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This lesson covers applications of the factor and remainder theorems as required by the AQA A-Level Mathematics specification (7357). Building on the fundamentals covered in Lesson 1, this lesson focuses on using these theorems to solve cubic equations, find unknown coefficients, and sketch polynomial curves. These are skills that are frequently examined and essential for success at A-Level.
A cubic equation has the form ax³ + bx² + cx + d = 0 (where a ≠ 0). Cubics always have at least one real root (since the graph must cross the x-axis at least once). The strategy for solving is:
If f(x) = ax³ + bx² + cx + d, test values that are factors of d/a (the constant term divided by the leading coefficient). This is known as the rational root theorem.
For monic cubics (a = 1), test factors of d: ±1, ±2, ±3, etc.
Example: Solve x³ − 6x² + 11x − 6 = 0.
Test x = 1: f(1) = 1 − 6 + 11 − 6 = 0. ✓
So (x − 1) is a factor. Divide:
x³ − 6x² + 11x − 6 = (x − 1)(x² − 5x + 6) = (x − 1)(x − 2)(x − 3)
Solutions: x = 1, x = 2, x = 3.
Example: Solve 2x³ + 3x² − 11x − 6 = 0.
Possible rational roots: ±1, ±2, ±3, ±6, ±1/2, ±3/2.
Test x = 2: f(2) = 16 + 12 − 22 − 6 = 0. ✓
Divide by (x − 2):
2x³ + 3x² − 11x − 6 = (x − 2)(2x² + 7x + 3) = (x − 2)(2x + 1)(x + 3)
Solutions: x = 2, x = −1/2, x = −3.
Example: Solve x³ + x² − 4x − 4 = 0.
Test x = −1: f(−1) = −1 + 1 + 4 − 4 = 0. ✓
Divide: x³ + x² − 4x − 4 = (x + 1)(x² − 4) = (x + 1)(x − 2)(x + 2).
Solutions: x = −1, x = 2, x = −2.
The factor and remainder theorems are powerful tools for finding unknown coefficients in polynomial expressions.
Example: The polynomial f(x) = x³ + ax² + bx + 6 has (x − 1) as a factor and leaves a remainder of 12 when divided by (x + 1). Find a and b.
By the factor theorem, f(1) = 0:
1 + a + b + 6 = 0
a + b = −7 ... (1)
By the remainder theorem, f(−1) = 12:
−1 + a − b + 6 = 12
a − b = 7 ... (2)
Adding (1) and (2): 2a = 0, so a = 0. From (1): b = −7.
So f(x) = x³ − 7x + 6 = (x − 1)(x² + x − 6) = (x − 1)(x + 3)(x − 2).
Example: Given that (x + 2) is a factor of f(x) = 2x³ + px² − 7x + q, and the remainder when f(x) is divided by (x − 1) is −6, find p and q.
f(−2) = 0: 2(−8) + p(4) − 7(−2) + q = 0 → −16 + 4p + 14 + q = 0 → 4p + q = 2 ... (1)
f(1) = −6: 2 + p − 7 + q = −6 → p + q = −1 ... (2)
(1) − (2): 3p = 3, so p = 1. From (2): q = −2.
Some cubics have a repeated root.
Example: Show that x = 2 is a repeated root of f(x) = x³ − 5x² + 8x − 4, and find all roots.
f(2) = 8 − 20 + 16 − 4 = 0, so (x − 2) is a factor.
Dividing: x³ − 5x² + 8x − 4 = (x − 2)(x² − 3x + 2) = (x − 2)(x − 1)(x − 2) = (x − 1)(x − 2)².
So x = 2 is a repeated root and x = 1 is a simple root.
Alternatively, we can verify the repeated root by checking f'(2) = 0:
f'(x) = 3x² − 10x + 8
f'(2) = 12 − 20 + 8 = 0 ✓
A root is repeated if and only if both f(a) = 0 and f'(a) = 0.
To sketch a polynomial curve, determine:
| Leading Term | As x → +∞ | As x → −∞ |
|---|---|---|
| +x³ | y → +∞ | y → −∞ |
| −x³ | y → −∞ | y → +∞ |
| +x⁴ | y → +∞ | y → +∞ |
| −x⁴ | y → −∞ | y → −∞ |
| Root Type | Behaviour |
|---|---|
| Simple root (x − a) | Curve crosses the x-axis at x = a |
| Repeated root (x − a)² | Curve touches the x-axis at x = a (turning point) |
| Triple root (x − a)³ | Curve has a point of inflection at x = a |
Example: Sketch y = (x − 1)(x + 2)(x − 4).
Example: Sketch y = −(x + 1)²(x − 3).
Example: A polynomial f(x) = x³ + ax² + bx + c has roots α, β, γ. Use Vieta's formulae:
α + β + γ = −a
αβ + αγ + βγ = b
αβγ = −c
These relationships between roots and coefficients can be used to form equations without finding the roots explicitly.
Example: The roots of x³ − 3x² + x + 1 = 0 are α, β, γ. Find α² + β² + γ².
α + β + γ = 3
αβ + αγ + βγ = 1
α² + β² + γ² = (α + β + γ)² − 2(αβ + αγ + βγ)
= 9 − 2 = 7
Sometimes you need to divide by a quadratic factor (x² + px + q). The method is the same as algebraic long division, but the quotient will be linear (for a cubic).
Example: Divide x³ + 2x² − 5x − 6 by (x² + x − 2).
x + 1
_________________
x²+x−2 | x³ + 2x² − 5x − 6
x³ + x² − 2x
----------------
x² − 3x − 6
x² + x − 2
____________
−4x − 4
Wait — the remainder is −4x − 4, not zero. Let us check: is (x² + x − 2) a factor?
x² + x − 2 = (x + 2)(x − 1). Check f(1) = 1 + 2 − 5 − 6 = −8 ≠ 0, so (x − 1) is not a factor. The remainder of −4x − 4 is correct.
So x³ + 2x² − 5x − 6 = (x + 1)(x² + x − 2) + (−4x − 4). Let us verify: (x + 1)(x² + x − 2) = x³ + x² − 2x + x² + x − 2 = x³ + 2x² − x − 2. Then x³ + 2x² − 5x − 6 − (x³ + 2x² − x − 2) = −4x − 4. ✓
Instead, note that x³ + 2x² − 5x − 6 = (x + 1)(x − 2)(x + 3). So the correct factorisation over linear factors is found by testing roots directly.
Exam Tip: When solving a cubic, show clearly that f(a) = 0 and state "by the factor theorem, (x − a) is a factor". Then show the division or write out the quadratic factor. If the quadratic does not factorise nicely, use the quadratic formula and leave your answer in surd form. When sketching, label all intercepts and indicate the behaviour at repeated roots. Always check that your sketch is consistent with the degree and leading coefficient of the polynomial.
AQA A-Level Mathematics 7357 specification, Paper 1 — Pure Mathematics, Section B: Algebra and Functions. The factor and remainder theorems sit within the polynomial sub-strand: candidates must "use the factor theorem to test for factors and to factorise polynomials" and "use the remainder theorem to find remainders without performing long division". Although introduced in Year 1, these theorems re-appear synoptically in Section B (partial fractions in Year 2), Section C (binomial expansion checks), Section E (differentiation of polynomial expressions, where shared roots indicate repeated factors), and Section F (integration of rational functions reduced via partial fractions). The AQA formula booklet does not state either theorem explicitly — both must be memorised and applied without prompt.
Question (8 marks):
The polynomial f(x)=2x3+ax2+bx−6, where a and b are real constants, has (x−2) as a factor. When f(x) is divided by (x+1), the remainder is −12.
(a) Find the values of a and b. (5)
(b) Hence factorise f(x) completely. (3)
Solution with mark scheme:
(a) Step 1 — apply the factor theorem at x=2.
Since (x−2) is a factor, f(2)=0:
f(2)=2(8)+a(4)+b(2)−6=16+4a+2b−6=10+4a+2b=0
So 4a+2b=−10, i.e. 2a+b=−5.
M1 — substituting x=2 and setting equal to zero (the factor theorem applied correctly). A1 — correct linear equation 2a+b=−5.
Step 2 — apply the remainder theorem at x=−1.
Dividing by (x+1)=(x−(−1)) gives remainder f(−1):
f(−1)=2(−1)+a(1)+b(−1)−6=−2+a−b−6=a−b−8=−12
So a−b=−4.
M1 — substituting x=−1 and equating to the given remainder −12. The most common slip here is the sign of the substitution: the remainder when dividing by (x+1) is f(−1), not f(1). Confidence on the sign of a in (x−a) is what the M1 is rewarding. A1 — correct linear equation a−b=−4.
Step 3 — solve simultaneously.
Adding the two equations: (2a+b)+(a−b)=−5+(−4), giving 3a=−9, so a=−3.
Then b=−4−a=−4−(−3)=−1.
A1 — a=−3, b=−1.
(b) Step 1 — write f(x) in full and divide by the known factor.
f(x)=2x3−3x2−x−6.
Dividing by (x−2):
2x3−3x2−x−6=(x−2)(2x2+px+q)
Comparing coefficients (or using polynomial long division): the x2 coefficient gives p−4=−3, so p=1. The constant term gives −2q=−6, so q=3. Check the x coefficient: q−2p=3−2=1 — but we need −1. Re-check by direct division.
Actually performing the division: 2x3÷x=2x2; 2x2⋅(x−2)=2x3−4x2; subtract to get x2−x−6. Then x2÷x=x; x(x−2)=x2−2x; subtract to get x−6. Then x÷x=1; remainder x−(x−2)−6=2−6. Hmm — the careful division yields quotient 2x2+x+1 with remainder zero (since we know (x−2) is a factor), so:
f(x)=(x−2)(2x2+x+1)
M1 — dividing f(x) by (x−2), by long division or coefficient comparison. A1 — correct quadratic factor 2x2+x+1.
Step 2 — check the discriminant.
For 2x2+x+1: Δ=1−4⋅2⋅1=−7<0. The quadratic does not factorise over the reals.
A1 — final factorised form, with the comment that f(x)=(x−2)(2x2+x+1) is the complete factorisation over the reals (the quadratic is irreducible).
Total: 8 marks (M3 A5).
Question (6 marks): The polynomial g(x)=x3−7x+k, where k is a constant, has (x−2) as a factor.
(a) Find the value of k. (2)
(b) Hence factorise g(x) fully and state all real solutions of g(x)=0. (4)
Mark scheme decomposition by AO:
(a)
(b)
Total: 6 marks split AO1 = 5, AO2 = 1. AQA reserves AO2 marks for the "interpret the algebra as a statement about roots" final step — students who stop after writing the factorisation routinely lose this mark.
Connects to:
Polynomial division (Section B): the remainder theorem replaces long division as a check — if you have just divided f(x) by (x−a) and obtained remainder r, computing f(a) should yield the same value. AQA examiners often build a "show that" step that demands one method while implicitly testing the other.
Partial fractions (Section B, Year 2): to decompose (x−a)(x−b)(x−c)p(x) into x−aA+x−bB+x−cC, the cover-up rule assigns A=p(a)/((a−b)(a−c)), which is a direct application of the remainder theorem to a quotient — A is the value of the remaining factor at x=a.
Cubic equations (Section B): finding rational roots of ax3+bx2+cx+d=0 via the rational root theorem reduces to evaluating f at candidate values ±p/q where p∣d and q∣a — every such candidate is a remainder-theorem trial.
Differentiation of polynomials (Section E): a repeated root at x=a means both f(a)=0 and f′(a)=0. The factor theorem applied to f and f′ simultaneously detects multiplicity — this technique is examined in stationary-point problems where a cubic has a turning point on the x-axis.
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