Applying the Binomial Theorem - Algebra 2

1. What is the coefficient of $x^4 y^2$ in the expansion of $(x + y)^6?$
   Answer: 15 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 $(x + y)^6$, row 6 of Pascal's Triangle is 1, 6, 15, 20, 15, 6, 1. The term $x^4 y^2$ has $x$ to power 4 and $y$ to power 2 (note: $4 + 2 = 6$), 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 $x^5 y$ or $x y^5$, Choice C (20) is the coefficient of $x^3 y^3$, 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 ($y^2$ means 3rd term), then read off the coefficient—systematic and foolproof!
2. Use Pascal's Triangle (row 5: $1,5,10,10,5,1$) to expand $(x+2)^5.$
   Answer: $x^5+10x^4+40x^3+80x^2+80x+32$ 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!
3. Find the expansion of $(x+2)^5$ using the Binomial Theorem (or Pascal's Triangle).
   Answer: $x^5+10x^4+40x^3+80x^2+80x+32$ 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 $ $(x + 2)^5$ $, we use row 5 of Pascal's Triangle (1, 5, 10, 10, 5, 1) and must handle powers of 2 carefully. The expansion is: $1\cdot x^5 + 5\cdot x^4\cdot(2) + 10\cdot x^3\cdot(2)^2 + 10\cdot x^2\cdot(2)^3 + 5\cdot x\cdot(2)^4 + 1\cdot(2)^5 = x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32$. Choice A correctly computes all powers of 2: $ $2^1$ = 2 $, $ $2^2$ = 4 $, $ $2^3$ = 8 $, $ $2^4$ = 16 $, $ $2^5$ = 32 $, and multiplies by Pascal's coefficients: $5\cdot 2 = 10$, $10\cdot 4 = 40$, $10\cdot 8 = 80$, $5\cdot 16 = 80$. 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 $ $(x + a)^n$ $, systematically compute $ $a^k$ $ for k = 1, 2, ..., n and multiply by Pascal's coefficients—the powers of a grow quickly but predictably!
4. What is the 4th term (counting from the first term as term 1) in the expansion of $(x+y)^6?$ (Pascal's Triangle row 6: $1,6,15,20,15,6,1$.)
   Answer: $20x^3y^3$ 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! In $(x+y)^6$ using row 6 (1,6,15,20,15,6,1), the 4th term is the coefficient 20 times $x^{6-3} y^3$, or $20x^3 y^3$, matching choice B. Choice A ($15x^4 y^2$) is actually the 3rd term—remember to count starting from term 1 as the $x^6$ 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 $2^n$ (row 4 sums to 16 = $2^4$). This build-as-you-go method means you never need to memorize rows—just construct them!
5. Use Pascal's Triangle (row 3: $1,3,3,1$) to expand $(2x+1)^3.$​
   Answer: $8x^3+12x^2+6x+1$ 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 $2^3$ = 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!
6. Find the expansion of $(x+2)^5$ using the Binomial Theorem (or Pascal's Triangle).​
   Answer: $x^5+10x^4+40x^3+80x^2+80x+32$ 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 $(x + 2)^5$, 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·$x^5$ + 5·$x^4$·(2) + 10·$x^3$·$(2)^2$ + 10·$x^2$·$(2)^3$ + 5·x·$(2)^4$ + 1·$(2)^5$ = $x^5$ + $10x^4$ + $40x^3$ + $80x^2$ + 80x + 32. Choice A correctly computes all powers of 2: $2^1$ = 2, $2^2$ = 4, $2^3$ = 8, $2^4$ = 16, $2^5$ = 32, and multiplies by Pascal's coefficients: 5·2 = 10, 10·4 = 40, 10·8 = 80, 5·16 = 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 $(x + a)^n$, systematically compute $a^k$ for k = 1, 2, ..., n and multiply by Pascal's coefficients—the powers of a grow quickly but predictably!
7. Use Pascal's Triangle (row 5: $1,5,10,10,5,1$) to expand $(a + b)^5$.
   Answer: $a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$ 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 row 5 ($1,5,10,10,5,1$) for $(a+b)^5$ gives $a^5 + 5a^4 b + 10a^3 b^2 + 10a^2 b^3 + 5a b^4 + b^5$, exactly choice A. Choice B swaps the 10 and 5 in the later terms, but the row is symmetric, so coefficients mirror from start to end. 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 $16 = 2^4$). This build-as-you-go method means you never need to memorize rows—just construct them!
8. Use the Binomial Theorem to compute $103^3$ by rewriting it as $(100+3)^3$.
   Answer: 1,092,727 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! Rewriting $103^3$ as $(100 + 3)^3$ uses row 3: $1*(100)^3 + 3*(100)^2*3 + 3*100*3^2 + 1*3^3 = 1,000,000 + 90,000 + 2,700 + 27 = 1,092,727$. Choice B correctly computes the value using the Binomial Theorem expansion. Choosing something like 1,093,727 (choice D) might result from adding an extra 1,000—ensure you calculate each term accurately, like $3*10,000*3=90,000$. 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 $16 = 2^4$). This build-as-you-go method means you never need to memorize rows—just construct them!
9. What is the 4th term (counting from the first term as 1st) in the expansion of $(x+y)^6?$
   Answer: $20x^3y^3$ 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!
10. What is the coefficient of $x^4 y^2$ in the expansion of $(x+y)^6$ (use Pascal's Triangle coefficients)?
    Answer: 15 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 is 1,6,15,20,15,6,1, and the $x^4 y^2$ term is for $y^2$ (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 $y^3$ instead—confirm the exponents: $x^4 y^2$ needs k=2, $C(6,2)=15$. 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 $2^n$ (row 4 sums to 16 = $2^4$). This build-as-you-go method means you never need to memorize rows—just construct them!
11. Find the expansion of $(2x-3)^3$ using the Binomial Theorem.
    Answer: $8x^3-36x^2+54x-27$ 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 (2x-3)³, we use row 3 of Pascal's Triangle (1, 3, 3, 1) and treat it as (2x + (-3))³. The expansion gives: 1·(2x)³·(-3)⁰ + 3·(2x)²·(-3)¹ + 3·(2x)¹·(-3)² + 1·(2x)⁰·(-3)³ = 8x³ + 3·4x²·(-3) + 3·2x·9 + 1·(-27) = 8x³ - 36x² + 54x - 27. Choice A correctly expands with proper coefficients: 8x³ - 36x² + 54x - 27. Choice B has -18x² instead of -36x² (error in calculating 3·4·(-3)), Choice C has wrong signs throughout, and Choice D starts with 4x³ instead of 8x³ (error in calculating (2x)³). 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!
12. What is the coefficient of $x^4 y^2$ in the expansion of $(x + y)^6$? (You may use Pascal's Triangle row 6: $1, 6, 15, 20, 15, 6, 1$.)
    Answer: 15 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: $1 \cdot x^4 + 4 \cdot x^3 y + 6 \cdot x^2 y^2 + 4 \cdot x y^3 + 1 \cdot y^4$. Each term has exponents summing to 4, and coefficients from the triangle make this work perfectly! In $(x+y)^6$ with row 6 (1,6,15,20,15,6,1), the coefficient for $x^4 y^2$ is the third one (15, for $y^2$), so choice B is correct. Choice C (20) is for $y^3$ instead—ensure the y-power matches the position in the row (k+1 for $y^k$). 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 $x^n$, 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: $[1\text{st coefficient}] \cdot x^n + [2\text{nd coefficient}] \cdot x^{n-1} y + [3\text{rd coefficient}] \cdot x^{n-2} y^2 + \dots$ The pattern is systematic and reliable!
13. What is the coefficient of $x^3y^3$ in the expansion of $(x+y)^6?$ (You may use Pascal's Triangle.)
    Answer: 20 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 $x^3y^3$ appears when both variables have power 3 (note: $3 + 3 = 6$), 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 $x^4y^2$ or $x^2y^4$, 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, $1+5=6$, $5+10=15$, $10+10=20$, $10+5=15$, $5+1=6$, 1—the middle coefficient is always the largest in each row!
14. Use the Binomial Theorem to compute $103^3$ by writing it as $(100+3)^3.$
    Answer: 1,092,727 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!
15. What is the coefficient of $x^3y^2$ in the expansion of $(x+y)^5?$ (You may use Pascal's Triangle row 5: $1,5,10,10,5,1$.)
    Answer: 10 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! In the expansion of (x+y)⁵, we need the coefficient of x³y². The general term in the expansion is C(5,k)·$x^{5-k}$·$y^k$, where C(5,k) is the binomial coefficient. For x³y², we have x³ = $x^{5-2}$ and y² = $y^2$, so k = 2. The coefficient is the 3rd number in row 5 of Pascal's Triangle (counting from position 0): 1, 5, 10, 10, 5, 1. The third position (k=2) gives us 10. Choice B correctly identifies coefficient as 10. Choice A (5) would be the coefficient of x⁴y¹, Choice C (20) is from row 6 not row 5, and Choice D (15) is also from row 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!
16. Use Pascal's Triangle (row 4: $1,4,6,4,1$) to expand $(x-2)^4.$
    Answer: $x^4-8x^3+24x^2-32x+16$ 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 $(x-2)^4$ using row 4 (1,4,6,4,1), expand to $x^4 + 4x^3(-2) + 6x^2(4) + 4x(-8) + 16$, simplifying to $x^4 -8x^3 +24x^2 -32x +16$, as in choice A. Choice C uses coefficients for $(x-1)^4$ instead—check the powers of the second term, like $(-2)^k$ for each. 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 = $2^4$). This build-as-you-go method means you never need to memorize rows—just construct them!
17. Use the Binomial Theorem to compute $103^3$ by rewriting it as $(100+3)^3$ and expanding.
    Answer: 1,092,727 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: $1 \cdot x^4 + 4 \cdot x^3 y + 6 \cdot x^2 y^2 + 4 \cdot x y^3 + 1 \cdot y^4$. Each term has exponents summing to 4, and coefficients from the triangle make this work perfectly! Rewriting $103^3$ as $(100+3)^3$ and expanding with row 3 (1,3,3,1) gives $100^3 + 3 \cdot 100^2 \cdot 3 + 3 \cdot 100 \cdot 3^2 + 3^3 = 1,000,000 + 90,000 + 2,700 + 27 = 1,092,727$, as in choice A. Choice B might come from forgetting to multiply by the coefficients, like adding 80,000 instead, but always apply the Pascal's row multipliers. 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: $[1\text{st coefficient}] \cdot x^n + [2\text{nd coefficient}] \cdot x^{n-1} y + [3\text{rd coefficient}] \cdot x^{n-2} y^2 + \dots$ The pattern is systematic and reliable!
18. In the expansion of $(x+y)^6$, what is the coefficient of $x^3 y^3$? (Pascal's Triangle row 6: $1,6,15,20,15,6,1$.)
    Answer: 20 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: $1 \cdot x^4 + 4 \cdot x^3 y + 6 \cdot x^2 y^2 + 4 \cdot x y^3 + 1 \cdot y^4$. Each term has exponents summing to 4, and coefficients from the triangle make this work perfectly! In $(x+y)^6$ with row 6 (1,6,15,20,15,6,1), the term $x^3 y^3$ uses the fourth coefficient, 20, as choice C states. Choice A (15) is for adjacent terms like $x^4 y^2$ or $x^2 y^4$, but for equal exponents in even $n$, it's the middle peak. 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]·$x^{n-2} y^2$ + ... The pattern is systematic and reliable!
19. Use Pascal's Triangle (row 3: $1,3,3,1$) to expand $(a+b)^3.$
    Answer: $a^3+3a^2b+3ab^2+b^3$ 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!
20. Expand $(x-2)^4$ using the Binomial Theorem (Pascal's Triangle row 4: $1,4,6,4,1$).​
    Answer: $x^4-8x^3+24x^2-32x+16$ 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·$x^4$ + 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) = -8, 6·4 = 24, 4·(-8) = -32, and $(-2)^4$ = 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), plus for even powers—this pattern is systematic and predictable!

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