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This lesson covers polynomial manipulation as required by the AQA A-Level Mathematics specification (7357). Polynomials are expressions involving powers of a variable with non-negative integer exponents. At A-Level, you must be able to add, subtract, multiply, and divide polynomials, apply the factor theorem and remainder theorem, and use algebraic long division and nested multiplication (Horner's method). These skills underpin much of pure mathematics and are essential for solving equations, sketching curves, and simplifying algebraic expressions.
A polynomial in x is an expression of the form:
aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀
where n is a non-negative integer (the degree of the polynomial) and aₙ, aₙ₋₁, ..., a₀ are constants called coefficients. The term aₙxⁿ (where aₙ ≠ 0) is the leading term and aₙ is the leading coefficient.
| Term | Meaning |
|---|---|
| Degree | The highest power of x with a non-zero coefficient |
| Monic polynomial | A polynomial whose leading coefficient is 1 |
| Constant term | The term a₀ (the value when x = 0) |
Examples:
Combine like terms (terms with the same power of x).
Example: (3x³ − 2x² + 4x − 1) + (x³ + 5x² − 3x + 7)
= 4x³ + 3x² + x + 6
Multiply every term in one polynomial by every term in the other, then collect like terms.
Example: (2x + 3)(x² − x + 4)
= 2x(x² − x + 4) + 3(x² − x + 4)
= 2x³ − 2x² + 8x + 3x² − 3x + 12
= 2x³ + x² + 5x + 12
Exam Tip: When multiplying polynomials, work systematically. Multiply each term in the first bracket by every term in the second bracket. It can help to use a grid method for more complex products.
To divide a polynomial f(x) by a divisor d(x), we use algebraic long division. This produces a quotient q(x) and a remainder r(x) such that:
f(x) = d(x) × q(x) + r(x)
where the degree of r(x) is less than the degree of d(x).
Example: Divide x³ + 2x² − 5x − 6 by (x + 3).
x² − x − 2
___________________
x+3 | x³ + 2x² − 5x − 6
x³ + 3x²
---------
−x² − 5x
−x² − 3x
---------
−2x − 6
−2x − 6
--------
0
So x³ + 2x² − 5x − 6 = (x + 3)(x² − x − 2) with remainder 0.
Example: Divide 2x³ − 3x² + x + 7 by (x − 2).
2x² + x + 3
___________________
x−2 | 2x³ − 3x² + x + 7
2x³ − 4x²
----------
x² + x
x² − 2x
--------
3x + 7
3x − 6
------
13
So 2x³ − 3x² + x + 7 = (x − 2)(2x² + x + 3) + 13.
The factor theorem states:
If f(a) = 0, then (x − a) is a factor of f(x).
Conversely, if (x − a) is a factor of f(x), then f(a) = 0.
This is a special case of the remainder theorem (see below) when the remainder is zero.
Example: Show that (x − 2) is a factor of f(x) = x³ − 6x² + 11x − 6.
f(2) = 8 − 24 + 22 − 6 = 0
Since f(2) = 0, (x − 2) is a factor of f(x) by the factor theorem. ∎
Example: Factorise f(x) = x³ − 6x² + 11x − 6 completely.
Since f(2) = 0, (x − 2) is a factor. Dividing:
x³ − 6x² + 11x − 6 = (x − 2)(x² − 4x + 3) = (x − 2)(x − 1)(x − 3)
Exam Tip: To find factors, try substituting x = ±1, ±2, ±3, ... (factors of the constant term divided by factors of the leading coefficient). Once you find one factor, use algebraic long division or inspection to find the remaining quadratic, then factorise or use the formula.
The remainder theorem states:
When a polynomial f(x) is divided by (x − a), the remainder is f(a).
More generally, when f(x) is divided by (ax − b), the remainder is f(b/a).
Example: Find the remainder when f(x) = 2x³ − x² + 3x − 4 is divided by (x − 1).
f(1) = 2 − 1 + 3 − 4 = 0
The remainder is 0 — so (x − 1) is actually a factor.
Example: Find the remainder when f(x) = x⁴ + 2x³ − x + 5 is divided by (x + 2).
f(−2) = 16 − 16 + 2 + 5 = 7
The remainder is 7.
Example: Find the remainder when f(x) = 4x³ − 2x + 1 is divided by (2x − 1).
f(1/2) = 4(1/8) − 2(1/2) + 1 = 1/2 − 1 + 1 = 1/2
The remainder is 1/2.
Nested multiplication (also called Horner's method or Horner's scheme) is an efficient way to evaluate a polynomial at a given value. It also provides a compact method for polynomial division by a linear factor.
Rewrite the polynomial in nested form by factoring out x repeatedly:
f(x) = 2x³ − 3x² + x − 5
= ((2x − 3)x + 1)x − 5
To evaluate f(4):
Start with 2
2 × 4 = 8; 8 − 3 = 5
5 × 4 = 20; 20 + 1 = 21
21 × 4 = 84; 84 − 5 = 79
So f(4) = 79.
Synthetic division is a shorthand for dividing by (x − a). Write down the coefficients and use a table.
Example: Divide 2x³ + 3x² − 8x + 3 by (x − 1) using synthetic division.
| 2 3 −8 3
1 | 2 5 −3
|____________________
2 5 −3 0
Reading off: quotient is 2x² + 5x − 3, remainder is 0.
So 2x³ + 3x² − 8x + 3 = (x − 1)(2x² + 5x − 3).
We can further factorise: 2x² + 5x − 3 = (2x − 1)(x + 3).
Therefore 2x³ + 3x² − 8x + 3 = (x − 1)(2x − 1)(x + 3).
The factor and remainder theorems can be used to find unknown coefficients in a polynomial.
Example: Given that (x + 2) is a factor of f(x) = x³ + kx² − x + 6, find k.
By the factor theorem, f(−2) = 0:
(−2)³ + k(−2)² − (−2) + 6 = 0
−8 + 4k + 2 + 6 = 0
4k = 0
k = 0
Example: When f(x) = 2x³ + ax² + bx − 3 is divided by (x − 1) the remainder is 4, and (x + 1) is a factor. Find a and b.
f(1) = 2 + a + b − 3 = 4 → a + b = 5 ... (1)
f(−1) = −2 + a − b − 3 = 0 → a − b = 5 ... (2)
Adding (1) and (2): 2a = 10, so a = 5. From (1): b = 0.
Exam Tip: When factorising a cubic, always start by testing small integer values (x = ±1, ±2, ±3) to find a root using the factor theorem. Once you have one factor, divide to find the quadratic factor. Always check your factorisation by expanding. In the exam, show the substitution f(a) = 0 clearly and state the factor theorem by name.
AQA 7357 specification, Paper 1 — Pure Mathematics, Section B: Algebra and Functions. The relevant sub-strands cover the manipulation of polynomials (addition, subtraction, multiplication and division), the factor theorem, the remainder theorem, and the binomial expansion of (a+b)n for positive integer n. Polynomial fluency is examined throughout AQA 7357/1 and 7357/2, and reappears synoptically in calculus (differentiating and integrating polynomial forms), partial fractions (where polynomial division is the first step when the numerator's degree is at least the denominator's), and proof questions where divisibility arguments rest on factor-theorem reasoning. The AQA formula booklet provides the binomial theorem statement and the expansion of (1+x)n for ∣x∣<1, but does not list the factor or remainder theorems — they must be memorised.
Question (8 marks):
The polynomial p(x)=2x3−5x2−14x+8.
(a) Find the remainder when p(x) is divided by (x−1). (2)
(b) Show that (x−4) is a factor of p(x). (2)
(c) Hence factorise p(x) completely into linear factors with integer coefficients. (4)
Solution with mark scheme:
(a) Step 1 — apply the remainder theorem.
By the remainder theorem, the remainder when p(x) is divided by (x−a) equals p(a). So evaluate p(1):
p(1)=2(1)3−5(1)2−14(1)+8=2−5−14+8=−9
M1 — correct application of the remainder theorem (substituting x=1 rather than performing long division). Long division also earns the M1, but is slower and more error-prone.
A1 — correct numerical answer −9. Sign errors here (+9, −1) are the typical loss point.
(b) Step 2 — apply the factor theorem.
By the factor theorem, (x−a) is a factor of p(x) if and only if p(a)=0. Evaluate:
p(4)=2(64)−5(16)−14(4)+8=128−80−56+8=0
M1 — substituting x=4 into p(x).
A1 — showing the result equals zero, and stating the conclusion: "since p(4)=0, by the factor theorem (x−4) is a factor of p(x)." The verbal justification is essential — examiners look for explicit invocation of the theorem, not just the arithmetic.
(c) Step 3 — divide p(x) by (x−4).
Using polynomial long division:
x−42x3−5x2−14x+8
2x3÷x=2x2. Multiply: 2x2(x−4)=2x3−8x2. Subtract: (−5x2)−(−8x2)=3x2. Bring down −14x.
3x2÷x=3x. Multiply: 3x(x−4)=3x2−12x. Subtract: (−14x)−(−12x)=−2x. Bring down +8.
−2x÷x=−2. Multiply: −2(x−4)=−2x+8. Subtract: 8−8=0.
So p(x)=(x−4)(2x2+3x−2).
M1 — attempting polynomial division (or equivalent inspection / synthetic division).
A1 — correct quadratic factor 2x2+3x−2.
Step 4 — factorise the quadratic.
2x2+3x−2 factorises as (2x−1)(x+2) (since (2x)(x)=2x2, (−1)(2)=−2, and (2x)(2)+(−1)(x)=4x−x=3x).
M1 — correct method for factorising the quadratic (cross method, formula, or completing the square).
A1 — final factorisation p(x)=(x−4)(2x−1)(x+2) with all three linear factors written explicitly.
Total: 8 marks (M4 A4, split as shown).
Question (6 marks): The polynomial q(x)=x3+ax2+bx−6 has (x−1) and (x+2) as factors.
(a) Show that a+b=5 and find a second linear equation in a and b. (3)
(b) Hence find the values of a and b, and factorise q(x) completely. (3)
Mark scheme decomposition by AO:
(a)
(b)
Total: 6 marks split AO1 = 4, AO2 = 2. This is a synoptic question — the AO2 marks are awarded for the structural reasoning that links Vieta's relations (sum and product of roots) to the constant term and leading coefficient.
Connects to:
Factor and remainder theorems (Section B): the factor theorem is the special case of the remainder theorem where the remainder is zero. Both rest on the same algebraic identity: p(x)=(x−a)q(x)+p(a). Recognising this single underlying structure unifies what students often treat as two separate techniques.
Binomial expansion (Section B): expanding (a+b)n via the binomial theorem produces a polynomial in a and b. Recognising (x+1)4=x4+4x3+6x2+4x+1 as a polynomial allows you to apply the factor theorem to it directly — for example, (x+1)4−16=0 has x=1 as a root because the polynomial value drops to zero there.
Partial fractions (Section B): decomposing p(x)r(x) into partial fractions requires that degr<degp. When this fails, polynomial long division is the preliminary step to extract the polynomial part. Without confident long division, partial-fraction questions stall before the partial-fraction step even begins.
Differentiation and integration of polynomials (Sections G and H): every polynomial differentiates and integrates term-by-term using the power rule. Factorising a polynomial first often makes the subsequent calculus easier — for example, ∫(x−1)(x−2)(x−3)dx is conceptually clean, but expanding first to x3−6x2+11x−6 may be needed for term-wise integration.
Proof by contradiction / direct proof (Section A): polynomial divisibility arguments appear in proof questions, e.g. "prove that n3−n is divisible by 6 for all integers n." Factorising n3−n=n(n−1)(n+1) — three consecutive integers — connects polynomial manipulation to number-theoretic reasoning.
Polynomial-manipulation questions on AQA 7357/1 split AO marks broadly toward AO1, with a meaningful AO2 component when the question requires linking factor-theorem reasoning to structural deductions (e.g. finding unknown coefficients):
| AO | Typical share | Earned by |
|---|---|---|
| AO1 (knowledge / procedure) | 60–70% | Performing polynomial division, applying the factor / remainder theorems by substitution, factorising quadratic quotients, expanding binomials |
| AO2 (reasoning / interpretation) | 20–30% | Stating the factor theorem explicitly when concluding, linking constant term to product of roots, choosing efficient method (theorem vs division), justifying sign of remainder |
| AO3 (problem-solving) | 5–15% | Multi-step questions where the factorisation is the gateway to a further task (solving a cubic inequality, sketching a curve, finding stationary points of a polynomial) |
Examiner-rewarded phrasing: "since p(a)=0, by the factor theorem (x−a) is a factor"; "by inspection / by polynomial division, p(x)=(x−a)(quadratic quotient)"; "comparing coefficients of x2 on both sides…". Phrases that lose marks: "(x−a) is a factor because it works" (no theorem invoked); leaving the cubic only partially factored when the question demands "fully" or "completely"; sign errors in the substitution p(−2) when the bracket is (x+2).
A specific AQA pattern to watch: questions phrased "factorise completely" demand all factors broken down to linear (or, if the discriminant is negative, irreducible quadratic) form. Stopping at (x−4)(2x2+3x−2) in the worked example above costs the final A1 even though the quadratic factor is correct — "completely" is a binding instruction.
Question: Find the remainder when f(x)=x4−3x2+2x−5 is divided by (x+2).
Grade C response (~150 words):
By the remainder theorem, the remainder is f(−2).
f(−2)=(−2)4−3(−2)2+2(−2)−5=16−12−4−5=−5.
So the remainder is −5.
Examiner commentary: Full marks (3/3). The candidate correctly identifies that the remainder theorem applies (substitution rather than long division), substitutes x=−2 accurately, and handles the negative-base powers cleanly. (−2)4=16 and (−2)2=4 are written with care — sign slips here are common. The arithmetic is laid out so the examiner can verify each term. Brief but complete: this is the model Grade C answer for a procedural question. The only stylistic improvement would be explicit invocation: "by the remainder theorem, the remainder when f(x) is divided by (x−a) is f(a), so with a=−2..."
Grade A response (~210 words):*
By the remainder theorem, when a polynomial f(x) is divided by (x−a), the remainder is f(a). Here the divisor is (x+2)=(x−(−2)), so we evaluate f(−2):
f(−2)=(−2)4−3(−2)2+2(−2)−5 =16−3(4)−4−5 =16−12−4−5 =−5
Therefore, the remainder when f(x) is divided by (x+2) is −5.
Examiner commentary: Full marks (3/3). Beyond the procedural correctness of the Grade C response, this candidate names the theorem explicitly, identifies that (x+2) rewrites as (x−(−2)) to match the standard form of the theorem, and uses display-math to make each step legible. The final sentence connects the numerical result back to the question stem. This is the kind of self-checking, theorem-aware writing that examiners flag for top-band performance. On longer questions where mark-scheme conditions are stricter, this verbal scaffolding earns AO2 marks for "communicating mathematical reasoning."
Question: The polynomial g(x)=2x3+x2−13x+6.
(a) Show that (x−2) is a factor of g(x). (2)
(b) Hence factorise g(x) completely. (4)
Grade B response (~250 words):
(a) g(2)=2(8)+4−26+6=16+4−26+6=0. So (x−2) is a factor.
(b) Divide g(x) by (x−2):
2x3+x2−13x+6=(x−2)(2x2+5x−3).
The quadratic factorises as (2x−1)(x+3).
So g(x)=(x−2)(2x−1)(x+3).
Examiner commentary: (a) earns 1/2 — the substitution and arithmetic are correct, but the candidate omits the theorem-name justification. The mark scheme requires "since g(2)=0, by the factor theorem (x−2) is a factor" for full marks. (b) earns 4/4 — division correct, quadratic factorisation correct, complete linear factorisation given. Total: 5/6. This is a characteristic Grade B loss: the maths is right, but the verbal frame is missing. An A* candidate writes the theorem name into the working.
Grade A response (~290 words):*
(a) By the factor theorem, (x−a) is a factor of g(x) if and only if g(a)=0. Substitute x=2:
g(2)=2(2)3+(2)2−13(2)+6=16+4−26+6=0
Since g(2)=0, by the factor theorem (x−2) is a factor of g(x).
(b) Divide g(x) by (x−2) using polynomial long division:
2x3÷x=2x2; multiply and subtract to get 5x2−13x.
5x2÷x=5x; multiply and subtract to get −3x+6.
−3x÷x=−3; multiply and subtract to get 0.
So g(x)=(x−2)(2x2+5x−3).
Factorise the quadratic by inspection: we need two numbers multiplying to 2⋅(−3)=−6 and summing to 5 — these are 6 and −1. Splitting the middle term:
2x2+6x−x−3=2x(x+3)−1(x+3)=(2x−1)(x+3)
Therefore g(x)=(x−2)(2x−1)(x+3).
Examiner commentary: Full marks (6/6). The candidate invokes the factor theorem by name in (a), making the AO2 communication mark secure. In (b), the polynomial division is laid out with each subtraction step shown, the quadratic factorisation uses the standard "split the middle term" technique with explicit reasoning, and the final factorisation honours "completely" by giving three linear factors. The factorisation could equivalently use the formula x=4−5±25+24=4−5±7, giving x=21 or x=−3, hence factors (2x−1) and (x+3).
Question: p(x)=x4+ax3+bx2+cx+d has roots 1, −1, 2, and −3.
(a) Express p(x) as a product of linear factors. (2)
(b) Hence find the values of a, b, c, and d. (4)
(c) Find the remainder when p(x) is divided by (x−4). (3)
Grade A response (~370 words):*
(a) Since the roots are 1, −1, 2, and −3, by the factor theorem the corresponding factors are (x−1), (x+1), (x−2), and (x+3). Because the leading coefficient of p(x) is 1:
p(x)=(x−1)(x+1)(x−2)(x+3)
(b) Expand systematically. First, pair conjugate-like factors:
(x−1)(x+1)=x2−1
(x−2)(x+3)=x2+x−6
Now multiply the two quadratics:
(x2−1)(x2+x−6)=x4+x3−6x2−x2−x+6=x4+x3−7x2−x+6
Comparing with p(x)=x4+ax3+bx2+cx+d:
a=1,b=−7,c=−1,d=6
(c) By the remainder theorem, the remainder when p(x) is divided by (x−4) is p(4). Use the factored form for efficiency:
p(4)=(4−1)(4+1)(4−2)(4+3)=3⋅5⋅2⋅7=210
So the remainder is 210.
Examiner commentary: Full marks (9/9). Part (a) earns 2/2 — factor theorem invoked, leading-coefficient assumption made explicit. Part (b) earns 4/4 — the strategic pairing of factors before expansion (recognising (x−1)(x+1) as a difference of squares) is the AO2-rewarded efficiency move. Each coefficient is identified by direct comparison. Part (c) earns 3/3 — using the factored form of p(x) to evaluate p(4) avoids a much longer expanded-form arithmetic and demonstrates structural awareness. A weaker candidate would substitute into the expanded polynomial p(x)=x4+x3−7x2−x+6 and likely lose a mark to arithmetic slips: 256+64−112−4+6=210 — correct but high-risk. The A* move is to choose the form that minimises arithmetic risk.
The errors that distinguish A from A* on polynomial-manipulation questions:
Sign errors with (x+a) vs (x−a). The factor theorem says (x−a) is a factor iff p(a)=0. So for (x+2), you substitute x=−2, not x=2. Substituting the wrong sign is the single most common factor-theorem error and converts a 4-mark question into 0 marks.
Polynomial long division alignment slips. When dividing a polynomial that has no x2 term, students sometimes drop the "missing" terms — failing to insert 0x2 as a placeholder, so the column-arithmetic shifts by one place. The cure: always write x4+0x3+ax2+bx+c in full before dividing.
Stopping at "factorised" rather than "completely factorised." If the question says "factorise completely", you must reduce every factor to either linear or irreducible-quadratic form. Leaving a factorable quadratic intact (e.g. (x−2)(x2+5x+6) when x2+5x+6=(x+2)(x+3)) loses the final A1.
Confusing the binomial theorem with the polynomial theorem. (x+1)4 expands via the binomial theorem to x4+4x3+6x2+4x+1. (x+1)(x3−1) expands by distribution. Mixing these — applying binomial coefficients to a non-binomial product — produces nonsense. The binomial theorem applies only to powers of a binomial.
Forgetting that the factor theorem requires polynomial input. The theorem applies to polynomials. f(x)=sinx has f(0)=0, but (x−0)=x is not a "factor" of sinx in the polynomial-algebra sense — at most you can write sinx=x⋅g(x) for some g, but the factor theorem statement does not apply.
Mishandling repeated roots. If p(x) has a double root at x=1, then (x−1)2 is a factor, not just (x−1). The factor theorem alone does not detect multiplicity — you also need p′(1)=0 (a Year-2 idea). At AS, repeated roots are usually flagged by the question stem ("x=2 is a root of multiplicity 2").
Constant term ≠ product of roots when leading coefficient ≠ 1. For monic p(x)=xn+…+d, the product of roots is (−1)nd. For non-monic p(x)=axn+…+d, the product of roots is (−1)nd/a. Forgetting the leading-coefficient division is a Vieta's-relations slip that propagates through any "find the missing root" question.
Three patterns repeatedly cost candidates marks on Paper 1 polynomial-division and factorisation questions.
This pattern is endemic to Paper 1 polynomial questions: the technique is well-known, but the conditions (sign of root, form of divisor, theorem-name justification) trip up otherwise-strong candidates.
Polynomial manipulation points directly toward several undergraduate trajectories:
Oxbridge interview prompt: "Prove that the polynomial x4+4 factors over the integers, even though it has no real roots. (Hint: try x4+4=(x2+2x+2)(x2−2x+2) — the Sophie Germain identity.) Now: when does x4+4y4 factor over the integers in two variables?"
A common A* enrichment topic is Horner's method — a re-parenthesisation of a polynomial that evaluates it using the minimum number of multiplications. It also provides a fast hand-calculation route for the remainder theorem and an elegant alternative to polynomial long division.
The method: rewrite p(x)=anxn+an−1xn−1+…+a1x+a0 as
p(x)=(…((anx+an−1)x+an−2)x+…+a1)x+a0
Each step is "multiply by x, add the next coefficient." For a degree-n polynomial, Horner uses n multiplications and n additions, compared with up to n(n+1)/2 multiplications for naive evaluation.
Worked example: Use Horner's method to evaluate p(x)=2x3−5x2−14x+8 at x=4.
Start with the leading coefficient 2. Multiply by 4, add −5: 2(4)+(−5)=3. Multiply by 4, add −14: 3(4)+(−14)=−2. Multiply by 4, add 8: −2(4)+8=0.
So p(4)=0 — confirming (x−4) is a factor (matching the worked example earlier).
Why this matters for synthetic division. The intermediate values 2,3,−2,0 are exactly the coefficients of the quotient when p(x) is divided by (x−4), plus the remainder. So Horner gives us:
p(x)=(x−4)(2x2+3x−2)+0
This is the synthetic division layout in disguise. Recognising the connection between Horner evaluation and synthetic division is exactly the structural insight that A* candidates display — and it converts a 4-mark "divide and find the quotient" question into a 30-second hand calculation.
A subtlety: Horner's method generalises beautifully to evaluating polynomials at matrices (in linear algebra) and at elements of arbitrary rings (in abstract algebra). Numerical analysts also use it because each step has minimal floating-point error compared with the naive expansion.
This content is aligned with the AQA A-Level Mathematics (7357) specification, Paper 1 — Pure Mathematics, Section B: Algebra and Functions. For the most accurate and up-to-date information, please refer to the official AQA specification document.
graph TD
A["Polynomial p(x)<br/>of degree n"] --> B{"Goal?"}
B -->|"Find remainder<br/>on division by (x-a)"| C["Remainder theorem:<br/>remainder = p(a)"]
B -->|"Test if (x-a)<br/>is a factor"| D["Factor theorem:<br/>(x-a) is a factor iff p(a) = 0"]
B -->|"Factorise completely"| E["Find one root by<br/>factor theorem"]
B -->|"Expand a binomial"| F["Binomial theorem:<br/>(a+b)^n via Pascal coefficients"]
C --> G["Substitute x = a<br/>and simplify"]
D --> G
E --> H["Divide p(x) by (x-a)<br/>via long or synthetic division"]
H --> I["Factorise the<br/>quotient (degree n-1)"]
I --> J{"Quotient<br/>factorable?"}
J -->|"Yes"| K["Repeat factor theorem<br/>or use quadratic formula"]
J -->|"No (irreducible)"| L["Stop: leave as<br/>irreducible factor"]
K --> M["Final form:<br/>product of linear<br/>and irreducible quadratic<br/>factors"]
L --> M
F --> N["Combine like terms<br/>and simplify"]
G --> O["Verify and<br/>state conclusion"]
style D fill:#27ae60,color:#fff
style M fill:#3498db,color:#fff