ORIGINAL WORKED EXAMPLE
Expand with a negative group
Simplify 3(2x−5)−2(x+4).
- Distribute: 6x−15−2x−8.
- Combine like terms: (6x−2x)+(−15−8).
- The result is 4x−23.
Answer 4x−23
GRE · Algebra · TOPIC LESSON
Algebraic expressions and factoring
Rewrite expressions without changing their value or silently enlarging their domain.
BY THE END, YOU CAN
- Distribute and collect only like terms.
- Factor polynomials and simplify rational expressions.
- Preserve restrictions and distinguish an expression from an equation.
BEFORE YOU START
- Signed arithmetic
- Exponent laws
Original educational lesson. Work through the examples before revealing the practice answers. This original lesson supports current GRE General Test study; it does not reproduce official questions, certify a writing score or predict test content.
LEARN THROUGH ORIGINAL PRACTICE
Original GRE-style worked examples and practice
All teaching passages, writing prompts, examples and exercises in this lesson are original. No official ETS questions, prompts or explanations are reproduced. Current structure and observed public sample metadata are separate from this original teaching; no official score or question probability is implied.
01 / UNDERSTAND & APPLY
Structure comes before manipulation
An expression represents a value; an equation asserts that two expressions are equal. Terms are pieces joined by addition or subtraction, whereas factors are multiplied. In 3x(x−2)+5, the first term is a product and the second is a constant. Naming the structure helps choose legal operations.
Like terms have identical variable parts, including exponents: 4x²−x²=3x², but 4x²−x cannot be combined into one term. Evaluate an expression by substituting into every occurrence and using parentheses around negative inputs.
02 / UNDERSTAND & APPLY
Distribute a multiplier to every term
The distributive law a(b+c)=ab+ac works with negative multipliers as well. A minus before a group means multiplication by −1, so every sign inside changes. Expanding two binomials requires every term of one to multiply every term of the other.
Collect terms only after all products are formed. A common mistake is dropping a middle term or distributing a square across a sum: (x+y)²=x²+2xy+y², not x²+y².
03 / UNDERSTAND & APPLY
Factoring reverses distribution
To factor, identify a greatest common factor of all terms and divide each term by it. The result should multiply back to the original expression. Factoring is useful because zeros of a product, common factors of a fraction, and sign changes become visible.
For example, 6x²−9x=3x(2x−3). Both 3 and x are common factors. Factoring out x is an identity even at x=0; dividing an equation by x would require checking x=0 separately.
ORIGINAL WORKED EXAMPLE
Use grouping
Factor ax+ay+3x+3y.
- Group a(x+y)+3(x+y).
- The common factor is the entire expression x+y.
- Factor to obtain (a+3)(x+y), and multiply back to check all four terms.
Answer (a+3)(x+y)
04 / UNDERSTAND & APPLY
Recognize special products, with the middle term intact
The difference of squares is u²−v²=(u−v)(u+v). A perfect-square trinomial is u²+2uv+v²=(u+v)², or u²−2uv+v²=(u−v)². A sum of squares generally does not factor into real linear factors by changing a sign.
For x²+bx+c, seek numbers with sum b and product c if integer factoring is available. When the leading coefficient is not 1, verify both the middle and constant terms; guessing factors from the constant alone is insufficient.
ORIGINAL WORKED EXAMPLE
Factor a quadratic
Factor 2x²+7x+3.
- A product (2x+1)(x+3) gives leading term 2x² and constant 3.
- Its middle terms are 6x+x=7x.
- This confirms the factorization; (2x+3)(x+1) would have middle term 5x instead.
Answer (2x+1)(x+3)
05 / UNDERSTAND & APPLY
Cancel factors, then keep the original exclusions
A rational expression is undefined where its denominator is zero. Record those restrictions before cancelling. If a factor cancels, the simplified formula matches the original only on the original domain. A removable factor does not make an originally excluded input valid.
In (x²−9)/(x−3), factor the numerator as (x−3)(x+3), then cancel for x≠3. The result is x+3 with x≠3. Cancelling x from (x+3)/x is illegal because the numerator is a sum, not a product containing x as a factor.
ORIGINAL WORKED EXAMPLE
Simplify and retain the domain
Simplify (x²−4)/(x²+x−6), stating restrictions.
- Factor: (x−2)(x+2)/[(x+3)(x−2)].
- The original denominator excludes x=−3 and x=2.
- Cancel x−2 for allowed x. The result is (x+2)/(x+3), with both exclusions retained.
Answer (x+2)/(x+3), x≠−3 and x≠2
06 / UNDERSTAND & APPLY
Exponent laws do not cancel domain restrictions
For nonzero x, xᵃxᵇ=xᵃ⁺ᵇ and xᵃ/xᵇ=xᵃ⁻ᵇ for integer exponents. For integer powers of a product, raise every factor. A negative exponent requires a reciprocal, so it excludes a zero base.
For example, (2x²y⁻¹)³=8x⁶/y³ with y≠0. Do not multiply the exponent of the coefficient by 3: the coefficient 2 itself is cubed. If simplifying produces x⁰, the original nonzero restriction may still apply.
ORIGINAL WORKED EXAMPLE
A power quotient
Simplify (6a⁴b⁻²)/(3a⁻¹b), with a,b nonzero.
- Divide coefficients: 6/3=2.
- Subtract powers of a: 4−(−1)=5; subtract powers of b: −2−1=−3.
- Write positive exponents: 2a⁵/b³.
Answer 2a⁵/b³
07 / UNDERSTAND & APPLY
Choose the form that exposes the question
An expanded form makes coefficients and term comparison easy. A factored form makes zeros and restrictions easy. A completed square makes a nonnegative part visible. There is no single “most simplified” form for every purpose.
For QC, subtract the quantities when useful and analyze the sign of the difference. If A−B=(x−2)², then A≥B for every real x, but equality at x=2 prevents concluding A is always strictly greater unless that value is excluded.
08 / UNDERSTAND & APPLY
Choose a method and answer the requested question
All examples and practice in this lesson are original. For quantitative comparison (QC), use A if Quantity A is greater, B if Quantity B is greater, C if they are equal, and D if the relationship cannot be determined from the information given. A, B or C must hold for every permitted case. To establish D, give two allowed cases producing different relationships; an inability to calculate is not itself evidence for D.
A single-answer question requires one choice. For multiple-answer questions, obey the stated selection count or select every correct option if instructed. Test each option separately; do not stop after finding one. Multiple-answer questions receive credit only when the complete correct set is selected, with no incorrect choices. Numeric entry requires the requested value, with units handled in the calculation; round only when the question directs you to. A single numeric-entry box accepts an integer or decimal; a pair of fraction boxes accepts an integer numerator and a nonzero integer denominator. Fractions need not be reduced, but reducing makes checks easier. Do not enter a percent sign when the requested number is a percentage. Preserve exact fractions until the last step.
A calculator can check arithmetic but cannot decide the domain, interpretation or denominator for you. Estimate sign and size first, enter parentheses carefully, and compare the result with the original conditions. Cancel factors or exploit structure when doing so is faster and more reliable than decimal entry.
09 / UNDERSTAND & APPLY
Original practice: valid identities and domains
Select all valid identities only over the stated domain.
ORIGINAL PRACTICE · 1
Numeric entry: Evaluate 2x²−3x+1 at x=−2.
Reveal answer & reasoning
15
Substitute with parentheses: 2(4)−3(−2)+1=8+6+1=15. Forgetting the negative sign in −3x gives a different result.
ORIGINAL PRACTICE · 2
Single answer: 5−2(3−x) simplifies to A −1−2x; B −1+2x; C 11−2x; D 2−x; E 3+2x.
Reveal answer & reasoning
B
Distribute −2 to both terms: 5−6+2x=−1+2x. A changes the sign of the second product incorrectly.
ORIGINAL PRACTICE · 3
Select all that apply: Which equal x²−16 for every real x? A (x−4)(x+4); B (x−4)²; C (x+4)²−8x−32; D x(x−16); E (x−2)(x+2).
Reveal answer & reasoning
A and C
A is the difference-of-squares identity. C expands to x²+8x+16−8x−32=x²−16. B retains a middle term, D has −16x, and E has constant −4.
ORIGINAL PRACTICE · 4
Numeric entry: If 4x²+bx+9=(2x+3)² for all real x, find b.
Reveal answer & reasoning
12
The cross term is 2·2x·3=12x. The coefficient is 12, not 6; both cross products contribute.
ORIGINAL PRACTICE · 5
Single answer: The original expression (x²−25)/(x−5) is undefined at which value? A −5; B 0; C 5; D 10; E none.
Reveal answer & reasoning
C
The denominator vanishes at 5. Cancellation gives x+5 only for x≠5; it cannot repair the original domain.
ORIGINAL PRACTICE · 6
QC: x≠0. Quantity A: (x³+x)/x. Quantity B: x²+1.
Reveal answer & reasoning
C
Factor x(x²+1) in the numerator and cancel permitted nonzero x. The equality holds for every allowed x, including negative inputs.
ORIGINAL PRACTICE · 7
QC: x is real. Quantity A: (x−2)². Quantity B: 0.
Reveal answer & reasoning
D
A is nonnegative, but at x=2 it equals B and at x=3 it exceeds B. Since the relationship can be equality or strict inequality, D is required.
ORIGINAL PRACTICE · 8
Numeric entry: For nonzero t, (t⁵/t⁻²)=tᵏ as an exponent-law identity. Find k.
Reveal answer & reasoning
7
Subtract the denominator exponent: 5−(−2)=7. Adding 5 and −2 incorrectly gives 3; division subtracts exponents.
KEEP THE BIG IDEAS
Quick revision
- Distinguish terms joined by addition from factors joined by multiplication.
- Factoring is an identity; division by a variable can discard a case.
- Preserve original denominator restrictions after cancellation.
- Use the algebraic form that reveals the needed feature.
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References & scope
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