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Algebra 2 Flashcards: Applying The Binomial Theorem

Study Applying The Binomial Theorem in Algebra 2 with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Applying The Binomial Theorem, giving you a quick way to review the definitions, rules, and examples that matter most for Algebra 2.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Algebra 2 Flashcards: Applying The Binomial Theorem

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QUESTION

Find the value of (94)\binom{9}{4}(49​).

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ANSWER

126126126. Using (94)=9!4!5!=126\binom{9}{4}=\frac{9!}{4!5!}=126(49​)=4!5!9!​=126.

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Flashcard 1: Find the value of (94)\binom{9}{4}(49​).

Answer: 126126126. Using (94)=9!4!5!=126\binom{9}{4}=\frac{9!}{4!5!}=126(49​)=4!5!9!​=126.

Flashcard 2: Expand (x+y)3(x+y)^3(x+y)3.

Answer: x3+3x2y+3xy2+y3x^3+3x^2y+3xy^2+y^3x3+3x2y+3xy2+y3. Using coefficients 1,3,3,11,3,3,11,3,3,1 from Pascal's triangle.

Flashcard 3: Expand (x−y)3(x-y)^3(x−y)3.

Answer: x3−3x2y+3xy2−y3x^3-3x^2y+3xy^2-y^3x3−3x2y+3xy2−y3. Alternating signs from (−y)k(-y)^k(−y)k with coefficients 1,3,3,11,3,3,11,3,3,1.

Flashcard 4: What is the coefficient of x5y3x^5y^3x5y3 in (x−y)8(x-y)^8(x−y)8?

Answer: −(83)=−56-\binom{8}{3}=-56−(38​)=−56. Coefficient is (83)=56\binom{8}{3}=56(38​)=56 with negative sign from (−y)3(-y)^3(−y)3.

Flashcard 5: Find the value of (101)\binom{10}{1}(110​).

Answer: 101010. Using (101)=10\binom{10}{1}=10(110​)=10.

Flashcard 6: What is the factorial definition of the binomial coefficient (nk)\binom{n}{k}(kn​)?

Answer: (nk)=n!k!(n−k)!\binom{n}{k}=\frac{n!}{k!(n-k)!}(kn​)=k!(n−k)!n!​. Uses factorial formula to calculate combinations.

Flashcard 7: State the Binomial Theorem formula for expanding (x+y)n(x+y)^n(x+y)n using binomial coefficients.

Answer: (x+y)n=∑k=0n(nk)xn−kyk(x+y)^n=\sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^k(x+y)n=∑k=0n​(kn​)xn−kyk. The general formula for binomial expansion with coefficients and powers.

Flashcard 8: Expand (x−y)3(x-y)^3(x−y)3.

Answer: x3−3x2y+3xy2−y3x^3-3x^2y+3xy^2-y^3x3−3x2y+3xy2−y3. Alternating signs from (−y)k(-y)^k(−y)k with coefficients 1,3,3,11,3,3,11,3,3,1.

Flashcard 9: Expand (x−y)4(x-y)^4(x−y)4.

Answer: x4−4x3y+6x2y2−4xy3+y4x^4-4x^3y+6x^2y^2-4xy^3+y^4x4−4x3y+6x2y2−4xy3+y4. Alternating signs from (−y)k(-y)^k(−y)k with coefficients 1,4,6,4,11,4,6,4,11,4,6,4,1.

Flashcard 10: What is the coefficient of x7y2x^7y^2x7y2 in (x+y)9(x+y)^9(x+y)9?

Answer: (92)=36\binom{9}{2}=36(29​)=36. The coefficient of x7y2x^7y^2x7y2 is (92)=36\binom{9}{2}=36(29​)=36.

Flashcard 11: What is the coefficient of x3y5x^3y^5x3y5 in (x+y)8(x+y)^8(x+y)8?

Answer: (85)=56\binom{8}{5}=56(58​)=56. The coefficient of x3y5x^3y^5x3y5 is (85)=56\binom{8}{5}=56(58​)=56.

Flashcard 12: What is the coefficient of x6y4x^6y^4x6y4 in (x+y)10(x+y)^{10}(x+y)10?

Answer: (104)=210\binom{10}{4}=210(410​)=210. The coefficient of x6y4x^6y^4x6y4 is (104)=210\binom{10}{4}=210(410​)=210.

Flashcard 13: What is the coefficient of x5y3x^5y^3x5y3 in (x−y)8(x-y)^8(x−y)8?

Answer: −(83)=−56-\binom{8}{3}=-56−(38​)=−56. Coefficient is (83)=56\binom{8}{3}=56(38​)=56 with negative sign from (−y)3(-y)^3(−y)3.

Flashcard 14: What is the coefficient of x4y4x^4y^4x4y4 in (x−y)8(x-y)^8(x−y)8?

Answer: (84)=70\binom{8}{4}=70(48​)=70. Coefficient is (84)=70\binom{8}{4}=70(48​)=70 with positive sign from (−y)4(-y)^4(−y)4.

Flashcard 15: Identify the coefficient of the x2y3x^2y^3x2y3 term in (x+y)5(x+y)^5(x+y)5.

Answer: (53)=10\binom{5}{3}=10(35​)=10. The coefficient of x2y3x^2y^3x2y3 is (53)=10\binom{5}{3}=10(35​)=10.

Flashcard 16: What is the coefficient of a4b2a^4b^2a4b2 in (a+b)6(a+b)^6(a+b)6?

Answer: (62)=15\binom{6}{2}=15(26​)=15. The coefficient of a4b2a^4b^2a4b2 is (62)=15\binom{6}{2}=15(26​)=15.

Flashcard 17: What is the coefficient of a2b4a^2b^4a2b4 in (a+b)6(a+b)^6(a+b)6?

Answer: (64)=15\binom{6}{4}=15(46​)=15. The coefficient of a2b4a^2b^4a2b4 is (64)=15\binom{6}{4}=15(46​)=15.

Flashcard 18: What is the coefficient pattern (Pascal row) for expanding (x+y)3(x+y)^3(x+y)3?

Answer: 1,3,3,11,3,3,11,3,3,1. Row 3 of Pascal's triangle.

Flashcard 19: Find the value of (82)\binom{8}{2}(28​).

Answer: 282828. Using (82)=8!2!6!=8⋅72=28\binom{8}{2}=\frac{8!}{2!6!}=\frac{8 \cdot 7}{2}=28(28​)=2!6!8!​=28⋅7​=28.

Flashcard 20: Find the value of (86)\binom{8}{6}(68​).

Answer: 282828. Using symmetry: (86)=(82)=28\binom{8}{6}=\binom{8}{2}=28(68​)=(28​)=28.

Flashcard 21: Find the value of (101)\binom{10}{1}(110​).

Answer: 101010. Using (101)=10\binom{10}{1}=10(110​)=10.

Flashcard 22: Find the value of (102)\binom{10}{2}(210​).

Answer: 454545. Using (102)=10⋅92=45\binom{10}{2}=\frac{10 \cdot 9}{2}=45(210​)=210⋅9​=45.

Flashcard 23: Find the value of (103)\binom{10}{3}(310​).

Answer: 120120120. Using (103)=10⋅9⋅86=120\binom{10}{3}=\frac{10 \cdot 9 \cdot 8}{6}=120(310​)=610⋅9⋅8​=120.

Flashcard 24: What is the coefficient sum of (x−y)n(x-y)^n(x−y)n (equivalently, evaluate at x=1,y=1x=1,y=1x=1,y=1)?

Answer: 000 if nnn is odd; 2n2^n2n if nnn is even. Substitute x=1x=1x=1 and y=−1y=-1y=−1 into (x−y)n(x-y)^n(x−y)n.

Flashcard 25: In (x+y)n(x+y)^n(x+y)n, what is the general term (the kkkth term) written in powers of xxx and yyy?

Answer: (nk)xn−kyk\binom{n}{k}x^{n-k}y^k(kn​)xn−kyk. The (k+1)(k+1)(k+1)th term in the binomial expansion.

Flashcard 26: Find the value of (94)\binom{9}{4}(49​).

Answer: 126126126. Using (94)=9!4!5!=126\binom{9}{4}=\frac{9!}{4!5!}=126(49​)=4!5!9!​=126.

Flashcard 27: Expand (x+y)2(x+y)^2(x+y)2.

Answer: x2+2xy+y2x^2+2xy+y^2x2+2xy+y2. Using coefficients 1,2,11,2,11,2,1 from Pascal's triangle.

Flashcard 28: Expand (x+y)3(x+y)^3(x+y)3.

Answer: x3+3x2y+3xy2+y3x^3+3x^2y+3xy^2+y^3x3+3x2y+3xy2+y3. Using coefficients 1,3,3,11,3,3,11,3,3,1 from Pascal's triangle.

Flashcard 29: Expand (x+y)4(x+y)^4(x+y)4.

Answer: x4+4x3y+6x2y2+4xy3+y4x^4+4x^3y+6x^2y^2+4xy^3+y^4x4+4x3y+6x2y2+4xy3+y4. Using coefficients 1,4,6,4,11,4,6,4,11,4,6,4,1 from Pascal's triangle.

Flashcard 30: In (x+y)n(x+y)^n(x+y)n, what is the exponent of xxx in the term containing yky^kyk?

Answer: n−kn-kn−k. Exponents of xxx and yyy must sum to nnn.