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Algebra 2 Flashcards: Extending Polynomial Identities To Complex Numbers

Study Extending Polynomial Identities To Complex Numbers 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 Extending Polynomial Identities To Complex Numbers, 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: Extending Polynomial Identities To Complex Numbers

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QUESTION

Rewrite x2+36x^2+36x2+36 as a product of two complex conjugate binomials.

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ANSWER

(x+6i)(x−6i)(x+6i)(x-6i)(x+6i)(x−6i). x2+36=x2+(6i)2x^2 + 36 = x^2 + (6i)^2x2+36=x2+(6i)2 factors as conjugate pair.

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Flashcard 1: Rewrite x2+36x^2+36x2+36 as a product of two complex conjugate binomials.

Answer: (x+6i)(x−6i)(x+6i)(x-6i)(x+6i)(x−6i). x2+36=x2+(6i)2x^2 + 36 = x^2 + (6i)^2x2+36=x2+(6i)2 factors as conjugate pair.

Flashcard 2: What is i27i^{27}i27?

Answer: i27=−ii^{27}=-ii27=−i. i27=i24⋅i3=1⋅(−i)=−ii^{27} = i^{24} \cdot i^3 = 1 \cdot (-i) = -ii27=i24⋅i3=1⋅(−i)=−i

Flashcard 3: Factor x2+6x+13x^2+6x+13x2+6x+13 over the complex numbers.

Answer: (x+3+2i)(x+3−2i)(x+3+2i)(x+3-2i)(x+3+2i)(x+3−2i). Complete the square: (x+3)2−9+13=(x+3)2+4(x+3)^2 - 9 + 13 = (x+3)^2 + 4(x+3)2−9+13=(x+3)2+4.

Flashcard 4: What is i10i^{10}i10?

Answer: i10=−1i^{10}=-1i10=−1. i10=i8⋅i2=1⋅(−1)=−1i^{10} = i^{8} \cdot i^2 = 1 \cdot (-1) = -1i10=i8⋅i2=1⋅(−1)=−1

Flashcard 5: What are the complex roots of x2+2x+5=0x^2+2x+5=0x2+2x+5=0?

Answer: x=−1+2i,−1−2ix=-1+2i,-1-2ix=−1+2i,−1−2i. Use quadratic formula with discriminant 4−20=−164 - 20 = -164−20=−16.

Flashcard 6: Factor x2+2x+5x^2+2x+5x2+2x+5 over the complex numbers.

Answer: (x+1+2i)(x+1−2i)(x+1+2i)(x+1-2i)(x+1+2i)(x+1−2i). Complete the square: (x+1)2−1+5=(x+1)2+4(x+1)^2 - 1 + 5 = (x+1)^2 + 4(x+1)2−1+5=(x+1)2+4.

Flashcard 7: What are the complex roots of x2−6x+13=0x^2-6x+13=0x2−6x+13=0?

Answer: x=3+2i,3−2ix=3+2i,3-2ix=3+2i,3−2i. Use quadratic formula with discriminant 36−52=−1636 - 52 = -1636−52=−16.

Flashcard 8: Rewrite x2+25x^2+25x2+25 as a product of two complex conjugate binomials.

Answer: (x+5i)(x−5i)(x+5i)(x-5i)(x+5i)(x−5i). x2+25=x2+(5i)2x^2 + 25 = x^2 + (5i)^2x2+25=x2+(5i)2 factors as conjugate pair.

Flashcard 9: Rewrite x2+49x^2+49x2+49 as a product of two complex conjugate binomials.

Answer: (x+7i)(x−7i)(x+7i)(x-7i)(x+7i)(x−7i). x2+49=x2+(7i)2x^2 + 49 = x^2 + (7i)^2x2+49=x2+(7i)2 factors as conjugate pair.

Flashcard 10: Rewrite x2+64x^2+64x2+64 as a product of two complex conjugate binomials.

Answer: (x+8i)(x−8i)(x+8i)(x-8i)(x+8i)(x−8i). x2+64=x2+(8i)2x^2 + 64 = x^2 + (8i)^2x2+64=x2+(8i)2 factors as conjugate pair.

Flashcard 11: Rewrite x2+100x^2+100x2+100 as a product of two complex conjugate binomials.

Answer: (x+10i)(x−10i)(x+10i)(x-10i)(x+10i)(x−10i). x2+100=x2+(10i)2x^2 + 100 = x^2 + (10i)^2x2+100=x2+(10i)2 factors as conjugate pair.

Flashcard 12: Rewrite x2+2x^2+2x2+2 as a product of two complex conjugate binomials.

Answer: (x+i2)(x−i2)(x+i\sqrt{2})(x-i\sqrt{2})(x+i2​)(x−i2​). x2+2=x2+(i2)2x^2 + 2 = x^2 + (i\sqrt{2})^2x2+2=x2+(i2​)2 factors as conjugate pair.

Flashcard 13: Rewrite x2+7x^2+7x2+7 as a product of two complex conjugate binomials.

Answer: (x+i7)(x−i7)(x+i\sqrt{7})(x-i\sqrt{7})(x+i7​)(x−i7​). x2+7=x2+(i7)2x^2 + 7 = x^2 + (i\sqrt{7})^2x2+7=x2+(i7​)2 factors as conjugate pair.

Flashcard 14: Rewrite x2+12x^2+12x2+12 as a product of two complex conjugate binomials.

Answer: (x+2i3)(x−2i3)(x+2i\sqrt{3})(x-2i\sqrt{3})(x+2i3​)(x−2i3​). 12=4⋅312 = 4 \cdot 312=4⋅3, so 12=23\sqrt{12} = 2\sqrt{3}12​=23​.

Flashcard 15: Factor x2+4x+8x^2+4x+8x2+4x+8 over the complex numbers.

Answer: (x+2+2i)(x+2−2i)(x+2+2i)(x+2-2i)(x+2+2i)(x+2−2i). Complete the square: (x+2)2−4+8=(x+2)2+4(x+2)^2 - 4 + 8 = (x+2)^2 + 4(x+2)2−4+8=(x+2)2+4.

Flashcard 16: Factor x2+10x+29x^2+10x+29x2+10x+29 over the complex numbers.

Answer: (x+5+2i)(x+5−2i)(x+5+2i)(x+5-2i)(x+5+2i)(x+5−2i). Complete the square: (x+5)2−25+29=(x+5)2+4(x+5)^2 - 25 + 29 = (x+5)^2 + 4(x+5)2−25+29=(x+5)2+4.

Flashcard 17: What are the complex roots of x2+4=0x^2+4=0x2+4=0?

Answer: x=2i,−2ix=2i,-2ix=2i,−2i. Solve x2=−4x^2 = -4x2=−4 to get x=±2ix = \pm 2ix=±2i.

Flashcard 18: What are the complex roots of x2−2x+2=0x^2-2x+2=0x2−2x+2=0?

Answer: x=1+i,1−ix=1+i,1-ix=1+i,1−i. Use quadratic formula with discriminant 4−8=−44 - 8 = -44−8=−4.

Flashcard 19: State the Complex Conjugate Root Theorem for polynomials with real coefficients.

Answer: If a+bia+bia+bi is a root, then a−bia-bia−bi is a root. Complex roots of real polynomials come in conjugate pairs.

Flashcard 20: What is the product (a+bi)(a−bi)(a+bi)(a-bi)(a+bi)(a−bi) equal to, simplified?

Answer: a2+b2a^2+b^2a2+b2. Multiplying complex conjugates eliminates the imaginary terms.

Flashcard 21: Identify the factorization of x2−2ax+(a2+b2)x^2-2ax+(a^2+b^2)x2−2ax+(a2+b2) over complex numbers.

Answer: (x−(a+bi))(x−(a−bi))(x-(a+bi))(x-(a-bi))(x−(a+bi))(x−(a−bi)). Standard form with roots a±bia \pm bia±bi.

Flashcard 22: What is the factored form of x2+2ax+(a2+b2)x^2+2ax+(a^2+b^2)x2+2ax+(a2+b2) over complex numbers?

Answer: (x−(−a+bi))(x−(−a−bi))(x-(-a+bi))(x-(-a-bi))(x−(−a+bi))(x−(−a−bi)). Standard form with roots −a±bi-a \pm bi−a±bi.

Flashcard 23: Rewrite x2+3x^2+3x2+3 as a product of two complex conjugate binomials.

Answer: (x+i3)(x−i3)(x+i\sqrt{3})(x-i\sqrt{3})(x+i3​)(x−i3​). x2+3=x2+(i3)2x^2 + 3 = x^2 + (i\sqrt{3})^2x2+3=x2+(i3​)2 factors as conjugate pair.

Flashcard 24: State the identity for factoring a sum of squares using iii.

Answer: a2+b2=(a+bi)(a−bi)a^2+b^2=(a+bi)(a-bi)a2+b2=(a+bi)(a−bi). Uses complex conjugates to factor sum of squares.

Flashcard 25: What is i3i^3i3?

Answer: i3=−ii^3=-ii3=−i. i3=i2⋅i=−1⋅i=−ii^3 = i^2 \cdot i = -1 \cdot i = -ii3=i2⋅i=−1⋅i=−i

Flashcard 26: Rewrite x2+4x^2+4x2+4 as a product of two complex conjugate binomials.

Answer: (x+2i)(x−2i)(x+2i)(x-2i)(x+2i)(x−2i). x2+4=x2+(2i)2x^2 + 4 = x^2 + (2i)^2x2+4=x2+(2i)2 factors as conjugate pair.

Flashcard 27: Rewrite x2+9x^2+9x2+9 as a product of two complex conjugate binomials.

Answer: (x+3i)(x−3i)(x+3i)(x-3i)(x+3i)(x−3i). x2+9=x2+(3i)2x^2 + 9 = x^2 + (3i)^2x2+9=x2+(3i)2 factors as conjugate pair.

Flashcard 28: Rewrite x2+1x^2+1x2+1 as a product of two complex conjugate binomials.

Answer: (x+i)(x−i)(x+i)(x-i)(x+i)(x−i). x2+1=x2+i2x^2 + 1 = x^2 + i^2x2+1=x2+i2 factors as conjugate pair.

Flashcard 29: Rewrite x2+16x^2+16x2+16 as a product of two complex conjugate binomials.

Answer: (x+4i)(x−4i)(x+4i)(x-4i)(x+4i)(x−4i). x2+16=x2+(4i)2x^2 + 16 = x^2 + (4i)^2x2+16=x2+(4i)2 factors as conjugate pair.

Flashcard 30: Factor x2−4x+5x^2-4x+5x2−4x+5 over the complex numbers.

Answer: (x−2+i)(x−2−i)(x-2+i)(x-2-i)(x−2+i)(x−2−i). Complete the square: (x−2)2−4+5=(x−2)2+1(x-2)^2 - 4 + 5 = (x-2)^2 + 1(x−2)2−4+5=(x−2)2+1.