Algebra 2 Quiz: Applying The Binomial Theorem
20 questions · exam conditions
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Applying The Binomial TheoremQuestion 1 of 20

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

a3+3a2b+3ab2+b3a^3+3a^2b+3ab^2+b^3
a3+3a3b+3ab2+b3a^3+3a^3b+3ab^2+b^3
a3+2a2b+2ab2+b3a^3+2a^2b+2ab^2+b^3
a3+3a2b+3a2b2+b3a^3+3a^2b+3a^2b^2+b^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

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

  1. a3+3a2b+3ab2+b3a^3+3a^2b+3ab^2+b^3 (correct answer)
  2. a3+3a3b+3ab2+b3a^3+3a^3b+3ab^2+b^3
  3. a3+2a2b+2ab2+b3a^3+2a^2b+2ab^2+b^3
  4. a3+3a2b+3a2b2+b3a^3+3a^2b+3a^2b^2+b^3
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! Using the given row 3 (1, 3, 3, 1), we expand (a+b)³: 1·a³b⁰ + 3·a²b¹ + 3·a¹b² + 1·a⁰b³ = a³ + 3a²b + 3ab² + b³. Each term follows the pattern: Pascal coefficient times a^(decreasing power) times b^(increasing power), with all exponents summing to 3. Choice A correctly expands with proper coefficients and powers: a³ + 3a²b + 3ab² + b³. Choice B incorrectly writes 3a³b for the second term (should be 3a²b), Choice C uses coefficients 2 instead of 3 for the middle terms, and Choice D has 3a²b² instead of 3ab² for the third 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 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 2

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 using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! The Binomial Theorem says (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 x^6, term 2 is 6x^5y, term 3 is 15x^4y^2, and term 4 is 20x^3y^3. Choice B correctly identifies this as 20x^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=6, 5+10=15, 10+10=20, 10+5=15, 5+1=6, 1—systematic and reliable!

Question 3

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

  1. 15
  2. 18
  3. 20 (correct answer)
  4. 10
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. For (x+y)6(x + y)^6, we need row 6 of Pascal's Triangle: 1, 6, 15, 20, 15, 6, 1. The term x3y3x^3y^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^4y^2 or x2y4x^2y^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!

Question 4

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 5

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 6

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

  1. 15x4y215x^4y^2
  2. 20x3y320x^3y^3 (correct answer)
  3. 15x2y415x^2y^4
  4. 6x5y6x^5y
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)⁶, row 6 of Pascal's Triangle is 1, 6, 15, 20, 15, 6, 1. The terms in order are: 1st term: 1·x⁶, 2nd term: 6·x⁵y, 3rd term: 15·x⁴y², 4th term: 20·x³y³, 5th term: 15·x²y⁴, 6th term: 6·xy⁵, 7th term: 1·y⁶. The 4th term is 20x³y³. Choice B correctly identifies the 4th term as 20x³y³. Choice A (15x⁴y²) is the 3rd term, Choice C (15x²y⁴) is the 5th term, and Choice D (6x⁵y) is the 2nd term—all are off by one or more positions in the expansion. 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 7

Use Pascal's Triangle (row 5: 1,5,10,10,5,11,5,10,10,5,1) to expand (x+2)5.(x+2)^5.

  1. x5+10x4+40x3+80x2+80x+32x^5+10x^4+40x^3+80x^2+80x+32 (correct answer)
  2. x5+10x4+80x3+80x2+40x+32x^5+10x^4+80x^3+80x^2+40x+32
  3. x5+5x4+10x3+10x2+5x+1x^5+5x^4+10x^3+10x^2+5x+1
  4. x5+5x4+40x3+80x2+80x+32x^5+5x^4+40x^3+80x^2+80x+32
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! The Binomial Theorem says (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 uses row 4 of Pascal's Triangle (1, 4, 6, 4, 1) to give: 1·x⁴ + 4·x³y + 6·x²y² + 4·xy³ + 1·y⁴. Each term has exponents summing to 4, and coefficients from the triangle make this work perfectly! For (x+2)⁵, we use row 5 (1, 5, 10, 10, 5, 1) and expand: 1·x⁵·2⁰ + 5·x⁴·2¹ + 10·x³·2² + 10·x²·2³ + 5·x¹·2⁴ + 1·x⁰·2⁵ = x⁵ + 5·x⁴·2 + 10·x³·4 + 10·x²·8 + 5·x·16 + 1·32 = x⁵ + 10x⁴ + 40x³ + 80x² + 80x + 32. Choice A correctly expands with proper coefficients and powers: x⁵ + 10x⁴ + 40x³ + 80x² + 80x + 32. Choice B incorrectly swaps the coefficients of x³ and x terms, Choice C shows the expansion of (x+1)⁵ not (x+2)⁵, and Choice D has the wrong coefficient for x⁴ (should be 10, not 5). 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 x^n, last has coefficient 1 and is y^n, middle terms use Pascal's row with decreasing x-powers and increasing y-powers, (4) Write it out: [1st coefficient]·x^n + [2nd coefficient]·x^(n-1)·y + [3rd coefficient]·x^(n-2)·y² + ... The pattern is systematic and reliable!

Question 8

Find the expansion of (x+2)5(x+2)^5 using the Binomial Theorem (or Pascal's Triangle).

  1. x5+10x4+40x3+80x2+80x+32x^5+10x^4+40x^3+80x^2+80x+32 (correct answer)
  2. x5+5x4+20x3+40x2+80x+32x^5+5x^4+20x^3+40x^2+80x+32
  3. x5+10x4+20x3+80x2+80x+32x^5+10x^4+20x^3+80x^2+80x+32
  4. x5+5x4+40x3+80x2+80x+16x^5+5x^4+40x^3+80x^2+80x+16
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 (x+2)5(x + 2)^5, we use row 5 of Pascal's Triangle (1, 5, 10, 10, 5, 1) and must handle powers of 2 carefully. The expansion is: 1⋅x5+5⋅x4⋅(2)+10⋅x3⋅(2)2+10⋅x2⋅(2)3+5⋅x⋅(2)4+1⋅(2)5=x5+10x4+40x3+80x2+80x+321\cdot x^5 + 5\cdot x^4\cdot(2) + 10\cdot x^3\cdot(2)^2 + 10\cdot x^2\cdot(2)^3 + 5\cdot x\cdot(2)^4 + 1\cdot(2)^5 = x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32. Choice A correctly computes all powers of 2: 21=22^1 = 2, 22=42^2 = 4, 23=82^3 = 8, 24=162^4 = 16, 25=322^5 = 32, and multiplies by Pascal's coefficients: 5⋅2=105\cdot 2 = 10, 10⋅4=4010\cdot 4 = 40, 10⋅8=8010\cdot 8 = 80, 5⋅16=805\cdot 16 = 80. Choice B incorrectly has 5x^4 and 20x^3, Choice C has 20x^3 instead of 40x^3, and Choice D has the wrong constant term (16 instead of 32). When expanding (x+a)n(x + a)^n, systematically compute aka^k for k = 1, 2, ..., n and multiply by Pascal's coefficients—the powers of a grow quickly but predictably!

Question 9

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 10

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 11

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 12

What is the coefficient of x4y2x^4 y^2 in the expansion of (x+y)6?(x + y)^6?

  1. 6
  2. 15 (correct answer)
  3. 20
  4. 10
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, xx-powers decrease from nn to 0, and yy-powers increase from 0 to nn. For (x+y)6(x + y)^6, row 6 of Pascal's Triangle is 1, 6, 15, 20, 15, 6, 1. The term x4y2x^4 y^2 has xx to power 4 and yy to power 2 (note: 4+2=64 + 2 = 6), which is the 3rd term in the expansion, so its coefficient is 15. Choice B correctly identifies this coefficient as 15, while Choice A (6) would be the coefficient of x5yx^5 y or xy5x y^5, Choice C (20) is the coefficient of x3y3x^3 y^3, and Choice D (10) doesn't appear in row 6. The expansion recipe using Pascal's Triangle: identify which term you need by looking at the y-power (y2y^2 means 3rd term), then read off the coefficient—systematic and foolproof!

Question 13

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

  1. 8x3−36x2+54x−278x^3-36x^2+54x-27 (correct answer)
  2. 8x3−18x2+54x−278x^3-18x^2+54x-27
  3. 8x3+36x2+54x+278x^3+36x^2+54x+27
  4. 8x3−36x2+27x−278x^3-36x^2+27x-27
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: 1x4+4x3y+6x2y2+4xy3+1y41 x^4 + 4 x^3 y + 6 x^2 y^2 + 4 x y^3 + 1 y^4. Each term has exponents summing to 4, and coefficients from the triangle make this work perfectly! For (2x−3)3(2x - 3)^3, apply the coefficients (1,3,3,1) to (2x)3+3(2x)2(−3)+3(2x)(−3)2+(−3)3(2x)^3 + 3(2x)^2(-3) + 3(2x)(-3)^2 + (-3)^3, yielding 8x3−36x2+54x−278x^3 - 36x^2 + 54x - 27, as in choice A. Choice B might result from miscalculating the signs or powers, like using +18x^2 instead of -36x^2, but always include the negative sign from the binomial. 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 x^n, last has coefficient 1 and is y^n, middle terms use Pascal's row with decreasing x-powers and increasing y-powers, (4) Write it out: [1[1st coefficient]⋅xn+[2] \cdot x^n + [2nd coefficient]⋅xn−1⋅y+[3] \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 14

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 15

Expand (x−2)4(x-2)^4 using the Binomial Theorem (Pascal's Triangle row 4: 1,4,6,4,11,4,6,4,1).

  1. x4−4x3+6x2−4x+1x^4-4x^3+6x^2-4x+1
  2. x4−8x3+12x2−32x+16x^4-8x^3+12x^2-32x+16
  3. x4−8x3+24x2+32x+16x^4-8x^3+24x^2+32x+16
  4. x4−8x3+24x2−32x+16x^4-8x^3+24x^2-32x+16 (correct answer)
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 (x - 2)^4, we treat this as (x + (-2))^4 and use row 4 of Pascal's Triangle (1, 4, 6, 4, 1). The expansion becomes: 1⋅x4+4⋅x3⋅(−2)+6⋅x2⋅(−2)2+4⋅x⋅(−2)3+1⋅(−2)4=x4−8x3+24x2−32x+161·x^4 + 4·x^3·(-2) + 6·x^2·(-2)^2 + 4·x·(-2)^3 + 1·(-2)^4 = x^4 - 8x^3 + 24x^2 - 32x + 16. Choice A correctly shows this with alternating signs (due to odd powers of -2 being negative) and proper coefficients: 4⋅(−2)=−84·(-2) = -8, 6⋅4=246·4 = 24, 4⋅(−8)=−324·(-8) = -32, and (−2)4=16(-2)^4 = 16. Choice B has the wrong sign on the -32x term, Choice C appears to use (x-1)^4 instead, and Choice D has 12x^2 instead of 24x^2. When expanding (x - a)^n, remember that negative terms create an alternating sign pattern: minus for odd powers of (−a)(-a), plus for even powers—this pattern is systematic and predictable!

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+5x2y3+10xy4+y5x^5+5x^4y+10x^3y^2+5x^2y^3+10xy^4+y^5
  2. x5+4x4y+6x3y2+4x2y3+xy4+y5x^5+4x^4y+6x^3y^2+4x^2y^3+xy^4+y^5
  3. x5+5x4y+10x3y2+10x2y3+5xy4+y5x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5 (correct answer)
  4. x5+5x4y+10x2y3+10x3y2+5xy4+y5x^5+5x^4y+10x^2y^3+10x^3y^2+5xy^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 (1,5,10,10,5,11, 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→05 \to 0) while increasing for y (0→50 \to 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=5n = 5, (2) Use row 5: 1,5,10,10,5,11, 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

Use the Binomial Theorem to compute 1033103^3 by writing it as (100+3)3.(100+3)^3.

  1. 1,090,000
  2. 1,092,727 (correct answer)
  3. 1,092,700
  4. 1,093,727
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! The Binomial Theorem says (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 uses row 4 of Pascal's Triangle (1, 4, 6, 4, 1) to give: 1·x⁴ + 4·x³y + 6·x²y² + 4·xy³ + 1·y⁴. Each term has exponents summing to 4, and coefficients from the triangle make this work perfectly! To compute 103³ = (100+3)³, we use row 3 of Pascal's Triangle (1, 3, 3, 1): (100+3)³ = 1·100³ + 3·100²·3 + 3·100·3² + 1·3³ = 1,000,000 + 3·10,000·3 + 3·100·9 + 27 = 1,000,000 + 90,000 + 2,700 + 27 = 1,092,727. Choice B correctly uses row 3 of Pascal's Triangle to get 1,092,727. Choice A (1,090,000) omits the last two terms (2,700 + 27), Choice C (1,092,700) omits just the final 27, and Choice D (1,093,727) has an arithmetic error adding 1,000 extra. 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 x^n, last has coefficient 1 and is y^n, middle terms use Pascal's row with decreasing x-powers and increasing y-powers, (4) Write it out: [1st coefficient]·x^n + [2nd coefficient]·x^(n-1)·y + [3rd coefficient]·x^(n-2)·y² + ... The pattern is systematic and reliable!

Question 18

Use the Binomial Theorem (coefficients from Pascal's Triangle row 4: 1,4,6,4,11,4,6,4,1) to expand (x+2)4.(x+2)^4.

  1. x4+8x3+12x2+32x+16x^4+8x^3+12x^2+32x+16
  2. x4+8x2+24x2+32x+16x^4+8x^2+24x^2+32x+16
  3. x4+8x3+24x2+32x+16x^4+8x^3+24x^2+32x+16 (correct answer)
  4. x4+4x3+6x2+4x+1x^4+4x^3+6x^2+4x+1
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+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+2)^4 using row 4 (1,4,6,4,1), expand to x4+4x3(2)+6x2(4)+4x(8)+16x^4 + 4x^3(2) + 6x^2(4) + 4x(8) + 16, simplifying to x4+8x3+24x2+32x+16x^4 +8x^3 +24x^2 +32x +16, matching choice A. Choice B is for (x+1)^4—verify by calculating powers of 2, like 22=42^2=4, 23=82^3=8, 24=162^4=16. 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=41+3=4, 3+3=63+3=6, 3+1=43+1=4, giving row '1, 4, 6, 4, 1.' Check: row sums to 2n2^n (row 4 sums to 16=2416 = 2^4). This build-as-you-go method means you never need to memorize rows—just construct them!

Question 19

Use Pascal's Triangle to expand (x+2)4.(x+2)^4.

  1. x4+8x3+24x2+32x+16x^4+8x^3+24x^2+32x+16 (correct answer)
  2. x4+4x3+24x2+32x+16x^4+4x^3+24x^2+32x+16
  3. x4+8x3+12x2+32x+16x^4+8x^3+12x^2+32x+16
  4. x4+8x3+24x2+16x+16x^4+8x^3+24x^2+16x+16
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! 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, 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+2)4(x + 2)^4, row 4 gives x4+4x3(2)+6x2(2)2+4x(2)3+(2)4=x4+8x3+24x2+32x+16x^4 + 4 x^3 (2) + 6 x^2 (2)^2 + 4 x (2)^3 + (2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16. Choice A correctly expands with proper coefficients and numerical applications from row 4. A distractor like choice B might undercalculate the second term—it's 4⋅x3⋅2=8x34 \cdot x^3 \cdot 2 = 8x^3, not 4x34x^3; always multiply the coefficient by the power of the constant. 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+[2] \cdot x^n + [2nd coefficient]⋅xn−1⋅y+[3] \cdot x^{n-1} \cdot y + [3rd coefficient]⋅xn−2⋅y2+…] \cdot x^{n-2} \cdot y^2 + \ldots The pattern is systematic and reliable!

Question 20

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!