Equivalent expressions and polynomials
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Learning objectives
Choose a form that reveals information instead of merely making the expression longer.
What equivalence means
Two expressions are equivalent when they have the same value for every input in the relevant domain. The forms may look different while encoding the same quantity. For example,
(x − 2)(x + 5)
The product exposes factors; the expanded form exposes coefficients. Neither is universally “better.” The requested information determines which form is useful.
A term is a part separated by addition or subtraction. A factor is a part multiplied by another part. In 3x(x − 2) + 5, the terms are 3x(x − 2) and 5; inside the first term, 3, x, and x − 2 are factors. This distinction controls what can be combined or canceled.
Operations and the factoring toolkit
Add and subtract: combine only like terms, which have the same variables raised to the same powers. Distribute a subtraction sign across every term in the polynomial being subtracted.
Multiply: each term in one factor must multiply each term in the other. Combine like terms only after the products are formed. Multiplying binomials is an application of distribution, not a separate rule that stops working when there are three terms.
Factor: reverse multiplication. First remove a greatest common factor; then inspect the remaining structure. A factorization is checked by multiplying back.
Structure | Useful identity or method |
|---|---|
Common factor | AB + AC |
Difference of squares | A2 − B2 |
Perfect-square trinomial | A2 ± 2AB + B2 |
Monic quadratic | x2 + sx + p |
Nonmonic quadratic | For ax2 + bx + c, split bx using two numbers with sum b and product ac, then group. |
Grouping | AB + AC + DB + DC |
Cubes | A3 − B3 |
A familiar‐looking pattern is not enough (A + B)2 contains the middle term 2AB. In contrast, A2 + B2 is not generally (A + B)2 and does not factor as (A − B)(A + B). Check both the signs and the middle coefficient before naming a pattern.
Unknown coefficients, factors, and remainders
An identity is true for every allowed value of the variable. After expanding both sides of a polynomial identity, the coefficients of corresponding powers must match. Include the constant term and any missing power, whose coefficient is zero.
For a polynomial P(x) and a constant r, division by x − r can be written
P(x)
where R is a constant remainder. Substituting x
For a divisor such as 2x − 3, use its zero x
Structure before expansion; domain before cancellation
A repeated expression can be treated as a single object. For instance, a polynomial in x2 may become a quadratic in u
Cancellation means dividing by a common nonzero factor. In
the factor x − 4 cancels only where it is nonzero. The original expression remains undefined at 4 even though x + 3 is defined there. You cannot cancel the x in (x + 3)/x: that x is only one term of the numerator, not a factor of the whole numerator.
The zero-product property needs a zero product. If AB
The shortest valid route depends on the target
Asked for coefficients? Expand or compare a few strategically chosen inputs.
Asked for zeros? Factor, then use the zero-product property.
Asked for a remainder or a factor condition? Evaluate at the divisor’s zero.
Asked for an equivalent form? Match structure and retain the domain.
Asked for a combination of quantities? Look for an identity before solving for each quantity.
Calculator judgment. Substituting a convenient number can eliminate an incorrect multiple-choice expression. Agreement at one or two inputs does not generally prove that two polynomials are identical. Algebra supplies the proof. Avoid test inputs that make every choice zero or accidentally hide a difference.
15 worked examples
1.01. Subtract an entire polynomial
Which expression is equivalent to (3x2 − 2x + 7) − (x2 + 5x − 4)? A) 2x2 + 3x + 3 B) 2x2 − 7x + 11
C) 2x2 + 3x + 11 D) 4x2 − 7x + 3
Show worked solutionHide worked solution for example 1.01
Recognize the structure. The minus sign applies to all three terms in the second parentheses.
Work it through. Distribute the subtraction, then collect like powers:
3x2 − 2x + 7 − x2 − 5x + 4
The constant becomes 7 + 4, because subtracting −4 means adding 4.
Answer: B, 2x2 − 7x + 11.
Check. At x
Avoid the trap. Changing only the first sign in the second polynomial gives the wrong linear and constant terms.
1.02. Multiply all four term pairs
Expand (2x − 3)(x + 5).
Show worked solutionHide worked solution for example 1.02
Recognize the structure. Each of the two terms in the first factor multiplies each term in the second.
Work it through. (2x − 3)(x + 5) = 2x2 + 10x − 3x − 15 = 2x2 + 7x − 15.
The two middle products are like terms; the quadratic term and constant are not.
Answer: 2x2 + 7x − 15.
Check. At x
Avoid the trap. Multiplying only the first terms and last terms omits 10x − 3x. A product of binomials usually has a middle term.
1.03. Remove the greatest common factor first
Factor 12x3y − 18x2y2 by taking out its greatest common monomial factor.
Show worked solutionHide worked solution for example 1.03
Recognize the structure. Use the greatest common numerical factor and the smaller exponent of each shared variable.
Work it through. The numerical greatest common factor is 6. Both terms contain x2 and y, so the common monomial is 6x2y: 12x3y − 18x2y2
The remaining terms have no further common nonconstant factor.
Answer: 6x2y(2x − 3y).
Check. Distributing produces 12x3y and −18x2y2, exactly the original terms.
Avoid the trap. Factoring out x3 would leave a negative power of x in the second term; it is not the greatest common monomial factor of this polynomial.
1.04. Recognize two complete squares
Factor 49a2 − 16b2.
Show worked solutionHide worked solution for example 1.04
Recognize the structure. The terms are (7a)2 and (4b)2, separated by subtraction.
Work it through. Use A2 − B2
49a2 − 16b2
Answer: (7a − 4b)(7a + 4b).
Check. The cross terms are 28ab and −28ab, so they cancel when the factors are multiplied.
Avoid the trap. (7a − 4b)2 would contain the extra term −56ab. A difference of squares uses opposite signs in the two factors.
1.05. Factor when the leading coefficient is not 1
Factor 6x2 + x − 2.
Show worked solutionHide worked solution for example 1.05
Recognize the structure. For middle-term splitting, the two numbers must multiply to 6(−2)
Work it through. Those numbers are 4 and −3. Split and group:
6x2 + x − 2
= 2x(3x + 2) − 1(3x + 2)
= (2x − 1)(3x + 2).
Answer: (2x − 1)(3x + 2).
Check. The middle products are 4x and −3x, giving the required +x.
Avoid the trap. Numbers that multiply only to c
1.06. Recognize a perfect‐square trinomial
Which expression is equivalent to 9x2 − 30x + 25? A) (3x − 5)2 B) (3x + 5)2 C) (9x − 5)2 D) (3x − 25)2
Show worked solutionHide worked solution for example 1.06
Recognize the structure. The first and last terms are squares; check whether the middle term is twice the product of their square roots.
Work it through. The roots of 9x2 and 25 are represented by 3x and 5. The cross term in (3x−5)2 is −2(3x)(5)
Answer: A, (3x − 5)2.
Check. At x
Avoid the trap. The sign of the middle term determines the sign inside the squared binomial. The final constant stays positive in either square.
1.07. Group first, then factor again
Factor x3 + 3x2 − 4x − 12 completely into real linear factors.
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Recognize the structure. The first two and last two terms share the same binomial after grouping.
Work it through. x3 + 3x2 − 4x − 12 = x2(x + 3) − 4(x + 3) = (x + 3)(x2 − 4)
= (x + 3)(x − 2)(x + 2).
The second stage is a difference of squares.
Answer: (x + 3)(x − 2)(x + 2).
Check. The product (x − 2)(x + 2) is x2 − 4, which multiplies by x + 3 to recover the original.
Avoid the trap. Stopping at (x + 3)(x2 − 4) leaves a factor that can still be factored over the reals.
1.08. Simplify without enlarging the domain
For x ≠ 4, simplify . State the restriction that must remain.
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Recognize the structure. The numerator factors into a product containing the entire denominator.
Work it through. The numbers −4 and 3 multiply to −12 and add to −1, so
= = x + 3, x ≠ 4.
The canceled factor was nonzero because of the stated restriction.
Answer: x + 3, with x ≠ 4.
Check. At x
Avoid the trap. A simpler formula does not fill a hole in the original domain. The graph is the line y
1.09. Rewrite a rational expression using a remainder
For x ≠ −2, write in the form x + a + . What is a + b?
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Recognize the structure. Match the numerator to a multiple of x + 2 plus a constant remainder.
Work it through. Multiplication shows
(x + 2)(x + 3)
The original numerator is 2 larger. Thus
So a
Answer: a + b
Check. Combine the two terms over x + 2 to recover x2 + 5x + 8.
Avoid the trap. A nonzero remainder is not discarded. The quotient alone would differ from the original expression by 2/(x + 2).
1.10. Use a cube identity without guessing signs
Factor x3 − 27 into a linear factor and a quadratic factor.
Show worked solutionHide worked solution for example 1.10
Recognize the structure. 27
Work it through. Apply A3 − B3
x3 − 27
The positive middle term in the quadratic is necessary for the unwanted x2 and x terms to cancel.
Answer: (x − 3)(x2 + 3x + 9).
Check. Multiplying gives x3 + 3x2 + 9x − 3x2 − 9x − 27
Avoid the trap. (x − 3)3 has extra terms. Also, x2 + 3x + 9 is not (x + 3)2, whose middle term would be 6x.
1.11. Match coefficients in a polynomial identity
For all real x, (ax − 2)(3x + b) = 12x2 + cx − 10, where a, b, and c are constants. What is a + b + c?
Show worked solutionHide worked solution for example 1.11
Recognize the structure. An identity requires the quadratic, linear, and constant coefficients to match.
Work it through. Expand the left side:
3ax2 + (ab − 6)x − 2b.
Matching the quadratic coefficients gives 3a
Answer: a + b + c = 4 + 5 + 14 = 23.
Check. (4x − 2)(3x + 5)
Avoid the trap. The coefficient of x is the sum of two cross products. It is not simply ab.
1.12. Use a factor to determine a different remainder
Let P(x) = 2x3 + kx2 − 7x + 6. If x − 2 is a factor of P(x), what is the remainder when P(x) is divided by x − 3?
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Recognize the structure. The factor condition gives P(2)
Work it through. First determine k:
P(2) = 16 + 4k − 14 + 6 = 8 + 4k = 0 k = −2.
Now evaluate at the zero of the new divisor:
P(3) = 54 − 18 − 21 + 6 = 21.
Answer: 21.
Check. For k
Avoid the trap. The zero remainder for division by x − 2 says nothing automatic about division by x − 3. Use the correct input for each divisor.
Factors with integer and positivity constraints
Use the factor theorem or coefficient comparison already taught above to obtain an equation, then apply the parameter’s domain. A real solution can fail an integer requirement, and zero is not a positive integer. Write the conditions beside your algebra before canceling or dividing.
Constant terms often give a divisibility test. If integer factors have constant terms r and s and their product is a nonzero integer C, then r and s are nonzero and C/r = s is an integer. Do not infer that every other coefficient divided by r is also an integer.
Root information gives another bridge. A quadratic with roots u and v and nonzero leading coefficient a has the following coefficients:
a(x − u)(x − v) = ax2 − a(u + v)x + auv.
b = −a(u + v); c = auv.
Integer roots are not required merely because a is an integer. Conversely, rational roots may impose divisibility conditions on a if the other coefficients must also be integers. Check the exact stated conditions. To show a value “could” work, exhibit allowed parameters. To show a quantity “must” be integral, derive it from integer quantities; one successful numerical example is not a proof.
Before dividing by a parameter, ask whether it can equal zero. If it can, test that case in the original equation and handle the nonzero case separately. A canceled factor can hide a whole family of solutions, just as canceling a denominator can hide a domain restriction.
Worked example 1: A factor condition gives a finite integer search
The polynomial x2 − kx + 18 has a factor x − n, where n is a positive integer and k is an integer. Find the smallest possible k.
Show worked solutionHide worked solution: A factor condition gives a finite integer search
Substitute the root n using the existing factor theorem.
n2 − kn + 18 = 0.
Because n is positive, division by n is allowed.
k = n + .
Since k and n are integers, 18/n must be an integer. Thus n is a positive divisor of 18: 1, 2, 3, 6, 9, or 18. The corresponding k values are 19, 11, 9, 9, 11, and 19. The smallest is 9, attained at n = 3 or n = 6.
x2 − 9x + 18 = (x − 3)(x − 6).
Allowing an arbitrary real n would change the optimization. The integer condition supplies the finite candidate list; it is not an optional check after choosing a convenient real value.
Worked example 2: Rational roots and integer coefficients
A quadratic has roots 1/2 and −3. All three coefficients a, b, c are integers, and its leading coefficient a is greater than 2. Find the smallest possible a and the associated b.
Show worked solutionHide worked solution: Rational roots and integer coefficients
Keep the leading coefficient when writing the factors.
a(x − )(x + 3) = ax2 + x − .
Both 5a/2 and −3a/2 must be integers. Since a is an integer, that requires a to be even. The smallest even integer greater than 2 is 4, giving b = 10 and c = −6.
4x2 + 10x − 6 = (2x − 1)(2x + 6).
The displayed product checks the roots and all coefficients. Taking a = 3 would meet a > 2 but make b nonintegral. Taking a = 2 would give integer coefficients but fail the strict inequality.
Worked example 3: Constant-term divisibility without assuming more
Suppose (2x + r)(x + s) has integer r and s and constant term −15. Must 15/r be an integer? Must the coefficient of x divided by r be an integer?
Show worked solutionHide worked solution: Constant-term divisibility without assuming more
Matching constants gives rs = −15, so r cannot be zero and 15/r = −s is necessarily an integer. The linear coefficient is 2s + r, whose quotient by r is 1 + 2s/r; no condition forces 2s/r to be integral.
For a counterexample, choose r = 3 and s = −5. The linear coefficient is −7, and −7/3 is not an integer, even though rs = −15. Prove a “must” statement from the factor relation; disprove an overclaim with one admissible counterexample.
1.13. Find a product without finding either variable
Real numbers x and y satisfy x + y = 11 and x2 + y2 = 65. What is xy?
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Recognize the structure. Squaring the known sum creates the missing product.
Work it through. Use (x + y)2
112 = 65 + 2xy 121 − 65 = 2xy.
Therefore 2xy
Answer: 28.
Check. The possible pair x
Avoid the trap. x2 +y2 is not (x+y)2. The difference between them is exactly the term needed in this problem.
1.14. Count distinct roots, not repeated factors
What is the sum of the distinct real solutions of x(x − 2)2(x + 3) = 0?
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Recognize the structure. The expression is already a zero product. Repetition of a factor does not create a new root value.
Work it through. Set each different factor equal to zero:
x
The distinct real solutions are 0, 2, and −3. Their sum is 0 + 2 − 3
Answer: −1.
Check. Each listed value makes at least one factor zero. No other input can make this finite product zero.
Avoid the trap. Counting the root 2 twice would answer a sum with multiplicity, not the requested sum of distinct solutions.
1.15. Treat a fourth‐degree expression as a quadratic in a block
Factor x4 − 10x2 + 9 completely into real linear factors.
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Recognize the structure. Only even powers appear. Treat x2 as a single variable before factoring again.
Work it through. Let u
u2 − 10u + 9
since −1 and −9 multiply to 9 and add to −10. Substitute x2 back, then use two differences of squares:
(x2 − 1)(x2 − 9)
Answer: (x − 1)(x + 1)(x − 3)(x + 3).
Check. The product of x2 − 1 and x2 − 9 is x4 − 10x2 + 9. At x
Avoid the trap. The factors in u are not yet factors in x. Replacing u − 1 by x − 1 would change the original expression.
What this section should change in your approach
Before expanding, scan for a common factor, a repeated block, a difference of squares, or a target identity. After factoring, ask whether the problem actually sets the product equal to zero. After canceling, carry the original restriction forward. These three checks prevent many avoidable errors in later chapters.
Section exit check
Attempt these five questions without the lesson or worked solutions. Give exact answers unless a decimal is requested. Open each model answer after attempting the question. These are learning checks, not an official score scale.
Mastery check
You should be able to explain why coefficients may be compared in Question 3, why Question 4 retains a restriction, and why Question 5 does not require finding u and v separately.