5.1 Algebraic Manipulation & Special Products

On the GMAT Focus Edition Quantitative Reasoning section you answer 21 Problem Solving questions in 45 minutes with no calculator. Algebraic manipulation sits in Official Guide Quantitative Review 2026–2027 Math Review 3.2, Algebra, Equalities, and Inequalities. Quant tests arithmetic and elementary algebra applied to Problem Solving. Euclidean geometry is not in this section, and Data Sufficiency lives in Data Insights.

The no-calculator rule is the design of the item, not a handicap. GMAT algebra is written so that brute-force expansion is slow and special products are fast. Recognizing a difference of squares inside two large integers, or recovering a^2 + b^2 from a known sum and product, often turns a 90-second grind into a 20-second product you can cancel.

The three classical quadratic identities

Memorize these three identities forwards (expand) and backwards (factor). Fluency means you do not pause to re-derive the middle term.

Difference of squares

$a^2 - b^2 = (a - b)(a + b)$

This is the single most useful factoring pattern on Quant. It applies to ordinary integers, monomials, and compound binomials.

• Numeric: $105^2 - 95^2 = (105 - 95)(105 + 95) = 10 \times 200 = 2{,}000$.
• Higher even power: $x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$.
• Nested radicals, with $x, y \ge 0$: $x - y = (\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})$.

$a^2 + b^2$ does not factor over the reals. A sum of squares is not a difference of squares.

Square of a sum and square of a difference

$(a + b)^2 = a^2 + 2ab + b^2$ $(a - b)^2 = a^2 - 2ab + b^2$

The middle term is twice the product. Dropping $2ab$ is the most common expansion error on the exam: $(a + b)^2$ is not $a^2 + b^2$.

The three identities are linked, and the links are as useful as the identities themselves:

$(a + b)^2 = (a - b)^2 + 4ab$ $(a - b)^2 = (a + b)^2 - 4ab$ $a^2 + b^2 = (a + b)^2 - 2ab = (a - b)^2 + 2ab$

Hidden difference of squares with large numbers

GMAT writers love integers that look too large to square by hand. They are almost always $a^2 - b^2$ in disguise. Do not expand $103^2$ as $10{,}609$ unless you have already ruled out the identity.

Worked example 1. Evaluate $103^2 - 97^2$.

$103^2 - 97^2 = (103 - 97)(103 + 97) = 6 \times 200 = 1{,}200$

Squaring both numbers and subtracting also gives $1{,}200$, but that is slower and more error-prone without a calculator.

Worked example 2. Evaluate $201^2 - 199^2$.

$(201 - 199)(201 + 199) = 2 \times 400 = 800$

Worked example 3. Evaluate $50^2 - 48^2$.

$(50 - 48)(50 + 48) = 2 \times 98 = 196$

Check: $50^2 = 2{,}500$ and $48^2 = 2{,}304$, and $2{,}500 - 2{,}304 = 196$. Same result, more work.

Worked example 4. Evaluate $999^2 - 1$.

Write $1$ as $1^2$:

$999^2 - 1^2 = (999 - 1)(999 + 1) = 998 \times 1{,}000 = 998{,}000$

Worked example 5. A consecutive-odd-pair shortcut. For any $n$,

$(n + 1)^2 - (n - 1)^2 = \bigl[(n + 1) - (n - 1)\bigr]\bigl[(n + 1) + (n - 1)\bigr] = (2)(2n) = 4n$

So $76^2 - 74^2 = 4 \times 75 = 300$. You never square $76$ or $74$.

Worked example 6. Factor $81x^4 - 16y^4$ completely over the integers.

$81x^4 - 16y^4 = (9x^2)^2 - (4y^2)^2 = (9x^2 - 4y^2)(9x^2 + 4y^2) = (3x - 2y)(3x + 2y)(9x^2 + 4y^2)$

Stop after one difference-of-squares step only if the remaining sum of squares is what the later cancel needs. If a linear factor is still sitting inside $9x^2 - 4y^2$, factor again.

Finding $a^2 + b^2$ from $a + b$ and $ab$

You often know the sum $S = a + b$ and the product $P = ab$ without knowing $a$ and $b$ separately. Do not solve the quadratic $t^2 - St + P = 0$ if the question only wants a symmetric expression.

Worked example. If $a + b = 9$ and $ab = 14$, find $a^2 + b^2$ and $a^3 + b^3$.

$a^2 + b^2 = 9^2 - 2(14) = 81 - 28 = 53$ $a^3 + b^3 = (a + b)(a^2 - ab + b^2) = 9(53 - 14) = 9 \times 39 = 351$

The compact cube formula gives the same $351$: $S^3 - 3SP = 729 - 3(9)(14) = 729 - 378 = 351$. Then

$a^4 + b^4 = 53^2 - 2(14)^2 = 2{,}809 - 392 = 2{,}417$

You never found $a$ and $b$. (They happen to be $7$ and $2$, and $7^2 + 2^2 = 53$, but that is a check, not the method.)

Sum and difference of cubes

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

SOAP mnemonic for the signs: the binomial factor has the Same sign as the original cube sum or difference; the trinomial's middle term has the Opposite sign; the last term is Always Positive.

Binomial cubes, useful when $S$ and $P$ are known:

$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + b^3 + 3ab(a + b)$ $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 = a^3 - b^3 - 3ab(a - b)$

Worked example. Factor $8x^3 - 27$.

$8x^3 - 27 = (2x)^3 - 3^3 = (2x - 3)\bigl(4x^2 + 6x + 9\bigr)$

Worked example. Factor $x^3 + 64$.

$x^3 + 64 = (x + 4)(x^2 - 4x + 16)$

Do not write $a^3 + b^3 = (a + b)^3$. Expanding $(a + b)^3$ produces two extra middle terms.

Factoring by grouping

Four-term polynomials usually factor by pairing terms that share a common binomial:

$ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y)$

Worked example. Factor $3x^3 - 6x^2 + 5x - 10$.

$3x^2(x - 2) + 5(x - 2) = (3x^2 + 5)(x - 2)$

Worked example. Factor $x^3 + 2x^2 - 4x - 8$.

$x^2(x + 2) - 4(x + 2) = (x^2 - 4)(x + 2) = (x - 2)(x + 2)(x + 2) = (x - 2)(x + 2)^2$

If the first pairing produces no common binomial, try pairing the first term with the third. The GMAT rarely requires more than one regrouping.

Quadratic-form substitution

An expression is quadratic in form when a substitution $u$ turns it into $Au^2 + Bu + C$.

Common substitutions on Quant:

• $u = x^2$ for $ax^4 + bx^2 + c$
• $u = x^3$ for $ax^6 + bx^3 + c$
• $u = 2^x$ for $4^x + k \cdot 2^x + c$, because $4^x = (2^x)^2$
• $u = \sqrt{x}$ for $x + k\sqrt{x} + c$, then $u \ge 0$

Worked example. Factor $x^4 - 13x^2 + 36$.

Let $u = x^2$. Then $u^2 - 13u + 36 = (u - 9)(u - 4)$. Back-substitute:

$(x^2 - 9)(x^2 - 4) = (x - 3)(x + 3)(x - 2)(x + 2)$

Worked example. Solve $4^x - 6 \cdot 2^x + 8 = 0$.

Let $u = 2^x$ with $u > 0$. Then $u^2 - 6u + 8 = 0$, so $(u - 4)(u - 2) = 0$, and $u = 4$ or $u = 2$. Thus $2^x = 4$ gives $x = 2$, and $2^x = 2$ gives $x = 1$.

Special-product reference

Traps to mark on scratch paper

• $(a + b)^2$ is not $a^2 + b^2$. The missing piece is $2ab$.
• $a^2 + b^2$ does not factor over the reals. Only $a^2 - b^2$ does.
• $a^3 + b^3$ is not $(a + b)^3$. SOAP gives the linear times quadratic factorization.
• After cancelling $x - 2$ from $(x^2 - 4)/(x - 2)$, the original expression is still undefined at $x = 2$. State the restriction.
• $a^4 - b^4$ factors twice. Stopping after $(a^2 - b^2)(a^2 + b^2)$ can hide a linear factor you needed to cancel.