3.2 Factoring Cubes, Polynomial Identities, and Binomial Expansions

3.2 Polynomial Identities and Binomial Expansions

[!NOTE] What is tested here, and what is not. Only the first part of this section is on the Next Generation Algebra II blueprint. The sum and difference of cubes sit inside AII-A.SSE.2, whose NYSED instructional note names "factoring the sum and difference of cubes" outright, and both formulas are printed on the reference sheet.

Polynomial identities and the Binomial Theorem are enrichment. NYSED’s Algebra II snapshot records that A-APR.4 (prove polynomial identities) moved out of Algebra II and became a plus standard, and A-APR.5 (Binomial Theorem) is not listed anywhere in the Algebra II blueprint. Expect no operational item that asks you to prove an identity or extract a term of a binomial expansion. Study the rest of this section for structural fluency and for precalculus readiness - not as exam preparation.

In elementary algebra, equations are typically conditional—they are true only for specific values of the variable (e.g., $2x + 4 = 10$ is true exclusively when $x = 3$). A polynomial identity, by contrast, is an algebraic statement that holds true for all possible values of the variables involved. Identities provide powerful shortcuts for factoring, expanding, and discovering relationships in number theory.

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Proving Polynomial Identities Algebraically

Identity proofs are not an Algebra II Regents item type under the Next Generation standards, but the discipline they teach transfers directly to the trigonometric-identity work in Section 9.3, which is tested under AII-F.TF.8. When you prove an identity, adhere to these conventions:

1. Work with one side at a time: Choose either the Left-Hand Side (LHS) or the Right-Hand Side (RHS)—typically the more complex side—and apply algebraic operations (distributing, expanding, combining like terms) until it is identical to the other side.
2. Alternative Method (Dual Simplification): Expand the LHS and RHS separately in parallel columns until both reduce to the exact same polynomial expression.
3. Never treat an identity like an equation to be solved: Do not move terms across the equals sign, do not add/subtract from both sides, and do not cross-multiply. Treating an unproven identity as an equation assumes the truth of the statement you are required to establish. This is exactly the rule NYSED applies to trigonometric identity verification in Section 9.3.

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The Pythagorean Triple Identity

A classic application of polynomial identities is the algebraic generation of Pythagorean triples—sets of three positive integers $(a, b, c)$ satisfying $a^2 + b^2 = c^2$.

$\text{Identity: } (x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2$

Rigorous Algebraic Proof

To prove this identity, expand both sides algebraically:

• Expand Left-Hand Side (LHS): $\text{LHS} = (x^2 + y^2)^2 = (x^2)^2 + 2(x^2)(y^2) + (y^2)^2 = x^4 + 2x^2y^2 + y^4$
• Expand Right-Hand Side (RHS): $\text{RHS} = (x^2 - y^2)^2 + (2xy)^2$ $\text{RHS} = (x^4 - 2x^2y^2 + y^4) + 4x^2y^2$ $\text{RHS} = x^4 + (-2x^2y^2 + 4x^2y^2) + y^4 = x^4 + 2x^2y^2 + y^4$
• Conclusion: $\text{Since LHS} = \text{RHS} = x^4 + 2x^2y^2 + y^4, \text{ the identity is proven. } \blacksquare$

Generating Integer Right Triangles

By assigning integer values to $x$ and $y$ such that $x > y > 0$, this identity produces the side lengths of a right triangle:

$a = x^2 - y^2, \quad b = 2xy, \quad c = x^2 + y^2$

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Sum and Difference of Two Cubes

While a sum of squares cannot be factored over the real numbers, both the sum and difference of two cubes can always be factored into a product of a linear binomial and a quadratic trinomial.

The Formulas

$\text{Sum of Cubes: } a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ $\text{Difference of Cubes: } a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

The SOAP Mnemonic

To remember the signs of the factors, use the acronym SOAP:

• S — Same: The binomial factor takes the same sign as the original expression ($+$ for sum, $-$ for difference).
• O — Opposite: The middle term of the trinomial takes the opposite sign ($-ab$ for sum, $+ab$ for difference).
• AP — Always Positive: The last term of the trinomial ($+b^2$) is always positive.

[!IMPORTANT] The quadratic trinomial factor $(a^2 - ab + b^2)$ or $(a^2 + ab + b^2)$ is irreducible over the real numbers. Its discriminant is always negative: $D = (-b)^2 - 4(1)(b^2) = -3b^2 < 0$. It will never factor further into real binomials!

Worked Example 1: Factoring $8x^3 - 27$

1. Identify the cube roots: $a = \sqrt[3]{8x^3} = 2x$ and $b = \sqrt[3]{27} = 3$.
2. Apply the difference of cubes template $(a - b)(a^2 + ab + b^2)$:
   • Binomial factor: $(2x - 3)$
   • First trinomial term: $a^2 = (2x)^2 = 4x^2$
   • Middle trinomial term: $ab = (2x)(3) = 6x$
   • Last trinomial term: $b^2 = 3^2 = 9$
3. Combine terms: $8x^3 - 27 = (2x - 3)(4x^2 + 6x + 9)$

Worked Example 2: Factoring $27x^3 + 64$

1. Identify the cube roots: $a = \sqrt[3]{27x^3} = 3x$ and $b = \sqrt[3]{64} = 4$.
2. Apply the sum of cubes template $(a + b)(a^2 - ab + b^2)$:
   • Binomial factor: $(3x + 4)$
   • First trinomial term: $a^2 = (3x)^2 = 9x^2$
   • Middle trinomial term: $-ab = -(3x)(4) = -12x$
   • Last trinomial term: $b^2 = 4^2 = 16$
3. Combine terms: $27x^3 + 64 = (3x + 4)(9x^2 - 12x + 16)$

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The Binomial Theorem and Pascal's Triangle

Expanding $(a + b)^n$ by repeated polynomial multiplication is tedious and prone to algebraic sign errors. The Binomial Theorem provides a systematic method using combinatorial coefficients.

Pascal's Triangle

Each number in Pascal's Triangle is the sum of the two numbers immediately above it. Row $n$ provides the expansion coefficients for a binomial raised to power $n$:

Row 0:                      1
Row 1:                   1     1
Row 2:                1     2     1
Row 3:             1     3     3     1
Row 4:          1     4     6     4     1
Row 5:       1     5    10    10     5     1
Row 6:    1     6    15    20    15     6     1

The Binomial Theorem Formula

For any positive integer $n$:

$(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n - k} b^k = \binom{n}{0}a^n + \binom{n}{1}a^{n - 1}b + \binom{n}{2}a^{n - 2}b^2 + \cdots + \binom{n}{n}b^n$

Where the combination formula $\binom{n}{k} = \frac{n!}{k!(n - k)!}$ represents the entry at position $k$ (0-indexed) in row $n$ of Pascal's Triangle.

Key Properties of Binomial Expansions

1. Number of terms: An expansion of degree $n$ always contains exactly $n + 1$ terms.
2. Exponent symmetry: In every individual term, the exponent of $a$ decreases from $n$ to $0$, while the exponent of $b$ increases from $0$ to $n$. The sum of the exponents in every term always equals $n$.
3. Alternating signs: When expanding $(a - b)^n$, treat the second term as $(-b)$. The signs of the terms will strictly alternate, starting with positive: $(+) - (+) - (+) \dots$

Full Worked Example: Expand $(2x - 1)^4$

1. Identify components: $a = 2x$, $b = -1$, $n = 4$.
2. Retrieve Row 4 coefficients from Pascal's Triangle: $1, ; 4, ; 6, ; 4, ; 1$.
3. Write out the expansion terms:
   • Term 1: $1 \cdot (2x)^4 \cdot (-1)^0 = 1 \cdot 16x^4 \cdot 1 = 16x^4$
   • Term 2: $4 \cdot (2x)^3 \cdot (-1)^1 = 4 \cdot 8x^3 \cdot (-1) = -32x^3$
   • Term 3: $6 \cdot (2x)^2 \cdot (-1)^2 = 6 \cdot 4x^2 \cdot 1 = 24x^2$
   • Term 4: $4 \cdot (2x)^1 \cdot (-1)^3 = 4 \cdot 2x \cdot (-1) = -8x$
   • Term 5: $1 \cdot (2x)^0 \cdot (-1)^4 = 1 \cdot 1 \cdot 1 = 1$
4. Combine all simplified terms: $(2x - 1)^4 = 16x^4 - 32x^3 + 24x^2 - 8x + 1$

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Finding a Specific Term in a Binomial Expansion

Extracting one term of an expansion without writing out the rest is a standard precalculus technique (again, enrichment rather than Algebra II Regents content). Use the general term formula:

$T_{r + 1} = \binom{n}{r} a^{n - r} b^r$

[!CAUTION] Notice that the formula uses $r = k - 1$ for the $k$-th term. For example, to find the 4th term, you must set $r = 3$.

Step-by-Step Model: Find the 4th Term of $(2x - 3)^5$

• Step 1: Identify parameters: $n = 5, \quad a = 2x, \quad b = -3$
• Step 2: Determine $r$ for the 4th term: $r = 4 - 1 = 3$
• Step 3: Compute the binomial coefficient $\binom{5}{3}$: $\binom{5}{3} = \frac{5!}{3!(5 - 3)!} = \frac{5 \cdot 4 \cdot 3!}{3! \cdot 2 \cdot 1} = 10$
• Step 4: Substitute into the term formula: $T_4 = \binom{5}{3} (2x)^{5 - 3} (-3)^3 = 10 \cdot (2x)^2 \cdot (-27)$
• Step 5: Simplify power and coefficients: $T_4 = 10 \cdot (4x^2) \cdot (-27) = 40x^2 \cdot (-27) = -1080x^2$
• Conclusion: The 4th term is $-1080x^2$.