2x. Choice B incorrectly swaps the middle coefficients (6 and 12), Choice C has 6x36x^3 instead of 8x38x^3 (forgetting 23=82^3 = 8), and Choice D is missing the constant term 1. When expanding (ax+b)n(ax + b)^n, remember to raise the entire term 'ax' to each power: (ax)k=akxk(ax)^k = a^k x^k—don't forget to compute aka^k for each term!

Question 3

What is the coefficient of x4y2x^4 y^2 in the expansion of (x+y)6(x+y)^6 (use Pascal's Triangle coefficients)?

  1. 6
  2. 20
  3. 15 (correct answer)
  4. 30
Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x+y)n(x + y)^n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! Pascal's Triangle is built with a beautiful pattern: each row starts and ends with 1, and each interior number equals the sum of the two numbers above it in the previous row. Row 0 is just '1', row 1 is '1, 1', row 2 is '1, 2, 1' (2=1+12 = 1+1), row 3 is '1, 3, 3, 1' (3=1+23 = 1+2, middle 3=2+13 = 2+1), and so on. Once you build the triangle, you have instant access to binomial coefficients! For (x+y)6(x + y)^6, row 6 is 1,6,15,20,15,6,1, and the x4y2x^4 y^2 term is for y2y^2 (3rd position), coefficient 15. Choice B correctly identifies the coefficient as 15 using row 6 of Pascal's Triangle. Choosing 20 (choice C) might be from picking for y3y^3 instead—confirm the exponents: x4y2x^4 y^2 needs k=2, C(6,2)=15C(6,2)=15. Quick Pascal's Triangle construction: write 1s down both edges. For interior numbers, add the two directly above. Example for row 4: edges are 1, then interior: 1+3=4, 3+3=6, 3+1=4, giving row '1, 4, 6, 4, 1.' Check: row sums to 2n2^n (row 4 sums to 16 = 242^4). This build-as-you-go method means you never need to memorize rows—just construct them!

Question 4

In the expansion of (x2+1x)8(x^2 + \frac{1}{x})^8, what is the coefficient of the x4x^4 term?

  1. 70 (correct answer)
  2. 56
  3. 28
  4. 84
Explanation: The general term is (8k)(x2)8−k(1x)k=(8k)x2(8−k)x−k=(8k)x16−2k−k=(8k)x16−3k\binom{8}{k}(x^2)^{8-k}\left(\frac{1}{x}\right)^k = \binom{8}{k}x^{2(8-k)}x^{-k} = \binom{8}{k}x^{16-2k-k} = \binom{8}{k}x^{16-3k}. For the x4x^4 term, we need 16−3k=416-3k=4, so 3k=123k=12 and k=4k=4. The coefficient is (84)=8!4!4!=8⋅7⋅6⋅54⋅3⋅2⋅1=70\binom{8}{4} = \frac{8!}{4!4!} = \frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1} = 70. Choice B is (83)\binom{8}{3}, choice C is (82)\binom{8}{2}, and choice D is (85)\binom{8}{5}, all common errors from solving the exponent equation incorrectly.

Question 5

When (3x−2y)4(3x - 2y)^4 is expanded using the Binomial Theorem, what is the coefficient of the x2y2x^2y^2 term?

  1. 216 (correct answer)
  2. 324
  3. 162
  4. 108
Explanation: Using the Binomial Theorem, the general term is (4k)(3x)4−k(−2y)k\binom{4}{k}(3x)^{4-k}(-2y)^k. For the x2y2x^2y^2 term, we need 4−k=24-k=2 and k=2k=2. So the term is (42)(3x)2(−2y)2=6⋅9x2⋅4y2=216x2y2\binom{4}{2}(3x)^2(-2y)^2 = 6 \cdot 9x^2 \cdot 4y^2 = 216x^2y^2. Choice B incorrectly uses 343^4 instead of 323^2. Choice C forgets to square the −2-2 coefficient. Choice D uses the wrong binomial coefficient.

Question 6

What is the coefficient of x2y3x^2 y^3 in the expansion of (x+y)5?(x+y)^5? (You may use Pascal's Triangle row 5: 1,5,10,10,5,11,5,10,10,5,1.)

  1. 5
  2. 10 (correct answer)
  3. 15
  4. 20
Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x+y)n(x + y)^n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! Pascal's Triangle is built with a beautiful pattern: each row starts and ends with 1, and each interior number equals the sum of the two numbers above it in the previous row. Row 0 is just '1', row 1 is '1, 1', row 2 is '1, 2, 1' ((2=1+1)(2 = 1+1)), row 3 is '1, 3, 3, 1' ((3=1+2)(3 = 1+2), middle (3=2+1)(3 = 2+1)), and so on. Once you build the triangle, you have instant access to binomial coefficients! In the expansion of (x+y)5(x+y)^5 using row 5 (1,5,10,10,5,1), the term with x2y3x^2 y^3 corresponds to the fourth coefficient (10) times x2y3x^2 y^3, since powers decrease for x and increase for y. Choice B correctly identifies this coefficient as 10. A common mistake, like in choice A (5), might come from picking the adjacent coefficient, but count carefully: the coefficient for y^k is the (k+1)th in the row. Quick Pascal's Triangle construction: write 1s down both edges. For interior numbers, add the two directly above. Example for row 4: edges are 1, then interior: 1+3=4, 3+3=6, 3+1=4, giving row '1, 4, 6, 4, 1.' Check: row sums to 2n2^n (row 4 sums to 16 = 242^4). This build-as-you-go method means you never need to memorize rows—just construct them!

Question 7

Use Pascal's Triangle (row 4: 1,4,6,4,11,4,6,4,1) to expand (x+y)4.(x+y)^4.

  1. x4+4x2y+6x2y2+4xy2+y4x^4+4x^2y+6x^2y^2+4xy^2+y^4
  2. x4+6x3y+4x2y2+6xy3+y4x^4+6x^3y+4x^2y^2+6xy^3+y^4
  3. x4+4x3y+4x2y2+4xy3+y4x^4+4x^3y+4x^2y^2+4xy^3+y^4
  4. x4+4x3y+6x2y2+4xy3+y4x^4+4x^3y+6x^2y^2+4xy^3+y^4 (correct answer)
Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x+y)n(x + y)^n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! The Binomial Theorem says (x+y)n(x + y)^n expands to a sum of n+1 terms where coefficients come from row n of Pascal's Triangle. For example, (x+y)4(x + y)^4 uses row 4 of Pascal's Triangle (1, 4, 6, 4, 1) to give: 1⋅x4+4⋅x3y+6⋅x2y2+4⋅xy3+1⋅y41 \cdot x^4 + 4 \cdot x^3 y + 6 \cdot x^2 y^2 + 4 \cdot x y^3 + 1 \cdot y^4. Each term has exponents summing to 4, and coefficients from the triangle make this work perfectly! Using the given row 4 (1,4,6,4,1), the expansion is 1⋅x4+4⋅x3y+6⋅x2y2+4⋅xy3+1⋅y41 \cdot x^4 + 4 \cdot x^3 y + 6 \cdot x^2 y^2 + 4 \cdot x y^3 + 1 \cdot y^4, matching choice A exactly. Choice B incorrectly uses 4 for the middle coefficient instead of 6, but remember, the coefficients are symmetric and peak in the middle for even powers. The expansion recipe using Pascal's Triangle: (1) Identify n (the exponent on the binomial), (2) Write or construct row n of Pascal's Triangle—you'll have n+1 numbers, (3) Create n+1 terms: first has coefficient 1 and is xnx^n, last has coefficient 1 and is yny^n, middle terms use Pascal's row with decreasing x-powers and increasing y-powers, (4) Write it out: [1st coefficient]⋅xn+[2nd coefficient]⋅xn−1y+[3rd coefficient]⋅xn−2y2+…[1\text{st coefficient}] \cdot x^n + [2\text{nd coefficient}] \cdot x^{n-1} y + [3\text{rd coefficient}] \cdot x^{n-2} y^2 + \dots The pattern is systematic and reliable!

Question 8

Use Pascal's Triangle (row 4: 1,4,6,4,11,4,6,4,1) to expand (x−y)4.(x-y)^4.

  1. x4+4x3y+6x2y2+4xy3+y4x^4+4x^3y+6x^2y^2+4xy^3+y^4
  2. x4−4x3y+6x2y2−4xy3+y4x^4-4x^3y+6x^2y^2-4xy^3+y^4 (correct answer)
  3. x4−4x3y+6x2y2+4xy3−y4x^4-4x^3y+6x^2y^2+4xy^3-y^4
  4. x4−4x2y2+6x3y−4xy3+y4x^4-4x^2y^2+6x^3y-4xy^3+y^4
Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x+y)n(x + y)^n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! The Binomial Theorem says (x+y)n(x + y)^n expands to a sum of n+1 terms where coefficients come from row n of Pascal's Triangle, which saves enormous time compared to multiplying out by hand! For example, (x+y)4(x + y)^4 uses row 4 of Pascal's Triangle (1, 4, 6, 4, 1) to give: 1⋅x4+4⋅x3y+6⋅x2y2+4⋅xy3+1⋅y41 \cdot x^4 + 4 \cdot x^3 y + 6 \cdot x^2 y^2 + 4 \cdot x y^3 + 1 \cdot y^4. Each term has exponents summing to 4, and coefficients from the triangle make this work perfectly! For (x−y)4(x - y)^4, row 4 applies with alternating signs: x4−4x3y+6x2y2−4xy3+y4x^4 - 4x^3 y + 6x^2 y^2 - 4x y^3 + y^4. Choice A correctly expands with proper coefficients, powers, and alternating signs from row 4. A common error, like in choice C, is messing up the last signs—it's +y4+y^4 since (−y)4(-y)^4 is positive; always track the power of the negative. The expansion recipe using Pascal's Triangle: (1) Identify n (the exponent on the binomial), (2) Write or construct row n of Pascal's Triangle—you'll have n+1 numbers, (3) Create n+1 terms: first has coefficient 1 and is xnx^n, last has coefficient 1 and is yny^n, middle terms use Pascal's row with decreasing x-powers and increasing y-powers, (4) Write it out: [1st coefficient]⋅xn+[2nd coefficient]⋅xn−1y+[3rd coefficient]⋅xn−2y2+…[1\text{st coefficient}] \cdot x^n + [2\text{nd coefficient}] \cdot x^{n-1} y + [3\text{rd coefficient}] \cdot x^{n-2} y^2 + \dots. The pattern is systematic and reliable!

Question 9

What is the coefficient of x2y3x^2 y^3 in the expansion of (x+y)5?(x+y)^5? (You may use Pascal's Triangle.)

  1. 20
  2. 5
  3. 10 (correct answer)
  4. 6
Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x+y)n(x + y)^n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! Pascal's Triangle is built with a beautiful pattern: each row starts and ends with 1, and each interior number equals the sum of the two numbers above it in the previous row. To find the coefficient of x2y3x^2 y^3 in (x+y)5(x + y)^5, we need the term where x has power 2 and y has power 3 (note: 2+3=52 + 3 = 5, which matches our exponent). In the expansion, this is the 4th term (counting from term 1), and row 5 of Pascal's Triangle is 1, 5, 10, 10, 5, 1—so the 4th coefficient is 10. Choice B correctly identifies this coefficient as 10, while Choice A (5) would be the coefficient of x4yx^4 y or xy4x y^4, and Choice C (20) doesn't appear in row 5 at all. Quick Pascal's Triangle construction: write 1s down both edges, then for interior numbers, add the two directly above—row 5 builds from row 4 (1, 4, 6, 4, 1) to get 1, 1+4=51+4=5, 4+6=104+6=10, 6+4=106+4=10, 4+1=54+1=5, 1!

Question 10

In the expansion of (2a−3b)5(2a - 3b)^5, which statement about the signs of the terms is correct?

  1. All terms are positive when arranged in standard order
  2. Terms alternate in sign, starting with positive, when arranged in standard order (correct answer)
  3. Terms alternate in sign, starting with negative, when arranged in standard order
  4. All terms are negative when arranged in standard order
Explanation: The expansion is ∑k=05(5k)(2a)5−k(−3b)k\sum_{k=0}^{5} \binom{5}{k}(2a)^{5-k}(-3b)^k. The sign of each term depends on (−3)k=(−1)k⋅3k(-3)^k = (-1)^k \cdot 3^k. For k=0k=0: (−1)0=+1(-1)^0 = +1 (positive). For k=1k=1: (−1)1=−1(-1)^1 = -1 (negative). For k=2k=2: (−1)2=+1(-1)^2 = +1 (positive). The pattern continues as +, -, +, -, +, -, so terms alternate starting with positive. Choice A ignores the negative signs, choice C gets the starting sign wrong, and choice D incorrectly assumes all terms are negative.

Question 11

Use Pascal's Triangle (row 3: 1,3,3,11,3,3,1) to expand (2x+1)3.(2x+1)^3.​

  1. 8x3+12x2+6x+18x^3+12x^2+6x+1 (correct answer)
  2. 8x3+6x2+12x+18x^3+6x^2+12x+1
  3. 6x3+12x2+6x+16x^3+12x^2+6x+1
  4. 8x3+12x2+6x8x^3+12x^2+6x
Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x + y)^n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! For (2x + 1)^3, we use row 3 of Pascal's Triangle (1, 3, 3, 1) but must be careful with the coefficient 2 on x. The expansion is: 1·(2x)^3 + 3·(2x)^2·(1) + 3·(2x)·(1)^2 + 1·(1)^3 = 8x^3 + 3·4x^2 + 3·2x + 1 = 8x^3 + 12x^2 + 6x + 1. Choice A correctly shows this expansion with proper handling of the 2: (2x)^3 = 8x^3, (2x)^2 = 4x^2, and 2x stays as 2x. Choice B incorrectly swaps the middle coefficients (6 and 12), Choice C has 6x^3 instead of 8x^3 (forgetting 232^3 = 8), and Choice D is missing the constant term 1. When expanding (ax + b)^n, remember to raise the entire term 'ax' to each power: (ax)^k = a^k·x^k—don't forget to compute a^k for each term!

Question 12

What is the coefficient of x2y3x^2y^3 in the expansion of (x+y)5?(x+y)^5? (You may use Pascal's Triangle.)​

  1. 5
  2. 10 (correct answer)
  3. 20
  4. 6
Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x + y)^n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! Pascal's Triangle is built with a beautiful pattern: each row starts and ends with 1, and each interior number equals the sum of the two numbers above it in the previous row. To find the coefficient of x^2y^3 in (x + y)^5, we need the term where x has power 2 and y has power 3 (note: 2 + 3 = 5, which matches our exponent). In the expansion, this is the 4th term (counting from term 1), and row 5 of Pascal's Triangle is 1, 5, 10, 10, 5, 1—so the 4th coefficient is 10. Choice B correctly identifies this coefficient as 10, while Choice A (5) would be the coefficient of x^4y or xy^4, and Choice C (20) doesn't appear in row 5 at all. Quick Pascal's Triangle construction: write 1s down both edges, then for interior numbers, add the two directly above—row 5 builds from row 4 (1, 4, 6, 4, 1) to get 1, 1+4=5, 4+6=10, 6+4=10, 4+1=5, 1!

Question 13

In the expansion of (x+y)6(x+y)^6, what is the coefficient of x3y3x^3 y^3 (use Pascal's Triangle row 6)?

  1. 30
  2. 20 (correct answer)
  3. 18
  4. 15
Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x+y)n(x + y)^n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! Pascal's Triangle is built with a beautiful pattern: each row starts and ends with 1, and each interior number equals the sum of the two numbers above it in the previous row. Row 0 is just '1', row 1 is '1, 1', row 2 is '1, 2, 1' (2 = 1+1), row 3 is '1, 3, 3, 1' (3 = 1+2, middle 3 = 2+1), and so on. Once you build the triangle, you have instant access to binomial coefficients! For (x+y)6(x + y)^6, row 6 is 1,6,15,20,15,6,1, and the x3y3x^3 y^3 term corresponds to the 4th coefficient (for y3y^3), which is 20. Choice C correctly identifies the coefficient as 20 using row 6 of Pascal's Triangle. Picking 15 (choice A) might mean choosing the adjacent coefficient—note that for equal exponents in even nn, it's the middle one: here, 20 for k=3k=3 in C(6,3)C(6,3). Quick Pascal's Triangle construction: write 1s down both edges. For interior numbers, add the two directly above. Example for row 4: edges are 1, then interior: 1+3=4, 3+3=6, 3+1=4, giving row '1, 4, 6, 4, 1.' Check: row sums to 2n2^n (row 4 sums to 16 = 242^4). This build-as-you-go method means you never need to memorize rows—just construct them!

Question 14

What is the 4th term (counting from the first term as term 1) in the expansion of (x+y)6(x+y)^6?

  1. 15x2y415x^2y^4
  2. 20x3y320x^3y^3 (correct answer)
  3. 15x4y215x^4y^2
  4. 6x3y36x^3y^3
Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x+y)n(x + y)^n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! The Binomial Theorem says (x+y)6(x + y)^6 expands using row 6 of Pascal's Triangle, which is 1, 6, 15, 20, 15, 6, 1. The 4th term (counting from term 1) has the 4th coefficient (20) and follows the pattern where x-powers decrease and y-powers increase: term 1 is x6x^6, term 2 is 6x5y6x^5y, term 3 is 15x4y215x^4y^2, and term 4 is 20x3y320x^3y^3. Choice B correctly identifies this as 20x3y320x^3y^3, with coefficient 20 from Pascal's Triangle and exponents summing to 6. Choice A (15x2y415x^2y^4) would be the 5th term, Choice C (15x4y215x^4y^2) is the 3rd term, and Choice D (6x3y36x^3y^3) uses the wrong coefficient. Quick Pascal's Triangle construction for row 6: start with row 5 (1, 5, 10, 10, 5, 1), then build row 6 as 1, 1+5=61+5=6, 5+10=155+10=15, 10+10=2010+10=20, 10+5=1510+5=15, 5+1=65+1=6, 1—systematic and reliable!

Question 15

What is the 4th term (counting from the first term as term 1) in the expansion of (x+y)6?(x+y)^6? (Pascal's Triangle row 6: 1,6,15,20,15,6,11,6,15,20,15,6,1.)

  1. 20x3y320x^3y^3 (correct answer)
  2. 15x3y315x^3y^3
  3. 20x4y220x^4y^2
  4. 15x4y215x^4y^2
Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x+y)n(x + y)^n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! Pascal's Triangle is built with a beautiful pattern: each row starts and ends with 1, and each interior number equals the sum of the two numbers above it in the previous row. Row 0 is just '1', row 1 is '1, 1', row 2 is '1, 2, 1' (2 = 1+1), row 3 is '1, 3, 3, 1' (3 = 1+2, middle 3 = 2+1), and so on. Once you build the triangle, you have instant access to binomial coefficients! In (x+y)6(x+y)^6 using row 6 (1,6,15,20,15,6,1), the 4th term is the coefficient 20 times x6−3y3x^{6-3} y^3, or 20x3y320x^3 y^3, matching choice B. Choice A (15x4y215x^4 y^2) is actually the 3rd term—remember to count starting from term 1 as the x6x^6 term. Quick Pascal's Triangle construction: write 1s down both edges. For interior numbers, add the two directly above. Example for row 4: edges are 1, then interior: 1+3=4, 3+3=6, 3+1=4, giving row '1, 4, 6, 4, 1.' Check: row sums to 2n2^n (row 4 sums to 16 = 242^4). This build-as-you-go method means you never need to memorize rows—just construct them!

Question 16

Use Pascal's Triangle (row 5: 1,5,10,10,5,11,5,10,10,5,1) to expand (x+y)5(x+y)^5.

  1. x5+5x4y+10x3y2+10x2y3+5xy4+y5x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5 (correct answer)
  2. x5+5x4y+10x2y3+10x3y2+5xy4+y5x^5+5x^4y+10x^2y^3+10x^3y^2+5xy^4+y^5
  3. x5+4x4y+6x3y2+4x2y3+xy4+y5x^5+4x^4y+6x^3y^2+4x^2y^3+xy^4+y^5
  4. x5+5x4y+10x3y2+5x2y3+10xy4+y5x^5+5x^4y+10x^3y^2+5x^2y^3+10xy^4+y^5
Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x+y)n(x + y)^n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! The Binomial Theorem says (x+y)n(x + y)^n expands to a sum of n+1 terms where coefficients come from row n of Pascal's Triangle, x-powers decrease from n to 0, and y-powers increase from 0 to n. For (x+y)5(x + y)^5, we use row 5 of Pascal's Triangle (1,5,10,10,5,1)(1, 5, 10, 10, 5, 1) to get: 1⋅x5+5⋅x4y+10⋅x3y2+10⋅x2y3+5⋅xy4+1⋅y51 \cdot x^5 + 5 \cdot x^4 y + 10 \cdot x^3 y^2 + 10 \cdot x^2 y^3 + 5 \cdot x y^4 + 1 \cdot y^5. Choice A correctly shows this expansion with all six terms having the right coefficients from Pascal's Triangle and powers that decrease for x (5→0) while increasing for y (0→5). Choice B incorrectly swaps the middle coefficients (has 5 and 10 instead of 10 and 10), while Choice C uses row 4 coefficients instead of row 5. The expansion recipe using Pascal's Triangle: (1) Identify n = 5, (2) Use row 5: (1,5,10,10,5,1)(1, 5, 10, 10, 5, 1), (3) Create 6 terms with decreasing x-powers and increasing y-powers, (4) Write it out systematically—the pattern is beautiful and reliable!

Question 17

What is the coefficient of x4y2x^4 y^2 in the expansion of (x+y)6(x + y)^6? (You may use Pascal's Triangle row 6: 1,6,15,20,15,6,11, 6, 15, 20, 15, 6, 1.)

  1. 30
  2. 15 (correct answer)
  3. 6
  4. 20
Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x+y)n(x + y)^n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! The Binomial Theorem says (x+y)n(x + y)^n expands to a sum of n+1 terms where coefficients come from row n of Pascal's Triangle, x-powers decrease from n to 0, and y-powers increase from 0 to n. For example, (x+y)4(x + y)^4 uses row 4 of Pascal's Triangle (1, 4, 6, 4, 1) to give: 1⋅x4+4⋅x3y+6⋅x2y2+4⋅xy3+1⋅y41 \cdot x^4 + 4 \cdot x^3 y + 6 \cdot x^2 y^2 + 4 \cdot x y^3 + 1 \cdot y^4. Each term has exponents summing to 4, and coefficients from the triangle make this work perfectly! In (x+y)6(x+y)^6 with row 6 (1,6,15,20,15,6,1), the coefficient for x4y2x^4 y^2 is the third one (15, for y2y^2), so choice B is correct. Choice C (20) is for y3y^3 instead—ensure the y-power matches the position in the row (k+1 for yky^k). The expansion recipe using Pascal's Triangle: (1) Identify n (the exponent on the binomial), (2) Write or construct row n of Pascal's Triangle—you'll have n+1 numbers, (3) Create n+1 terms: first has coefficient 1 and is xnx^n, last has coefficient 1 and is yny^n, middle terms use Pascal's row with decreasing x-powers and increasing y-powers, (4) Write it out: [1st coefficient]⋅xn+[2nd coefficient]⋅xn−1y+[3rd coefficient]⋅xn−2y2+…[1\text{st coefficient}] \cdot x^n + [2\text{nd coefficient}] \cdot x^{n-1} y + [3\text{rd coefficient}] \cdot x^{n-2} y^2 + \dots The pattern is systematic and reliable!

Question 18

Use Pascal's Triangle (row 4: 1,4,6,4,11,4,6,4,1) to expand (a+b)4.(a+b)^4.

  1. a4+4a2b+6a2b2+4ab3+b4a^4+4a^2b+6a^2b^2+4ab^3+b^4
  2. a4+5a3b+10a2b2+5ab3+b4a^4+5a^3b+10a^2b^2+5ab^3+b^4
  3. a4+4a3b+6a2b2+ab3+b4a^4+4a^3b+6a^2b^2+ab^3+b^4
  4. a4+4a3b+6a2b2+4ab3+b4a^4+4a^3b+6a^2b^2+4ab^3+b^4 (correct answer)
Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x+y)n(x + y)^n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! The Binomial Theorem says (a+b)n(a + b)^n expands to a sum of n+1 terms where coefficients come from row n of Pascal's Triangle, a-powers decrease from n to 0, and b-powers increase from 0 to n. For (a+b)4(a + b)^4, we use row 4 (given as 1,4,6,4,11, 4, 6, 4, 1) to build: 1⋅a4+4⋅a3b+6⋅a2b2+4⋅ab3+1⋅b41 \cdot a^4 + 4 \cdot a^3 b + 6 \cdot a^2 b^2 + 4 \cdot a b^3 + 1 \cdot b^4. Choice A correctly shows all five terms with proper coefficients and powers summing to 4 in each term. Choice B has a missing coefficient (just ab3ab^3 instead of 4ab34ab^3), Choice C uses row 5 coefficients (5,10,55, 10, 5) instead of row 4, and Choice D has a typo with a^2b instead of a^3b in the second term. The expansion recipe using Pascal's Triangle: (1) Identify n = 4, (2) Use row 4: 1,4,6,4,11, 4, 6, 4, 1, (3) Create 5 terms with decreasing a-powers and increasing b-powers, (4) Write systematically—this method never fails!

Question 19

In the expansion of (x−y)4,(x-y)^4, what is the coefficient of x2y2x^2y^2?

  1. −6-6
  2. 66 (correct answer)
  3. −4-4
  4. 44
Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x + y)^n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! Pascal's Triangle is built with a beautiful pattern: each row starts and ends with 1, and each interior number equals the sum of the two numbers above it in the previous row. Row 0 is just '1', row 1 is '1, 1', row 2 is '1, 2, 1' (2 = 1+1), row 3 is '1, 3, 3, 1' (3 = 1+2, middle 3 = 2+1), and so on. Once you build the triangle, you have instant access to binomial coefficients! For (x-y)⁴ = (x+(-y))⁴, we use row 4 of Pascal's Triangle (1, 4, 6, 4, 1). The x²y² term comes from the 3rd term: 6·x²·(-y)² = 6·x²·y² (since (-y)² = y²). The coefficient is +6 because the even power of (-y) gives a positive result. Choice B correctly identifies coefficient as 6. Choice A (-6) incorrectly assumes the negative sign carries through, Choice C (-4) and Choice D (4) use the wrong Pascal's Triangle coefficient (4 instead of 6). Quick Pascal's Triangle construction: write 1s down both edges. For interior numbers, add the two directly above. Example for row 4: edges are 1, then interior: 1+3=4, 3+3=6, 3+1=4, giving row '1, 4, 6, 4, 1.' Check: row sums to 2^n (row 4 sums to 16 = 2⁴). This build-as-you-go method means you never need to memorize rows—just construct them!

Question 20

What is the coefficient of x3y3x^3 y^3 in the expansion of (x+y)6?(x+y)^6? (You may use Pascal's Triangle.)

  1. 15
  2. 20 (correct answer)
  3. 10
  4. 18
Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x + y)^n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! Pascal's Triangle is built with a beautiful pattern: each row starts and ends with 1, and each interior number equals the sum of the two numbers above it in the previous row. For (x + y)^6, we need row 6 of Pascal's Triangle: 1, 6, 15, 20, 15, 6, 1. The term x3y3x^3 y^3 appears when both variables have power 3 (note: 3+3=63 + 3 = 6), which is the middle term (4th term) in the expansion, so its coefficient is 20. Choice C correctly identifies this coefficient as 20, while Choice A (15) would be the coefficient of x4y2x^4 y^2 or x2y4x^2 y^4, and Choice D (10) doesn't appear in row 6 at all. Quick Pascal's Triangle construction: row 6 builds from row 5 (1, 5, 10, 10, 5, 1) by adding adjacent pairs: 1, 1+5=61+5=6, 5+10=155+10=15, 10+10=2010+10=20, 10+5=1510+5=15, 5+1=65+1=6, 1—the middle coefficient is always the largest in each row!