Lesson 2: Polynomials and Factorisation

A polynomial in x is an expression built from powers of x with numerical coefficients, like 4x³ − 2x² + 7x − 1. NCERT Class IX introduced factorisation; Class X makes the idea precise by connecting the zeroes of a polynomial to its coefficients and to its graph. This lesson follows that NCERT path: zeroes and the factor theorem, the relationships between zeroes and coefficients for quadratics and cubics, and division of polynomials.

Zeroes and the factor theorem

A real number k is a zero of p(x) if p(k) = 0. Geometrically, the zeroes are the x-coordinates where the graph y = p(x) crosses the x-axis — a linear polynomial crosses once, a quadratic at most twice, a cubic at least once.

Factor theorem / remainder fact: dividing p(x) by (x − k) leaves remainder p(k). In particular, (x − k) is a factor of p(x) exactly when p(k) = 0.

Worked example 1 (basic). Find the zeroes of p(x) = x² − 3x − 4 and verify the coefficient relationships. Factorise: x² − 3x − 4 = (x − 4)(x + 1), so the zeroes are α = 4, β = −1. Check: α + β = 3 = −(−3)/1 = −b/a ✓ and αβ = −4 = c/a ✓.

The zero–coefficient relationships (NCERT box)

For a quadratic ax² + bx + c with zeroes α and β:

• α + β = −b/a
• αβ = c/a

For a cubic ax³ + bx² + cx + d with zeroes α, β, γ:

• α + β + γ = −b/a
• αβ + βγ + γα = c/a
• αβγ = −d/a

These come from writing the polynomial as a(x − α)(x − β) [or a(x − α)(x − β)(x − γ)] and comparing coefficients — worth doing once yourself.

Worked example 2 (intermediate). Verify that 2, 3 and −1 are zeroes of p(x) = x³ − 4x² + x + 6, then check all three relationships. p(2) = 8 − 16 + 2 + 6 = 0 ✓; p(3) = 27 − 36 + 3 + 6 = 0 ✓; p(−1) = −1 − 4 − 1 + 6 = 0 ✓. Sum: 2 + 3 + (−1) = 4 = −(−4)/1 ✓. Pairwise: (2)(3) + (3)(−1) + (−1)(2) = 6 − 3 − 2 = 1 = c/a ✓. Product: 2 × 3 × (−1) = −6 = −d/a = −6 ✓. So p(x) = (x − 2)(x − 3)(x + 1).

Worked example 3 (NCERT exemplar style). If α and β are the zeroes of x² − 6x + k and α² + β² = 40, find k. α + β = 6 and αβ = k. Use the identity α² + β² = (α + β)² − 2αβ: 40 = 36 − 2k, so 2k = −4 and k = −2. (Indeed the polynomial becomes x² − 6x − 2, whose zeroes 3 ± √11 satisfy the condition.)

Division of polynomials

Dividing p(x) by g(x) gives p(x) = g(x) × q(x) + r(x) with deg r < deg g. If (x − k) is the divisor, the remainder is just p(k) — the remainder theorem in action.

Division example. Divide 2x² + 3x + 1 by x + 2. Long division gives quotient 2x − 1, remainder 3. Check: (x + 2)(2x − 1) + 3 = 2x² + 3x − 2 + 3 = 2x² + 3x + 1 ✓. And p(−2) = 8 − 6 + 1 = 3 ✓, as promised.

Competency application: designing a garden bed

A landscaper designs a rectangular bed whose length is 3 m more than its width, with area 40 m². If the width is x, then x(x + 3) = 40, i.e. x² + 3x − 40 = 0. Factorising: (x + 8)(x − 5) = 0, so x = 5 or x = −8. Both are *zeroes* of the polynomial, but a width cannot be negative — the bed is 5 m by 8 m. Board questions love this: the algebra gives two answers, and the *context* rejects one. A polynomial used backwards also appears in pricing: if a quadratic cost model C(x) = x² − 12x + 45 for x items has C(x) = 45, solving gives x(x − 12) = 0 — the break points of the model.

Common mistakes

• Sign errors in the sum of zeroes. For x² − 3x − 4, α + β = −b/a = −(−3) = +3, not −3. The formula has a built-in minus sign.
• Using the relationships before making the leading coefficient explicit. For 2x² − 5x + 3, αβ = c/a = 3/2 — students who skip a write 3.
• Claiming every zero found by factorisation needs the graph. The graph interpretation is a check, not a substitute: y = x² − 3x − 4 crosses the axis at x = 4 and x = −1, matching the algebra.
• In division, reporting the remainder with degree too high. If you divide by a quadratic, the remainder must be linear or constant; if it still contains x², the division is unfinished.

Exam Experience

Section A — Multiple choice (1 mark each)

Q1. A zero of the polynomial 3x + 5 is: (a) 5/3 (b) −5/3 (c) 3/5 (d) −3/5. Answer: (b) −5/3. Setting 3x + 5 = 0 gives x = −5/3.

Q2. If one zero of x² − 4x + k is 2, then k equals: (a) 2 (b) 4 (c) −4 (d) 0. Answer: (b) 4. p(2) = 4 − 8 + k = 0, so k = 4.

Section B — Short answer (3 marks) Q3. Find the value of k such that x = 2 is a zero of p(x) = 2x² − 3x + k, and verify your answer. Marking scheme: 1 mark for substituting x = 2: 2(4) − 3(2) + k = 0; 1 mark for 8 − 6 + k = 0 giving k = −2; 1 mark for verification: with k = −2, p(2) = 8 − 6 − 2 = 0 ✓.

Section C — Long answer (5 marks) Q4. If α and β are the zeroes of the polynomial 2x² − 5x + 3, find the values of: (i) 1/α + 1/β (2 marks) (ii) α³ + β³ (3 marks). Marking scheme: α + β = 5/2 and αβ = 3/2 (1 mark). (i) 1/α + 1/β = (α + β)/(αβ) = (5/2)/(3/2) = 5/3 (1 mark). (ii) α³ + β³ = (α + β)³ − 3αβ(α + β) (1 mark for the identity) = 125/8 − 3 × (3/2) × (5/2) (1 mark) = 125/8 − 90/8 = 35/8 (1 mark).

Going deeper: building a polynomial from its zeroes

The zero–coefficient relationships run backwards. Given zeroes α and β, the quadratic is x² − (α + β)x + αβ. For zeroes 3 and −2: x² − (1)x + (−6) = x² − x − 6. For a cubic with zeroes α, β, γ it is x³ − (α+β+γ)x² + (αβ+βγ+γα)x − αβγ; for zeroes 2, 3, −1 this gives x³ − 4x² + x + 6 — exactly the polynomial of Worked example 2, seen from the other direction. Any nonzero multiple (2x² − 2x − 12, say) has the same zeroes; CBSE accepts the monic form unless a leading coefficient is specified.

Going deeper: a full division, and what the remainder knows

Divide p(x) = x³ − 3x² + 5x − 3 by (x − 1) by long division: quotient x² − 2x + 3, remainder 0 — so (x − 1) is a factor, and 1 is a zero. Now divide the *same* p(x) by (x + 2): instead of redoing the division, use the remainder theorem: remainder = p(−2) = −8 − 12 − 10 − 3 = −33. One substitution replaces a page of long division. Moral: long division when you need the quotient; remainder theorem when you need only the remainder; factor theorem (remainder 0?) when you need to test a possible zero.

Going deeper: how many zeroes can a polynomial have?

A polynomial of degree n has at most n real zeroes — the graph of a quadratic crosses the axis at most twice, a cubic at most three times, and so on. Combined with the factor theorem this gives a proof strategy used in exemplar problems: to show a cubic is exactly (x − a)(x − b)(x − c) times a constant, verify three zeroes and compare leading coefficients; there is no room for anything else.

Final drills

Drill 1 — shifted zeroes. If α and β are the zeroes of x² − 6x + 8, find the quadratic whose zeroes are α + 2 and β + 2. First, α + β = 6 and αβ = 8 (indeed α, β = 2, 4). New sum = (α + 2) + (β + 2) = 10. New product = (α + 2)(β + 2) = αβ + 2(α + β) + 4 = 8 + 12 + 4 = 24. Answer: x² − 10x + 24 — whose zeroes are 4 and 6 ✓. Shifting both zeroes by k shifts the sum by 2k but the product in a way worth computing fresh each time.

Drill 2 — reciprocal zeroes. If one zero of x² − 5x + k is the reciprocal of the other, then α × (1/α) = 1 = k/1, so k = 1. Generally, "one zero is the reciprocal of the other" is code for *product of zeroes = 1*; "one zero is the negative of the other" is code for *sum = 0* (so b = 0). Translating verbal conditions into sum/product statements is the whole game in this chapter's harder questions.