Polynomial Identities (Advanced & Applications)

1) (x + y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴.

True

False

Binomial coefficients 1-4-6-4-1 come from Pascal’s triangle for n = 4. Each term follows x^(4−k) y^k with the corresponding coefficient. This ensures all mixed products appear with correct multiplicity.

Explanation

Binomial coefficients 1-4-6-4-1 come from Pascal’s triangle for n = 4. Each term follows x^(4−k) y^k with the corresponding coefficient. This ensures all mixed products appear with correct multiplicity.

2) If a + b = 5 and ab = 6, find a² + b².

7

11

13

19

Using (a+b)² = a²+2ab+b²: 25 = a²+b²+12. Therefore a²+b² = 25-12 = 13. This uses the identity to find the sum of squares without needing to solve for a and b individually. Option A gives 7, option B gives 11, option D gives 19.

Explanation

Using (a+b)² = a²+2ab+b²: 25 = a²+b²+12. Therefore a²+b² = 25-12 = 13. This uses the identity to find the sum of squares without needing to solve for a and b individually. Option A gives 7, option B gives 11, option D gives 19.

3) What does (a + b) cubed + (a - b) cubed equal?

2a³ + 2b³

6a²b + 2b³

2a³ + 6ab²

2a³ - 6ab²

(a+b)³ = a³+3a²b+3ab²+b³ and (a-b)³ = a³-3a²b+3ab²-b³. Adding: 3a²b cancels, b³ cancels, a³ doubles, 3ab² doubles. Result = 2a³+6ab².

Explanation

(a+b)³ = a³+3a²b+3ab²+b³ and (a-b)³ = a³-3a²b+3ab²-b³. Adding: 3a²b cancels, b³ cancels, a³ doubles, 3ab² doubles. Result = 2a³+6ab².

4) What does (a + b) squared + (a - b) squared equal?

2ab

2(a² - b²)

4ab

2(a² + b²)

(a+b)² = a²+2ab+b² and (a-b)² = a²-2ab+b². Adding: the 2ab and -2ab cancel, leaving 2a²+2b² = 2(a²+b²). The cross terms cancel completely, leaving only the squared terms doubled.

Explanation

(a+b)² = a²+2ab+b² and (a-b)² = a²-2ab+b². Adding: the 2ab and -2ab cancel, leaving 2a²+2b² = 2(a²+b²). The cross terms cancel completely, leaving only the squared terms doubled.

5) What is (x - y) to the fourth power?

X⁴ + 4x³y + 6x²y² + 4xy³ + y⁴

X⁴ - 4x³y + 6x²y² - 4xy³ + y⁴

X⁴ - 4x³y - 6x²y² + 4xy³ + y⁴

X⁴ - 4x³y + 6x²y² + 4xy³ - y⁴

Pascal coefficients for n=4 are 1,4,6,4,1 with alternating signs for (x-y)⁴: x⁴-4x³y+6x²y²-4xy³+y⁴. Signs alternate starting positive because (-y) to even powers is positive and to odd powers is negative. Option A has all positive signs. Options C and D have wrong sign patterns.

Explanation

Pascal coefficients for n=4 are 1,4,6,4,1 with alternating signs for (x-y)⁴: x⁴-4x³y+6x²y²-4xy³+y⁴. Signs alternate starting positive because (-y) to even powers is positive and to odd powers is negative. Option A has all positive signs. Options C and D have wrong sign patterns.

6) What does (a + b) cubed - (a - b) cubed equal?

2a³ + 2b³

6a²b - 2b³

2a³ - 6ab²

6a²b + 2b³

(a+b)³ = a³+3a²b+3ab²+b³ and (a-b)³ = a³-3a²b+3ab²-b³. Subtracting: a³ cancels, 3a²b doubles to 6a²b, 3ab² cancels, b³ doubles to 2b³. Result = 6a²b+2b³.

Explanation

(a+b)³ = a³+3a²b+3ab²+b³ and (a-b)³ = a³-3a²b+3ab²-b³. Subtracting: a³ cancels, 3a²b doubles to 6a²b, 3ab² cancels, b³ doubles to 2b³. Result = 6a²b+2b³.

7) What does (a + b + c) squared equal?

A² + b² + c² + 2(ab + bc + ca)

A² + b² + c²

A² + b² + c² - 2(ab + bc + ca)

2a² + 2b² + 2c²

Squaring creates square terms a², b², c² and cross terms ab, bc, ca each appearing twice from two multiplication paths. The complete expansion is a²+b²+c²+2(ab+bc+ca). This holds for all real values of a, b, and c without exception.

Explanation

Squaring creates square terms a², b², c² and cross terms ab, bc, ca each appearing twice from two multiplication paths. The complete expansion is a²+b²+c²+2(ab+bc+ca). This holds for all real values of a, b, and c without exception.

8) (a + b)³ − (a − b)³ = 6a²b + 2b³.

True

False

Expanding both cubes shows odd-powered b terms double, while even-powered ones cancel. This produces a simplified expression involving only terms with a²b and b³. Such patterns frequently appear with conjugate expressions.

Explanation

Expanding both cubes shows odd-powered b terms double, while even-powered ones cancel. This produces a simplified expression involving only terms with a²b and b³. Such patterns frequently appear with conjugate expressions.

9) (x + 1)³ − (x − 1)³ = 6x² + 2.

True

False

Apply a³ − b³ = (a − b)(a² + ab + b²) with a = x + 1 and b = x − 1. Evaluate a − b = 2 and a² + ab + b² = 3x² + 1. Multiplying yields 6x² + 2, showing how conjugates simplify cleanly.

Explanation

Apply a³ − b³ = (a − b)(a² + ab + b²) with a = x + 1 and b = x − 1. Evaluate a − b = 2 and a² + ab + b² = 3x² + 1. Multiplying yields 6x² + 2, showing how conjugates simplify cleanly.

10) (a − b)(a² + ab + b²) = a³ − b³.

True

False

Multiplying the binomial and trinomial causes the middle terms to cancel. The result a³ − b³ matches the difference-of-cubes identity exactly. This factorization is essential in algebraic simplification.

Explanation

Multiplying the binomial and trinomial causes the middle terms to cancel. The result a³ − b³ matches the difference-of-cubes identity exactly. This factorization is essential in algebraic simplification.

11) Expand (2x + 3y)³.

8x³ + 12x²y + 6xy² + 27y³

8x³ + 36x²y + 54xy² + 27y³

8x³ + 18x²y + 27y³

8x³ + 24x²y + 36xy² + 27y³

Using the identity (a + b)³ = a³ + 3a²b + 3ab² + b³ makes expansion straightforward.

Substituting a = 2x and b = 3y shows how each term becomes a combination of powers of x and y.

This produces 8x³ + 36x²y + 54xy² + 27y³ by following the binomial pattern.

Explanation

Using the identity (a + b)³ = a³ + 3a²b + 3ab² + b³ makes expansion straightforward.

Substituting a = 2x and b = 3y shows how each term becomes a combination of powers of x and y.

This produces 8x³ + 36x²y + 54xy² + 27y³ by following the binomial pattern.

12) (x + y)³ + (x − y)³ = 2x³ + 6xy².

True

False

Expanding each cube reveals that +3x²y and −3x²y cancel. The terms +3xy² from both expansions add together, doubling their coefficient. This creates 2x³ + 6xy², showing symmetry in conjugate binomials

Explanation

Expanding each cube reveals that +3x²y and −3x²y cancel. The terms +3xy² from both expansions add together, doubling their coefficient. This creates 2x³ + 6xy², showing symmetry in conjugate binomials

13) (a + b + c)³ = a³ + b³ + c³ + 3(a + b)(b + c)(c + a).

True

False

The proposed formula is incorrect. The true identity is a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca). Comparing expansions shows the given expression does not match the real cubic identity.

Explanation

The proposed formula is incorrect. The true identity is a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca). Comparing expansions shows the given expression does not match the real cubic identity.

14) If a³ + b³ = 343 and a + b = 7, find ab.

0

12

14

16343 = 7³ − 3ab(7) ⇒ 343 = 343 − 21ab ⇒ ab = 14.

Use the identity: a³ + b³ = (a + b)³ − 3ab(a + b).

Substituting values gives 343 = 343 − 21ab, so ab = 14.

This demonstrates how cube sums relate directly to the product ab. Substituting the known values creates a simple equation involving ababab. Solving for ababab shows how cube sums relate directly to the product of variables.

Explanation

Use the identity: a³ + b³ = (a + b)³ − 3ab(a + b).

Substituting values gives 343 = 343 − 21ab, so ab = 14.

This demonstrates how cube sums relate directly to the product ab. Substituting the known values creates a simple equation involving ababab. Solving for ababab shows how cube sums relate directly to the product of variables.

15) Simplify (x² − y²)².

X⁴ − 2x²y² + y⁴

X⁴ + y⁴

X⁴ + 2x²y² + y⁴

X² − y²

Treat the expression as a binomial square (a − b)² with a = x² and b = y².

Apply a² − 2ab + b² to obtain x⁴ − 2x²y² + y⁴.

The structure reflects a symmetric fourth-degree polynomial. This yields a fourth-degree polynomial showing symmetry between xxx and yyy.

Explanation

Treat the expression as a binomial square (a − b)² with a = x² and b = y².

Apply a² − 2ab + b² to obtain x⁴ − 2x²y² + y⁴.

The structure reflects a symmetric fourth-degree polynomial. This yields a fourth-degree polynomial showing symmetry between xxx and yyy.

16) Expand (x + y + z)².

X² + y² + z² + 2(xy + yz + zx)

X² + y² + z² − 2(xy + yz + zx)

X² + y² + z²

2(x² + y² + z²)

When squaring a trinomial, each variable multiplies with every other, producing both square terms and pairwise products. This happens because xy, yz, and zx all appear in two multiplication paths. Thus, the full expansion is x² + y² + z² + 2(xy + yz + zx). This expansion reflects the structure of multiplying identical trinomials.

Explanation

When squaring a trinomial, each variable multiplies with every other, producing both square terms and pairwise products. This happens because xy, yz, and zx all appear in two multiplication paths. Thus, the full expansion is x² + y² + z² + 2(xy + yz + zx). This expansion reflects the structure of multiplying identical trinomials.

17) Verify the identity (a + b + c)² = a² + b² + c² + 2(ab + bc + ca).

False

True

Sometimes true

True only if a + b + c = 0

Expanding (a + b + c)(a + b + c) shows every pair ab, bc, ca appears twice.

This produces the required 2(ab + bc + ca), confirming the identity.

The formula holds for all real values of a, b, and c.

Explanation

Expanding (a + b + c)(a + b + c) shows every pair ab, bc, ca appears twice.

This produces the required 2(ab + bc + ca), confirming the identity.

The formula holds for all real values of a, b, and c.

18) Evaluate (x + 2)³ − (x − 2)³.

12x² + 16

8x² + 12

16x² + 12

Use a³ − b³ = (a − b)(a² + ab + b²) with a = x + 2 and b = x − 2.

Compute a − b = 4, then evaluate a² + ab + b², which simplifies to 3x² + 4.

Multiplying gives 12x² + 16, showing how symmetric cube terms cancel neatly.

Explanation

Use a³ − b³ = (a − b)(a² + ab + b²) with a = x + 2 and b = x − 2.

Compute a − b = 4, then evaluate a² + ab + b², which simplifies to 3x² + 4.

Multiplying gives 12x² + 16, showing how symmetric cube terms cancel neatly.

19) Simplify (x² + 3)².

X⁴ + 9

X⁴ + 3x² + 6

X⁴ + 6x² + 9

X⁴ − 6x² + 9

Use the square identity (a + b)² = a² + 2ab + b². Here, a = x² and b = 3, so after squaring each term and multiplying the middle product, you obtain x⁴ + 6x² + 9. This method ensures all cross-terms appear correctly.

Explanation

Use the square identity (a + b)² = a² + 2ab + b². Here, a = x² and b = 3, so after squaring each term and multiplying the middle product, you obtain x⁴ + 6x² + 9. This method ensures all cross-terms appear correctly.

20) Expand (x − 2y)³.

X³ − 6x²y + 12xy² − 8y³

X³ − 8y³

X³ + 6x²y − 12xy² + 8y³

X³ − 3x²y + 3xy² − y³

The formula (a − b)³ = a³ − 3a²b + 3ab² − b³ applies here with a = x and b = 2y. Each substituted term preserves the alternating signs required by the identity. This creates x³ − 6x²y + 12xy² − 8y³.

Explanation

The formula (a − b)³ = a³ − 3a²b + 3ab² − b³ applies here with a = x and b = 2y. Each substituted term preserves the alternating signs required by the identity. This creates x³ − 6x²y + 12xy² − 8y³.