SK Tuitions • Class 10 Mathematics

Polynomials: Virtual Teacher + Solved Question Bank

A complete CBSE-focused learning module covering concepts, graphs, zeroes, coefficient relations, transformed zeroes, polynomial division, competency questions, HOTS and Olympiad-style reasoning.

95Main question-bank entries

30MCQs

65Subjective & case-study sets

5Extra challenges

Part 1 • Virtual Teacher

Learn Polynomials from First Principles

Build the chapter in the same order a strong classroom lesson would: expression → polynomial → value → zeroes → graph → coefficient relations → formation → transformations → division → parameters.

1. Algebraic Expressions and the Meaning of a Polynomial

An algebraic expression combines numbers, variables and operations. A polynomial in one variable x is an expression of the form

a_nxⁿ + a_n−1xⁿ⁻¹ + ··· + a₁x + a₀

where the exponents of x are non-negative integers and the coefficients a₀, a₁, … are real numbers.

Variable

The letter whose value may change. In 3x²−5x+7, the variable is x.

Terms

3x², −5x and 7 are the three terms.

Coefficients

3 and −5 multiply powers of x. The sign is part of the coefficient.

Constant term

7 contains no variable; it is 7x⁰.

Leading coefficient

The coefficient of the highest-degree term; here it is 3.

Degree

The highest exponent with a non-zero coefficient; here it is 2.

Valid examples

p(x)=3x²−5x+7

q(x)=x³+2x²−x−4

Expressions that are not polynomials in x

1x+2 = x⁻¹+2

Negative exponent.

√x+3 = x^1/2+3

Fractional exponent.

x⁻²+5

Negative exponent.

Core rule

For a polynomial in x, every power of x must be 0, 1, 2, 3, … . Coefficients themselves may be positive, negative, fractional or irrational.

2. Degree and Types of Polynomials

To identify the degree, first write terms in descending powers and then locate the highest power with a non-zero coefficient.

5x⁴−3x²+x−7

Degree = 4

Type by degree	General idea	Example
Constant	Degree 0	5
Linear	Degree 1	2x+3
Quadratic	Degree 2	x²−5x+6
Cubic	Degree 3	x³−4x
Quartic (enrichment)	Degree 4	x⁴−5x²+4

Classification by number of non-zero terms is different: one term = monomial, two = binomial, three = trinomial. Thus x³−4x is both a cubic polynomial and a binomial.

Zero polynomial

The zero polynomial p(x)=0 has no highest non-zero term. In the usual Class 10 treatment, its degree is not defined.

3. Value of a Polynomial

To calculate p(a), substitute x=a everywhere and simplify carefully.

Example

If p(x)=2x²−3x+1, find p(2).

p(2)=2(2)²−3(2)+1

=8−6+1

Sign Alert

When substituting a negative number, brackets are essential.

(−2)²=4

but −2²=−(2²)=−4

So write p(−2) using (−2) in every substituted position.

4. Zeroes of a Polynomial

A number α is a zero of p(x) if

p(α)=0.

For p(x)=x−3:

p(3)=3−3=0

Therefore, 3 is a zero.

In this chapter, the words zero and root are commonly used for values of x that make the polynomial zero. When we write p(x)=0 as an equation, those same values are the solutions of the equation.

5. Zero of a Linear Polynomial

Let p(x)=ax+b, where a≠0. A zero makes p(x)=0:

ax+b=0

ax=−b

x=−ba

Zero of ax+b = −b/a

The result is derived from solving the corresponding linear equation; it is not a formula to memorise without meaning.

Visual Learning

Graphical Meaning of Zeroes

For y=p(x), every real zero occurs where the graph meets or touches the x-axis, because points on the x-axis have y=0.

Linear: one intersection

A non-constant linear polynomial has exactly one real zero.

Quadratic: two zeroes

Two intersections → two distinct real zeroes.

Quadratic: one repeated zero

One touch → one distinct real zero (repeated algebraically).

Quadratic: no real zero

No x-axis intersection → no real zero.

Cubic: three intersections

A cubic can have up to three real zeroes.

Zero Counter

Count only the marked x-axis intersections—not y-axis intersections or turning points.

Common Mistake

The y-intercept is the point obtained at x=0. It is not automatically a zero. A zero must satisfy y=p(x)=0 and therefore lies on the x-axis.

Core Board Concept

Quadratic Zeroes and Coefficients

6. Deriving the Relations

Consider p(x)=ax²+bx+c with zeroes α and β. A quadratic with those zeroes can be written as

p(x)=a(x−α)(x−β).

=a[x²−(α+β)x+αβ]

=ax²−a(α+β)x+aαβ.

Compare this term-by-term with ax²+bx+c:

−a(α+β)=b

α+β=−ba

aαβ=c

αβ=ca

Sum of zeroesα+β = −b/a

Product of zeroesαβ = c/a

Sign Check

The minus sign belongs to the sum formula. The product is c/a, not −c/a.

7. Finding Zeroes and Verifying the Relations

Example: Find the zeroes of 2x²−7x+3 and verify the relations.

2x²−7x+3

=2x²−6x−x+3

=2x(x−3)−1(x−3)

=(2x−1)(x−3).

Zeroes: α=12, β=3.

Sum check

α+β=12+3=72

−ba=−−72=72

Product check

αβ=12×3=32

ca=32

Verified.

8. Expressions Involving the Zeroes

When only a symmetric expression in α and β is required, coefficient relations are often faster than solving for the roots.

α²+β²=(α+β)²−2αβ

1α+1β=α+βαβ

provided αβ≠0

α³+β³=(α+β)³−3αβ(α+β)

αβ+βα=α²+β²αβ

provided αβ≠0

Why this step?

For α²+β², the values α+β and αβ come directly from a, b and c. Calculating α and β separately can waste time and introduce surd arithmetic.

HOTS / Olympiad Insight

Higher powers such as α⁴+β⁴ can be built from lower symmetric expressions. For example, α⁴+β⁴=(α²+β²)²−2(αβ)².

9. Forming a Polynomial from Given Zeroes

If the zeroes are α and β, then every non-zero scalar multiple

k(x−α)(x−β), k≠0

has the same zeroes. The monic choice takes k=1:

x²−(α+β)x+αβ.

If the zeroes are 2 and 3:

(x−2)(x−3)=x²−5x+6.

Other valid polynomials with the same zeroes include 2x²−10x+12 and −3x²+15x−18.

From sum S and product P

Monic polynomial = x²−Sx+P.

Irrational-zero example: zeroes 2+√3 and 2−√3 occur as a conjugate pair.

Sum=(2+√3)+(2−√3)=4.

Product=(2+√3)(2−√3)=4−3=1.

Therefore the monic polynomial is x²−4x+1.

10. Transformed Zeroes

Do not jump directly to expansion. First compute the new sum and new product.

Suppose α, β are old zeroes and the new zeroes are α+1 and β+1.

New Sum

(α+1)+(β+1)=α+β+2.

New Product

(α+1)(β+1)=αβ+α+β+1.

Then form

x²−(new sum)x+(new product).

New zeroes	New sum	New product
α+1, β+1	S+2	P+S+1
2α, 2β	2S	4P
1/α, 1/β	S/P	1/P
α², β²	S²−2P	P²

Here S=α+β and P=αβ. Reciprocal formulas require P≠0.

11. Cubic Relations — Enrichment / Olympiad Foundation

This extension is useful for higher-level reasoning and should not be confused with the core Class 10 board requirement on quadratic zeroes.

For ax³+bx²+cx+d with zeroes α, β, γ:

α+β+γ = −b/a

αβ+βγ+γα = c/a

αβγ = −d/a

Method Module

Polynomial Division Algorithm

12. Meaning of the Division Algorithm

If p(x) is divided by a non-zero polynomial g(x), then

p(x)=g(x)q(x)+r(x)

p(x) — dividend
g(x) — divisor
q(x) — quotient
r(x) — remainder

Either r(x)=0, or

deg r(x) < deg g(x).

13. Polynomial Long Division — Step by Step

Example: Divide 2x³+3x²−11x−6 by x−2.

Dividend: 2x³+3x²−11x−6

Divisor: x−2

2x³+3x²−11x−6 − (2x³−4x²) 7x²−11x − (7x²−14x) 3x−6 − (3x−6) 0

Divide the first term of the current dividend by the first term of the divisor.
Write that term in the quotient.
Multiply the whole divisor by the quotient term.
Subtract carefully, preserving signs.
Bring down the next term and repeat.
Stop when the remainder has lower degree than the divisor.

Quotient = 2x²+7x+3, Remainder = 0.

Verification

(x−2)(2x²+7x+3)+0

=2x³+3x²−11x−6.

Exam Tip

Before dividing, arrange terms in descending powers. If a power is missing, insert it with coefficient 0—for example 2x³−5x+3 becomes 2x³+0x²−5x+3.

14. Missing Coefficients and Parameter Problems

Direct substitution is often fastest. If 2 is a zero of 2x²+kx−6, find k.

p(2)=0

2(2)²+k(2)−6=0

8+2k−6=0

2k+2=0

k=−1.

Use zero relations when a relationship between roots is given. If one root is twice the other, set the roots α and 2α, then use both

α+2α=−ba

2α²=ca.

Why this step?

A known numerical zero gives an immediate equation p(a)=0. A ratio, sum, product or transformation of roots usually points to the zero-coefficient relations.

15. Polynomial vs Quadratic Equation

Polynomial expression

p(x)=x²−5x+6

This names an expression/function.

Quadratic equation

x²−5x+6=0

This asks for values of x that make the expression zero.

The zeroes of the polynomial are exactly the solutions of the corresponding equation p(x)=0.

16. Connecting Factorisation and the Graph

p(x)=x²−5x+6

=(x−2)(x−3).

Therefore p(2)=0 and p(3)=0. On the graph, those values appear as x-axis intersections (2,0) and (3,0).

Revision Module

Polynomials Formula & Method Revision Sheet

Linear polynomial ax+bZero = −b/a

Quadratic ax²+bx+cα+β = −b/a

Quadratic ax²+bx+cαβ = c/a

Polynomial from zeroesk(x−α)(x−β), k≠0

Monic from sum S, product Px²−Sx+P

Division algorithmp=gq+rr=0 or deg r<deg g

HOTS identityα²+β²=(α+β)²−2αβ

HOTS identity1/α+1/β=(α+β)/(αβ)αβ≠0

HOTS identityα³+β³=(α+β)³−3αβ(α+β)

Strategy

How to Choose the Correct Method

Asked whether a is a zero?

Compute p(a). If the result is 0, a is a zero.

Need actual quadratic zeroes?

Factorise when practical, then set each factor equal to zero.

Only sum/product requested?

Use −b/a and c/a. Do not solve the roots unnecessarily.

Need a polynomial from zeroes?

Use (x−α)(x−β), or x²−Sx+P for the monic form.

Zeroes are transformed?

Find their new sum and new product first; then form the polynomial.

Dividing polynomials?

Arrange descending powers, insert zero coefficients, divide, then verify p=gq+r.

Unknown coefficient + known zero?

Usually substitute the zero directly into p(x)=0.

Unknown coefficient + root relation?

Translate the relation into variables and use sum/product formulas.

Interactive Lab

Explore a Polynomial

Change a, b and c in p(x)=ax²+bx+c and watch the graph, discriminant and zeroes respond.

Coefficient a 1 Coefficient b -4 Coefficient c 3

Current polynomialx² − 4x + 3

Discriminant b²−4ac4

Real zeroes1, 3

α+β = −b/a4

αβ = c/a3

If a=0, the explorer temporarily becomes linear and the quadratic sum/product formulas are not applicable.

Error Prevention

10 Common Mistakes in Polynomials

1. Polynomial vs equation

p(x)=x²−5x+6 is an expression/function; x²−5x+6=0 is an equation.

2. Invalid powers

x⁻¹ and x^1/2 make an expression non-polynomial in x.

3. Wrong degree

Degree is the highest exponent with a non-zero coefficient, not the number of terms.

4. Negative substitution

Use brackets: (−2)²=4.

5. Zero vs y-intercept

Zeroes come from x-axis intersections, not the y-axis.

6. Wrong sum sign

For ax²+bx+c, α+β=−b/a.

7. Wrong product sign

αβ=c/a, not −c/a.

8. Solving roots unnecessarily

Use coefficient relations directly when only symmetric expressions are needed.

9. Missing zero coefficients

Write x³−5x+2 as x³+0x²−5x+2 before long division.

10. Transformed roots

Compute the new sum and product carefully before forming the new polynomial.

Solved Practice

95-Entry Polynomials Question Bank

Difficulty rises from foundation to Olympiad-style reasoning. Use filters to focus on a question type or concept. Every solution remains in the HTML and can be opened independently.

Solutions Viewed 0 / 95

Section A

30 MCQs • 1 Mark Each

1 MarkLevel 1NCERT TypeDefinition

Which of the following is a polynomial in x?

A3x² − 5x + 7

B1x + 2

C√x + 3

Dx⁻² + 5

View Solution

Correct answer: A — 3x² − 5x + 7

3x² – 5x + 7 has powers 2, 1 and 0, all non-negative integers.

The other expressions contain either a negative power or a fractional power of x.

1 MarkLevel 1NCERT TypeDegree

The degree of 5x⁴ − 3x² + x − 7 is:

View Solution

Correct answer: C — 4

Highest power of x with a non-zero coefficient = 4.

1 MarkLevel 1NCERT TypeCoefficients

In 3x³ − 7x² + 4x − 9, the coefficient of x² is:

B−7

D−9

View Solution

Correct answer: B — −7

The sign belongs to the coefficient.

Coefficient of x² = −7.

1 MarkLevel 1NCERT TypeValue of Polynomial

If p(x)=2x²−3x+1, then p(−2) equals:

B−1

D15

View Solution

Correct answer: D — 15

p(−2)=2(−2)²−3(−2)+1

=2(4)+6+1

=15

1 MarkLevel 1NCERT TypeLinear Zero

The zero of 3x−12 is:

A−4

D12

View Solution

Correct answer: B — 4

3x−12=0

3x=12

x=4

1 MarkLevel 1NCERT TypeZeroes

Which number is a zero of x²−5x+6?

View Solution

Correct answer: A — 2

p(2)=2²−5(2)+6=4−10+6=0

Therefore 2 is a zero.

1 MarkLevel 2Graph BasedGraphical Zeroes

The graph shown intersects the x-axis at two distinct points. How many real zeroes does the represented polynomial have?

View Solution

Correct answer: C — 2

A real zero is the x-coordinate of an x-axis intersection. The curve has two such intersections.

1 MarkLevel 2CBSE PatternCoefficient Relations

For 2x²−7x+3, the sum of its zeroes is:

A−7/2

B72

C32

View Solution

Correct answer: B — 7/2

α+β=−ba

=−−72=72

1 MarkLevel 2CBSE PatternCoefficient Relations

If α and β are the zeroes of 2x²−5x−3, then αβ is:

A32

B52

C−52

D−32

View Solution

Correct answer: D — −3/2

αβ=ca=−32

Q10

1 MarkLevel 2NCERT Exemplar TypeFormation of Polynomial

A monic quadratic polynomial whose zeroes are 2 and 3 is:

Ax²−5x+6

Bx²+5x+6

Cx²−x−6

Dx²+x−6

View Solution

Correct answer: A — x² − 5x + 6

p(x)=(x−2)(x−3)

=x²−5x+6

Q11

1 MarkLevel 3HOTSTransformed Zeroes

The zeroes of x²−5x+6 are α, β. A polynomial with zeroes 1/α, 1/β is:

Ax²−5x+1

B6x²+5x+1

C6x²−5x+1

Dx²−6x+5

View Solution

Correct answer: C — 6x² − 5x + 1

α+β=5, αβ=6

1α+1β=α+βαβ=56

1αβ=16

x²−56x+16=0

Multiplying by 6: 6x²−5x+1

Q12

1 MarkLevel 2NCERT TypePolynomial Division

If a cubic polynomial is divided by a quadratic polynomial, the remainder can have degree at most:

View Solution

Correct answer: B — 1

The remainder must have degree strictly less than the degree of the divisor.

deg(divisor)=2 ⇒ deg(remainder)<2

Maximum possible degree = 1.

Q13

1 MarkLevel 2CBSE PatternParameter

If 2 is a zero of 2x²+kx−6, then k equals:

A−1

C−2

View Solution

Correct answer: A — −1

p(2)=0

2(2)²+2k−6=0

8+2k−6=0

2k+2=0 ⇒ k=−1

Q14

1 MarkLevel 2ConceptualDegree

At the usual Class 10 level, the degree of the zero polynomial is:

C−1

DNot defined

View Solution

Correct answer: D — Not defined

Every coefficient in the zero polynomial is zero, so there is no highest non-zero power. Its degree is therefore not defined in the usual school treatment.

Q15

1 MarkLevel 2ConceptualTypes of Polynomial

The expression x³−4x is best described as:

AA quadratic trinomial

BA linear binomial

CA cubic binomial

DA cubic monomial

View Solution

Correct answer: C — A cubic binomial

Its degree is 3, so it is cubic. It has two non-zero terms, so it is also a binomial.

Q16

1 MarkLevel 2NCERT Exemplar TypeFormation of Polynomial

If the sum and product of the zeroes of a monic quadratic polynomial are 6 and 8 respectively, the polynomial is:

Ax²+6x+8

Bx²−6x+8

Cx²−8x+6

Dx²+8x−6

View Solution

Correct answer: B — x² − 6x + 8

p(x)=x²−(sum)x+(product)

=x²−6x+8

Q17

1 MarkLevel 3HOTSExpressions in Zeroes

If α, β are zeroes of x²−5x+6, then α²+β² equals:

A25

B19

C11

D13

View Solution

Correct answer: D — 13

α+β=5, αβ=6

α²+β²=(α+β)²−2αβ

=25−12=13

Q18

1 MarkLevel 3HOTSExpressions in Zeroes

If α, β are zeroes of 2x²−5x+2, then 1/α+1/β is:

A52

B25

View Solution

Correct answer: A — 5/2

α+β=52, αβ=22=1

1α+1β=α+βαβ=52

Q19

1 MarkLevel 2Graph BasedGraphical Zeroes

A parabola touches the x-axis at exactly one point and turns back. The corresponding quadratic polynomial has:

ANo real zero

BOne distinct real zero

CTwo distinct real zeroes

DThree real zeroes

View Solution

Correct answer: B — One distinct real zero

The touching point gives one distinct x-coordinate at which p(x)=0. Algebraically, it is a repeated real zero.

Q20

1 MarkLevel 3Assertion–ReasonDefinition

Assertion (A): x⁻¹+2 is not a polynomial in x.
Reason (R): In a polynomial, every exponent of the variable must be a non-negative integer.

ABoth A and R are true, and R is the correct explanation of A

BBoth A and R are true, but R is not the correct explanation of A

CA is true but R is false

DA is false but R is true

View Solution

Correct answer: A — Both A and R are true, and R is the correct explanation of A

Both statements are true, and the reason directly explains the assertion because x⁻¹ has exponent −1.

Q21

1 MarkLevel 3HOTSTransformed Zeroes

The zeroes of x²−3x+2 are α and β. A monic polynomial whose zeroes are 2α and 2β is:

Ax²−3x+8

Bx²−6x+4

Cx²−6x+8

Dx²+6x+8

View Solution

Correct answer: C — x² − 6x + 8

α+β=3, αβ=2

New sum=2α+2β=2(3)=6

New product=(2α)(2β)=4(2)=8

Required polynomial=x²−6x+8

Q22

1 MarkLevel 2ConceptualPolynomial Division

Which statement is always true in the polynomial division algorithm p(x)=g(x)q(x)+r(x)?

Adeg r(x) > deg g(x)

Bdeg q(x) < deg r(x) always

Cr(x) must be a constant

Ddeg r(x) < deg g(x), unless r(x)=0

View Solution

Correct answer: D — deg r(x) < deg g(x), unless r(x)=0

Either r(x)=0, or the degree of the remainder is smaller than the degree of the divisor g(x).

Q23

1 MarkLevel 4HOTSParameter

One zero of kx²−3x+2 is twice the other. The value of k is:

A−1

View Solution

Correct answer: B — 1

Let the zeroes be α and 2α.

3α=3k ⇒ α=1k

2α²=2k

2(1k)²=2k

k=1

Q24

1 MarkLevel 3Olympiad EnrichmentCubic Relations

For ax³+bx²+cx+d with zeroes α, β, γ, which relation is correct?

Aα+β+γ=ba

Bαβ+βγ+γα=−ca

Cαβγ=−da

Dαβγ=da

View Solution

Correct answer: C — αβγ = −d/a

αβγ=−da

This is the cubic counterpart of comparing the constant term after expanding a(x−α)(x−β)(x−γ).

Q25

1 MarkLevel 2Graph BasedGraphical Zeroes

Which graph feature determines the zeroes of a polynomial?

Ax-axis intersections or touches

By-axis intersection only

Chighest point only

Dslope at x=0 only

View Solution

Correct answer: A — x-axis intersections or touches

At a zero, p(x)=0, so y=0. Thus the point must lie on the x-axis.

Q26

1 MarkLevel 3Error AnalysisCoefficient Relations

A student says that the sum of the zeroes of 4x²+3x−1 is 3/4. What is the correct sum?

A34

B−34

C−14

D14

View Solution

Correct answer: B — −3/4

α+β=−ba

=−34

The student forgot the negative sign.

Q27

1 MarkLevel 3CBSE PatternPolynomial Division

The remainder when x³−4x+3 is divided by x−1 is:

C−1

View Solution

Correct answer: A — 0

p(1)=1−4+3=0

Equivalently, direct division gives remainder 0.

Q28

1 MarkLevel 3ConceptualFormation of Polynomial

If x²−5x+6 has zeroes 2 and 3, which polynomial has exactly the same zeroes?

Ax²−5x+3

Bx²+5x+6

C3x²−5x+6

D−3x²+15x−18

View Solution

Correct answer: D — −3x² + 15x − 18

Any non-zero scalar multiple has the same zeroes.

−3(x²−5x+6)=−3x²+15x−18

Q29

1 MarkLevel 4HOTSTransformed Zeroes

The zeroes of x²−4x+1 are α and β. The monic polynomial whose zeroes are α+1 and β+1 is:

Ax²−6x+6

Bx²−4x+6

Cx²+6x+6

Dx²−6x+1

View Solution

Correct answer: A — x² − 6x + 6

α+β=4, αβ=1

New sum=(α+1)+(β+1)=6

New product=(α+1)(β+1)=1+4+1=6

Required polynomial=x²−6x+6

Q30

1 MarkLevel 5OlympiadExpressions in Zeroes

If α and β are the zeroes of x²−x−1, then α³+β³ equals:

View Solution

Correct answer: C — 4

α+β=1, αβ=−1

α³+β³=(α+β)³−3αβ(α+β)

=1³−3(−1)(1)=4

Section B

15 Two-Mark Questions

Q31

2 MarksLevel 1NCERT TypeDegree & Coefficients

For p(x)=7x⁵−4x³+2x−9, state (i) its degree and (ii) its constant term.

View Solution

Highest power with non-zero coefficient = 5.

Therefore, degree = 5.

Constant term = −9.

Q32

2 MarksLevel 1NCERT TypeValue of Polynomial

Evaluate p(−3) for p(x)=x²+4x−5.

View Solution

p(−3)=(−3)²+4(−3)−5

=9−12−5

=−8.

Q33

2 MarksLevel 1NCERT TypeChecking a Zero

Check whether 2 is a zero of p(x)=x³−3x²+4.

View Solution

p(2)=2³−3(2)²+4

=8−12+4

=0.

Hence, 2 is a zero of p(x).

Q34

2 MarksLevel 1NCERT TypeLinear Zero

Find the zero of 5x+7.

View Solution

5x+7=0

5x=−7

x=−75.

Q35

2 MarksLevel 2NCERT TypeQuadratic Zeroes

Find the zeroes of x²−7x+12.

View Solution

x²−7x+12

=x²−3x−4x+12

=x(x−3)−4(x−3)

=(x−3)(x−4).

(x−3)(x−4)=0

Therefore, x=3 or x=4.

Q36

2 MarksLevel 2CBSE PatternCoefficient Relations

The zeroes of x²−5x+4 are 1 and 4. Verify the sum and product relations.

View Solution

α=1, β=4, a=1, b=−5, c=4

α+β=1+4=5

−ba=−−51=5

Thus, α+β=−b/a.

αβ=(1)(4)=4

ca=41=4

Thus, αβ=c/a.

Q37

2 MarksLevel 2NCERT TypeFormation of Polynomial

Form a monic quadratic polynomial whose zeroes are −2 and 5.

View Solution

p(x)=(x−(−2))(x−5)

=(x+2)(x−5)

=x²−3x−10.

Q38

2 MarksLevel 2NCERT Exemplar TypeFormation of Polynomial

The sum of the zeroes of a quadratic polynomial is 7 and their product is 10. Form the monic polynomial.

View Solution

Required polynomial=x²−(sum)x+(product)

=x²−7x+10.

Q39

2 MarksLevel 2NCERT TypePolynomial Division

Divide x²+5x+6 by x+2.

View Solution

Dividend: x²+5x+6

Divisor: x+2

x²+5x+6− (x²+2x)3x+6− (3x+6)0

Quotient=x+3, Remainder=0.

Q40

2 MarksLevel 2CBSE PatternParameter

If 3 is a zero of x²+kx−12, find k.

View Solution

Since 3 is a zero, p(3)=0.

3²+3k−12=0

9+3k−12=0

3k−3=0

k=1.

Q41

2 MarksLevel 2Graph BasedGraphical Zeroes

Study the graph of the linear polynomial. State (i) the number of zeroes and (ii) the zero shown by the graph.

View Solution

The line meets the x-axis once.

Number of zeroes = 1.

x-coordinate of intersection = 2.

Therefore, the zero is 2.

Q42

2 MarksLevel 2ConceptualDefinition & Degree

From the following, identify all polynomials in x and state their degrees: 3x²+1, x⁻¹+4, 7−2x.

View Solution

3x²+1 is a polynomial; degree = 2.

x⁻¹+4 is not a polynomial because −1 is a negative exponent.

7−2x is a polynomial; degree = 1.

Q43

2 MarksLevel 2CBSE PatternCoefficient Relations

Without finding the zeroes, find their product for 3x²−10x+3.

View Solution

a=3, b=−10, c=3

αβ=ca=33=1.

Q44

2 MarksLevel 3HOTSExpressions in Zeroes

If α, β are the zeroes of 2x²+x−6, find 1/α+1/β without finding α and β.

View Solution

α+β=−12, αβ=−62=−3

1α+1β=α+βαβ

=−1/2−3=16.

Why this step?

The requested expression depends only on α+β and αβ, so solving separately for the roots would add unnecessary work.

Q45

2 MarksLevel 3Competency BasedDivision Algorithm

A polynomial is written as p(x)=(x−1)(x²+2x+3)+5. If p(x) is divided by x−1, state the quotient and remainder and justify that the division algorithm is satisfied.

View Solution

Comparing with p(x)=g(x)q(x)+r(x):

g(x)=x−1, q(x)=x²+2x+3, r(x)=5.

deg r=0<1=deg g.

Hence the representation satisfies the polynomial division algorithm.

Section C

15 Three-Mark Questions

Q46

3 MarksLevel 2CBSE PatternZeroes & Coefficients

Find the zeroes of 2x²−7x+3 and verify the relationship between the zeroes and coefficients.

View Solution

2x²−7x+3

=2x²−6x−x+3

=2x(x−3)−1(x−3)

=(2x−1)(x−3).

(2x−1)(x−3)=0

α=12, β=3.

Verification of sum

α+β=12+3=72

−ba=−−72=72.

Verification of product

αβ=12×3=32

ca=32.

Both relations are verified.

Q47

3 MarksLevel 2NCERT TypeZeroes & Coefficients

Find the zeroes of x²−8x+15 and verify their sum and product.

View Solution

x²−8x+15=(x−3)(x−5)

Zeroes: α=3, β=5.

α+β=3+5=8

−ba=−−81=8.

αβ=3×5=15

ca=151=15.

Hence both coefficient relations are verified.

Q48

3 MarksLevel 2CBSE PatternParameter

If −2 is a zero of 3x²+kx−2, find k. Then find the other zero.

View Solution

p(−2)=0

3(−2)²+k(−2)−2=0

12−2k−2=0

10−2k=0 ⇒ k=5.

Polynomial becomes 3x²+5x−2.

=(3x−1)(x+2).

Zeroes are −2 and 13.

Q49

3 MarksLevel 2NCERT Exemplar TypeFormation of Polynomial

Form a quadratic polynomial with zeroes 32 and −4, having integer coefficients and the smallest positive integral leading coefficient.

View Solution

p(x)=(x−32)(x+4)

=x²+52x−6.

Multiplying by 2 to clear fractions:

p(x)=2x²+5x−12.

The smallest positive integral leading coefficient is 2.

Q50

3 MarksLevel 3HOTSTransformed Zeroes

If α, β are the zeroes of x²−5x+6, form a monic quadratic polynomial whose zeroes are α+2 and β+2.

View Solution

α+β=5, αβ=6.

New sum

(α+2)+(β+2)=α+β+4=9.

New product

(α+2)(β+2)=αβ+2(α+β)+4

=6+10+4=20.

Required polynomial=x²−9x+20.

Q51

3 MarksLevel 3HOTSTransformed Zeroes

The zeroes of 2x²−7x+3 are α and β. Form a polynomial with integer coefficients whose zeroes are 1/α and 1/β.

View Solution

α+β=72, αβ=32.

New sum

1α+1β=α+βαβ=7/23/2=73.

New product

1αβ=23.

x²−73x+23.

Multiplying by 3: 3x²−7x+2.

Q52

3 MarksLevel 3Graph BasedGraphical Zeroes

The graph represents p(x)=x²−4x+3.

From the graph, state the number of real zeroes and their values. Then verify them algebraically.

View Solution

The graph meets the x-axis at x=1 and x=3.

Therefore, number of real zeroes = 2.

Algebraically: x²−4x+3=(x−1)(x−3).

(x−1)(x−3)=0 ⇒ x=1 or x=3.

The graphical and algebraic results agree.

Q53

3 MarksLevel 3CBSE PatternPolynomial Division

Divide x³−6x²+11x−6 by x−1 and verify the division algorithm.

View Solution

Dividend: x³−6x²+11x−6

Divisor: x−1

x³−6x²+11x−6 − (x³−x²) −5x²+11x − (−5x²+5x) 6x−6 − (6x−6) 0

Quotient=x²−5x+6, Remainder=0.

Verification

(x−1)(x²−5x+6)+0

=x³−6x²+11x−6.

Hence the division algorithm is verified.

Q54

3 MarksLevel 3CBSE PatternParameter

The sum of the zeroes of kx²−10x+4 is 5. Find k and the product of the zeroes.

View Solution

α+β=−ba=10k.

10k=5 ⇒ 10=5k ⇒ k=2.

αβ=ca=42=2.

Q55

3 MarksLevel 3Competency BasedZeroes

The zeroes of a monic quadratic polynomial differ by 3, their sum is 7 and their product is 10. Determine the zeroes and hence write the polynomial.

View Solution

Let the zeroes be α and β with α+β=7, αβ=10.

Numbers with sum 7 and product 10 are 2 and 5.

|5−2|=3, so the difference condition is also satisfied.

Zeroes: 2 and 5.

Polynomial=(x−2)(x−5)=x²−7x+10.

Q56

3 MarksLevel 3HOTSExpressions in Zeroes

If α and β are the zeroes of 3x²−5x+1, find α²+β² without solving the quadratic.

View Solution

α+β=53, αβ=13.

α²+β²=(α+β)²−2αβ

=(53)²−2(13)

=259−69=199.

Why this step?

The identity converts the expression into the sum and product of the zeroes, which are available immediately from the coefficients.

Q57

3 MarksLevel 4HOTSFormation of Polynomial

If α and β are zeroes of x²−5x+6, form the monic quadratic polynomial whose zeroes are α+β and αβ.

View Solution

α+β=5, αβ=6.

The new zeroes are therefore 5 and 6.

Required polynomial=(x−5)(x−6)

=x²−11x+30.

Q58

3 MarksLevel 3CBSE PatternPolynomial Division

Divide 2x³−5x+3 by x+2. Show the missing x² term explicitly.

View Solution

Write dividend as 2x³+0x²−5x+3.

2x³+0x²−5x+3 − (2x³+4x²) −4x²−5x − (−4x²−8x) 3x+3 − (3x+6) −3

Quotient=2x²−4x+3, Remainder=−3.

Q59

3 MarksLevel 4Olympiad EnrichmentCubic Relations

The cubic polynomial 2x³−3x²−11x+6 has zeroes 3, 12 and −2. Verify all three cubic zero-coefficient relations.

View Solution

α+β+γ=3+12−2=32.

−ba=−−32=32.

αβ+βγ+γα=32−1−6=−112.

ca=−112.

αβγ=3×12×(−2)=−3.

−da=−62=−3.

All three relations are verified. This is enrichment beyond the core quadratic relation.

Q60

3 MarksLevel 4Competency BasedPolynomial Model

A rectangular display has length x+2 metres and width x−1 metres. (i) Write its area polynomial. (ii) Find the zeroes of that polynomial. (iii) Which zero can occur at the boundary of a physically meaningful width?

View Solution

Area=(x+2)(x−1)

=x²+x−2.

x²+x−2=(x+2)(x−1).

Zeroes: x=−2 and x=1.

For physical dimensions, x+2≥0 and x−1≥0; the relevant boundary for non-negative width is x=1. The algebraic zero x=−2 makes the length zero but the width negative.

Section D

20 Four-Mark Questions

Q61

4 MarksLevel 3CBSE PatternExpressions in Zeroes

If α and β are the zeroes of 2x²−5x−3, find: (i) α²+β² and (ii) 1/α+1/β, without finding α and β.

View Solution

α+β=52, αβ=−32.

(i) α²+β²

α²+β²=(α+β)²−2αβ

=(52)²−2(−32)

=254+3=374.

(ii) Reciprocal sum

1α+1β=α+βαβ

=5/2−3/2=−53.

Q62

4 MarksLevel 3CBSE PatternTransformed Zeroes

The zeroes of 3x²−7x+2 are α and β. Form a quadratic polynomial with integer coefficients whose zeroes are α+1 and β+1.

View Solution

α+β=73, αβ=23.

New sum

(α+1)+(β+1)=73+2=133.

New product

(α+1)(β+1)=αβ+(α+β)+1

=23+73+1=4.

Monic form: x²−133x+4.

Multiplying by 3: 3x²−13x+12.

Q63

4 MarksLevel 3NCERT Exemplar TypeTransformed Zeroes

If α, β are the zeroes of x²−4x+2, form the monic quadratic polynomial whose zeroes are 2α−1 and 2β−1.

View Solution

α+β=4, αβ=2.

New sum

(2α−1)+(2β−1)=2(α+β)−2=8−2=6.

New product

(2α−1)(2β−1)=4αβ−2(α+β)+1

=8−8+1=1.

Required polynomial=x²−6x+1.

Q64

4 MarksLevel 3HOTSTransformed Zeroes

The zeroes of 2x²−3x−2 are α and β. Form a polynomial with integer coefficients whose zeroes are 1/α and 1/β. Verify your result by first factorising the original polynomial.

View Solution

α+β=32, αβ=−1.

Using relations

New sum=α+βαβ=3/2−1=−32.

New product=1/(αβ)=−1.

x²+32x−1.

Multiplying by 2: 2x²+3x−2.

Verification by factorisation

2x²−3x−2=(2x+1)(x−2).

Original zeroes: −12, 2.

Reciprocals: −2, 12.

(x+2)(x−12)=0 ⇒ 2x²+3x−2.

Q65

4 MarksLevel 4HOTSParameter

The zeroes of kx²−7x+6 are in the ratio 2:3. Find k.

View Solution

Let the zeroes be 2t and 3t.

2t+3t=5t=7k

t=75k.

(2t)(3t)=6t²=6k

t²=1k.

(75k)²=1k

4925k²=1k

49=25k

k=4925.

Why this step?

The ratio describes the two roots but not their actual values, so introduce a common scale t and use both the sum and product relations.

Q66

4 MarksLevel 3CBSE PatternParameter

The zeroes of x²−kx+12 differ by 1. Find k if both zeroes are positive.

View Solution

Let the positive zeroes be α and β.

α+β=k, αβ=12, |α−β|=1.

(α−β)²=(α+β)²−4αβ

1=k²−48

k²=49

k=±7.

Because both zeroes are positive, their sum must be positive. Hence k=7.

Q67

4 MarksLevel 4HOTSMissing Coefficients

The monic cubic polynomial p(x)=x³+ax²+bx+6 has zeroes 1 and −2. Find a and b.

View Solution

Let the third zero be γ.

Enrichment: for a monic cubic, αβγ=−constant term.

(1)(−2)γ=−6

−2γ=−6 ⇒ γ=3.

Sum of zeroes=1−2+3=2=−a

Therefore, a=−2.

Sum of pairwise products=(1)(−2)+(−2)(3)+(3)(1)

=−2−6+3=−5=b.

Therefore, b=−5.

p(x)=x³−2x²−5x+6.

Q68

4 MarksLevel 3CBSE PatternPolynomial Division

Divide 2x⁴−3x³−11x²+12x+9 by x²−3. State the quotient and remainder.

View Solution

Dividend: 2x⁴−3x³−11x²+12x+9

Divisor: x²−3

2x⁴−3x³−11x²+12x+9 − (2x⁴−6x²) −3x³−5x²+12x+9 − (−3x³+9x) −5x²+3x+9 − (−5x²+15) 3x−6

Quotient=2x²−3x−5.

Remainder=3x−6.

deg(remainder)=1<2=deg(divisor).

Q69

4 MarksLevel 3Graph BasedGraphical Zeroes

The graph below represents a cubic polynomial proportional to (x+2)(x−1)(x−3).

(i) State its three zeroes. (ii) Which axis intersections determine them? (iii) What is the maximum number of real zeroes a cubic can have?

View Solution

The x-axis intersections occur at x=−2, x=1 and x=3.

Therefore, the zeroes are −2, 1 and 3.

Only x-axis intersections determine zeroes because p(x)=0 means y=0.

A cubic polynomial can have at most 3 real zeroes.

Q70

4 MarksLevel 4HOTSMissing Coefficients

Find k and m if p(x)=x³+kx²+mx−6 is divisible by both x−2 and x+1.

View Solution

Since x−2 is a factor, p(2)=0:

8+4k+2m−6=0

2k+m=−1. …(1)

Since x+1 is a factor, p(−1)=0:

−1+k−m−6=0

k−m=7. …(2)

From (2), m=k−7.

Substitute in (1): 2k+k−7=−1

3k=6 ⇒ k=2.

m=2−7=−5.

Thus k=2, m=−5.

Q71

4 MarksLevel 4HOTSExpressions in Zeroes

If α, β are the zeroes of 2x²−7x+3, find α/β+β/α without finding the zeroes.

View Solution

α+β=72, αβ=32.

αβ+βα=α²+β²αβ.

α²+β²=(α+β)²−2αβ

=494−3=374.

Therefore, α²+β²αβ=37/43/2=376.

Q72

4 MarksLevel 4OlympiadTransformed Zeroes

The zeroes of 3x²−5x+1 are α and β. Form a quadratic polynomial with integer coefficients whose zeroes are α² and β².

View Solution

S=α+β=53, P=αβ=13.

New sum

α²+β²=S²−2P

=259−69=199.

New product

α²β²=P²=19.

x²−199x+19.

Multiplying by 9: 9x²−19x+1.

Q73

4 MarksLevel 4HOTSFormation of Polynomial

If α and β are the zeroes of 2x²−5x+2, form a polynomial with integer coefficients whose zeroes are α+β and αβ.

View Solution

α+β=52, αβ=1.

New zeroes: 52 and 1.

New sum=72.

New product=52.

x²−72x+52.

Multiplying by 2: 2x²−7x+5.

Q74

4 MarksLevel 4HOTSParameter & Transformed Zeroes

The zeroes α, β of x²−6x+m satisfy α²+β²=20. Find m and then form a polynomial whose zeroes are 1/α and 1/β.

View Solution

α+β=6, αβ=m.

α²+β²=(α+β)²−2αβ

20=36−2m

2m=16 ⇒ m=8.

Reciprocal zeroes

New sum=α+βαβ=68=34.

New product=1αβ=18.

x²−34x+18.

Multiplying by 8: 8x²−6x+1.

Q75

4 MarksLevel 5OlympiadParameter

One zero of (k+1)x²−6x+k is twice the other. Find all possible values of k.

View Solution

Let the zeroes be t and 2t.

3t=6k+1 ⇒ t=2k+1.

2t²=kk+1.

2(2k+1)²=kk+1

8(k+1)²=kk+1

8=k(k+1)

k²+k−8=0.

k=−1±√332.

Neither value equals −1, so both keep the polynomial quadratic.

Q76

4 MarksLevel 5OlympiadTransformed Zeroes

The zeroes α, β of a quadratic polynomial satisfy α+β=5 and αβ=6. Form a polynomial whose zeroes are α/β and β/α.

View Solution

New sum

αβ+βα=α²+β²αβ.

α²+β²=5²−2(6)=13.

New sum=136.

New product

(αβ)(βα)=1.

x²−136x+1.

Multiplying by 6: 6x²−13x+6.

Q77

4 MarksLevel 3CBSE PatternDivision Algorithm

For the division in Q68, verify explicitly that p(x)=g(x)q(x)+r(x).

View Solution

g(x)=x²−3

q(x)=2x²−3x−5

r(x)=3x−6.

(x²−3)(2x²−3x−5)+(3x−6)

=2x⁴−3x³−5x²−6x²+9x+15+3x−6

=2x⁴−3x³−11x²+12x+9

=p(x).

Hence the division algorithm is verified.

Q78

4 MarksLevel 4HOTSMissing Coefficients

Find a and b if 2x³+ax²+bx+6 is exactly divisible by x²−x−2. Also find the quotient.

View Solution

x²−x−2=(x−2)(x+1).

Therefore p(2)=0 and p(−1)=0.

p(2)=16+4a+2b+6=0

2a+b=−11. …(1)

p(−1)=−2+a−b+6=0

a−b=−4. …(2)

From (2), b=a+4.

2a+a+4=−11 ⇒ 3a=−15 ⇒ a=−5.

b=−1.

p(x)=2x³−5x²−x+6.

=(x²−x−2)(2x−3).

Quotient=2x−3.

Q79

4 MarksLevel 4Graph BasedGraphical Reasoning

Consider p(x)=x²+2x+5.

(i) How many real zeroes are visible? (ii) Show algebraically why the graph cannot meet the x-axis. (iii) State the y-intercept and explain why it is not a zero.

View Solution

From the graph, there are 0 real zeroes.

p(x)=x²+2x+5

=(x+1)²+4.

Since (x+1)²≥0, p(x)≥4>0 for every real x. Hence the graph cannot meet the x-axis.

At x=0, p(0)=5, so the y-intercept is (0,5).

A zero requires y=0, so a y-axis intersection with y=5 is not a zero.

Q80

4 MarksLevel 5OlympiadTransformed Zeroes

A monic quadratic polynomial has zeroes α and β satisfying α+β=5 and α²+β²=13. Find αβ and form the monic polynomial whose zeroes are α³ and β³.

View Solution

α²+β²=(α+β)²−2αβ

13=25−2αβ

2αβ=12 ⇒ αβ=6.

New sum

α³+β³=(α+β)³−3αβ(α+β)

=125−3(6)(5)=125−90=35.

New product

α³β³=(αβ)³=6³=216.

Required polynomial=x²−35x+216.

Section E

10 Five-Mark Questions

Q81

5 MarksLevel 4HOTSRoots + Transformed Roots

The zeroes of 2x²−7x+3 are α and β. (i) Find the zeroes. (ii) Form a polynomial with integer coefficients whose zeroes are 1/α+1 and 1/β+1. (iii) Verify the transformed zeroes directly.

View Solution

2x²−7x+3=(2x−1)(x−3).

α=12, β=3.

New sum

(1α+1)+(1β+1)=α+βαβ+2.

α+β=72, αβ=32.

New sum=7/23/2+2=73+2=133.

New product

(1α+1)(1β+1)=1αβ+α+βαβ+1

=23+73+1=4.

x²−133x+4.

Required integer-coefficient polynomial: 3x²−13x+12.

Direct verification

For α=12: 1α+1=2+1=3.

For β=3: 1β+1=13+1=43.

(x−3)(x−43)=0 ⇒ 3x²−13x+12.

Q82

5 MarksLevel 4CBSE PatternParameter + Polynomial Formation

A quadratic polynomial kx²−6x+m has two positive zeroes that differ by 1 and whose product is 2. Find k and m, find the zeroes, and write the polynomial in factorised form.

View Solution

Let the zeroes be α and β.

αβ=2, |α−β|=1.

(α+β)²=(α−β)²+4αβ

=1+8=9.

Because both zeroes are positive, α+β=3.

For kx²−6x+m: α+β=6k.

6k=3 ⇒ k=2.

αβ=mk=2 ⇒ m=2k=4.

Polynomial=2x²−6x+4.

=2(x²−3x+2)

=2(x−1)(x−2).

Therefore, the zeroes are 1 and 2.

Q83

5 MarksLevel 4HOTSPolynomial Division + Zeroes

Divide x⁴−5x²+4 by x²−1. Verify the division algorithm and hence find all real zeroes of the quartic polynomial.

View Solution

Write the dividend as x⁴+0x³−5x²+0x+4.

x⁴−5x²+4 − (x⁴−x²) −4x²+4 − (−4x²+4) 0

Quotient=x²−4, Remainder=0.

Verification

(x²−1)(x²−4)+0

=x⁴−5x²+4.

Zeroes

p(x)=(x²−1)(x²−4)

=(x−1)(x+1)(x−2)(x+2).

Real zeroes: −2, −1, 1, 2.

Q84

5 MarksLevel 4Olympiad EnrichmentCubic Relations

A monic cubic polynomial has the form p(x)=x³+ax²+bx−12. One zero is 2 and the sum of the other two zeroes is 5. Find a and b, determine all three zeroes, and factorise p(x).

View Solution

Let the zeroes be 2, β and γ.

β+γ=5.

For a monic cubic x³+ax²+bx−12, the product of the three zeroes is 12.

2βγ=12 ⇒ βγ=6.

β and γ have sum 5 and product 6.

Therefore, β=2 and γ=3.

All zeroes are 2, 2 and 3.

Sum=2+2+3=7=−a ⇒ a=−7.

Pairwise product=2·2+2·3+2·3=4+6+6=16=b.

p(x)=x³−7x²+16x−12.

=(x−2)²(x−3).

Q85

5 MarksLevel 5OlympiadRational Transformed Zeroes

If α and β are the zeroes of 3x²−4x−2, form a quadratic polynomial with integer coefficients whose zeroes are (α+1)/(α−1) and (β+1)/(β−1).

View Solution

S=α+β=43, P=αβ=−23.

New sum

α+1α−1+β+1β−1

=(α+1)(β−1)+(β+1)(α−1)(α−1)(β−1).

Numerator=2αβ−2=2P−2.

Denominator=αβ−α−β+1=P−S+1.

New sum=2P−2P−S+1

=−4/3−2−2/3−4/3+1=−10/3−1=103.

New product

(α+1)(β+1)(α−1)(β−1)=P+S+1P−S+1

=−2/3+4/3+1−1=5/3−1=−53.

Required monic form: x²−103x−53.

Multiplying by 3: 3x²−10x−5.

Why this step?

For complicated transformed roots, calculate the new sum and new product symbolically first; only then substitute S=α+β and P=αβ.

Q86

5 MarksLevel 5OlympiadHigher Powers of Zeroes

If α and β are the zeroes of x²−5x+3: (i) find α⁴+β⁴; (ii) form the monic quadratic polynomial whose zeroes are α²+1 and β²+1.

View Solution

S=α+β=5, P=αβ=3.

α²+β²=S²−2P=25−6=19.

(i) Fourth powers

α⁴+β⁴=(α²+β²)²−2α²β²

=19²−2(3²)

=361−18=343.

(ii) New polynomial

New sum=(α²+1)+(β²+1)=19+2=21.

New product=(α²+1)(β²+1)

=α²β²+(α²+β²)+1

=9+19+1=29.

Required polynomial=x²−21x+29.

Q87

5 MarksLevel 5OlympiadParameter + Transformed Zeroes

The zeroes α and β of kx²−7x+6 are in the ratio 2:3. Find k, determine α and β, and form a polynomial with integer coefficients whose zeroes are α+1 and β+1.

View Solution

Let α=2t and β=3t.

5t=7k ⇒ t=75k.

6t²=6k ⇒ t²=1k.

(75k)²=1k ⇒ k=4925.

t=75×49/25=57.

α=107, β=157.

Shifted zeroes

α+1=177, β+1=227.

New sum=397, new product=37449.

x²−397x+37449.

Multiplying by 49: 49x²−273x+374.

Q88

5 MarksLevel 4Graph BasedMultiplicity & Graph

A polynomial is p(x)=(x+2)(x−1)²(x−3).

(i) List its distinct real zeroes. (ii) At which zero does the graph touch rather than cross the x-axis? (iii) State the degree. (iv) Expand p(x).

View Solution

From the factorised form, the distinct zeroes are −2, 1 and 3.

The factor (x−1)² has even multiplicity, so the graph touches the x-axis at x=1.

Total degree=1+2+1=4.

(x−1)²=x²−2x+1.

(x+2)(x−3)=x²−x−6.

p(x)=(x²−2x+1)(x²−x−6)

=x⁴−3x³−3x²+11x−6.

This multiplicity discussion is useful enrichment for interpreting polynomial graphs.

Q89

5 MarksLevel 4HOTSDivision Algorithm + Missing Coefficients

A polynomial p(x)=2x⁴+ax³+bx²+cx+6 satisfies p(x)=(x²−x−2)(2x²+x−3)+4x. Find a, b and c and verify the division algorithm.

View Solution

(x²−x−2)(2x²+x−3)

=2x⁴+x³−3x²−2x³−x²+3x−4x²−2x+6

=2x⁴−x³−8x²+x+6.

Adding remainder 4x:

p(x)=2x⁴−x³−8x²+5x+6.

Therefore, a=−1, b=−8, c=5.

Verification

Divisor=x²−x−2, quotient=2x²+x−3, remainder=4x.

deg(4x)=1<2=deg(divisor).

Divisor × Quotient + Remainder reproduces p(x), so the algorithm is verified.

Q90

5 MarksLevel 5OlympiadMultiple Root Relations

The non-zero zeroes α, β of a monic quadratic polynomial satisfy α+β=5 and 1/α+1/β=5/6. (i) Find αβ. (ii) Form the original polynomial. (iii) Find α³+β³. (iv) Form a polynomial whose zeroes are α/β and β/α.

View Solution

1α+1β=α+βαβ.

56=5αβ ⇒ αβ=6.

Original polynomial

x²−5x+6.

Cubic power sum

α³+β³=(α+β)³−3αβ(α+β)

=125−90=35.

Ratio zeroes

New sum=α²+β²αβ.

α²+β²=25−12=13.

New sum=136, new product=1.

x²−136x+1.

Multiplying by 6: 6x²−13x+6.

Section F

5 Case Studies • 4 Sub-Questions Each

Q91

4 MarksLevel 3Case StudyArea Polynomial

Case Study 1 — Rectangular Garden

A school designs a rectangular garden whose length is x+4 metres and width is x−1 metres. Its area is represented by A(x).

x + 4x − 1Garden

(a) Write A(x) in expanded polynomial form.

(b) Find the zeroes of A(x).

(c) Verify the sum and product of the zeroes using coefficients.

(d) Which zero represents the boundary where the width becomes zero? Explain why both algebraic zeroes are not equally useful in the physical model.

View Solution

(a) Area polynomial

A(x)=(x+4)(x−1)

=x²+3x−4.

(b) Zeroes

A(x)=(x+4)(x−1)=0

x=−4 or x=1.

(c) Coefficient verification

a=1, b=3, c=−4.

Sum of zeroes=−4+1=−3.

−ba=−31=−3.

Product=(−4)(1)=−4.

ca=−41=−4.

(d) Interpretation

The width x−1 becomes zero at x=1. At x=−4 the length becomes zero, but the width is −5 m, which is not a valid physical width. Algebraic zeroes describe the polynomial; a real-world model can impose extra domain restrictions.

Q92

4 MarksLevel 3Case StudyBridge Arch Graph

Case Study 2 — Decorative Arch

The height profile of a decorative arch is modelled, on a chosen coordinate grid, by h(x)=−(x−2)(x−8)=−x²+10x−16.

(a) Find h(5).

(b) State the zeroes from the factorised form and graph.

(c) Verify their sum and product from the coefficients of −x²+10x−16.

(d) What horizontal distance separates the two x-axis contact points?

View Solution

(a) Height at x=5

h(5)=−(5−2)(5−8)

=−(3)(−3)=9.

(b) Zeroes

−(x−2)(x−8)=0 ⇒ x=2 or x=8.

(c) Relations

a=−1, b=10, c=−16.

Sum=2+8=10.

−ba=−10−1=10.

Product=2×8=16.

ca=−16−1=16.

(d) Horizontal separation

8−2=6 units.

Q93

4 MarksLevel 3Case StudyPackaging Polynomial

Case Study 3 — Packaging Design Trial

During a packaging design trial, a dimension score is modelled by P(x)=x²−7x+12. The design team tests several values of x.

x	2	3	4	5
P(x)	2	0	0	2

(a) Verify P(5).

(b) Use the table and factorisation to identify the zeroes.

(c) Find the sum and product of the zeroes directly from the coefficients.

(d) If every zero is increased by 1 for a revised design, form the new monic quadratic polynomial.

View Solution

(a)

P(5)=25−35+12=2.

(b)

P(x)=x²−7x+12=(x−3)(x−4).

Zeroes are 3 and 4, matching the table entries where P(x)=0.

(c)

Sum=−ba=−−71=7.

Product=ca=121=12.

(d) Shifted zeroes

New zeroes: 4 and 5.

New polynomial=(x−4)(x−5)=x²−9x+20.

Q94

4 MarksLevel 4Case StudyProfit Curve

Case Study 4 — School Fair Profit Model

A simplified profit index for a school fair stall is R(x)=−x²+9x−20, where the zero level represents break-even.

(a) Factorise R(x).

(b) Identify the two break-even x-values.

(c) Verify the sum and product of the zeroes from the coefficients.

(d) Evaluate R(4.5) and explain what its sign means relative to the zero level in this mathematical model.

View Solution

(a)

R(x)=−(x²−9x+20)

=−(x−4)(x−5).

(b)

R(x)=0 at x=4 and x=5.

(c)

a=−1, b=9, c=−20.

−ba=−9−1=9=4+5.

ca=−20−1=20=4×5.

(d)

R(4.5)=−(4.5−4)(4.5−5)

=−(0.5)(−0.5)=0.25.

The positive value means the curve lies above the zero/break-even line between the two roots. This is an interpretation of the stated simplified model, not a general business-profit formula.

Q95

4 MarksLevel 4Case StudyNumber Pattern

Case Study 5 — Algebraic Number Pattern

A class studies the sequence generated by T(n)=n²−5n+6 for integer values of n.

n	1	2	3	4	5
T(n)	2	0	0	2	6

(a) Verify T(1).

(b) Which entries reveal the zeroes of T(n)?

(c) Use coefficient relations to find the sum and product of those zeroes.

(d) A new pattern is produced by shifting each zero one unit to the right. Form its monic quadratic polynomial.

View Solution

(a)

T(1)=1−5+6=2.

(b)

T(2)=0 and T(3)=0, so the zeroes are 2 and 3.

(c)

Sum=−−51=5=2+3.

Product=61=6=2×3.

(d)

Shifted zeroes: 3 and 4.

Required polynomial=(x−3)(x−4)

=x²−7x+12.

No questions match this filter.

Beyond the Main 95

Can You Solve These Without Looking at the Formula Sheet?

Five ungraded HOTS/Olympiad challenges. These are additional and are not included in the 95-entry count.

Challenge 1

Level 5 Ungraded Challenge

If α and β are the zeroes of x²−6x+4, form the monic quadratic polynomial whose zeroes are α²+β and β²+α.

View Solution

S=α+β=6, P=αβ=4.

α²+β²=S²−2P=36−8=28.

New sum

(α²+β)+(β²+α)=28+6=34.

New product

(α²+β)(β²+α)

=α²β²+α³+β³+αβ.

α³+β³=S³−3PS=216−72=144.

New product=16+144+4=164.

Required polynomial=x²−34x+164.

Challenge 2

Level 5 Ungraded Challenge

The zeroes of x²+kx+12 differ by 2. Find all possible real values of k.

View Solution

Let the zeroes be α and β.

α+β=−k, αβ=12, (α−β)²=4.

(α−β)²=(α+β)²−4αβ.

4=k²−48.

k²=52.

k=±√52=±2√13.

Challenge 3

Level 5 Ungraded Challenge

Find a and b if x³−3x²+ax+b is exactly divisible by x²−1. Also find the quotient.

View Solution

x²−1=(x−1)(x+1), so p(1)=0 and p(−1)=0.

p(1)=1−3+a+b=0 ⇒ a+b=2. …(1)

p(−1)=−1−3−a+b=0 ⇒ −a+b=4. …(2)

Adding (1) and (2): 2b=6 ⇒ b=3.

Then a=−1.

p(x)=x³−3x²−x+3.

=(x²−1)(x−3).

Quotient=x−3.

Challenge 4

Level 5 Ungraded Challenge

A cubic graph crosses the x-axis at x=−2, x=1 and x=4, and its leading coefficient is 1. Find the polynomial in factorised and expanded form. Then find its y-intercept.

View Solution

A monic cubic with zeroes −2, 1 and 4 is

p(x)=(x+2)(x−1)(x−4).

(x−1)(x−4)=x²−5x+4.

p(x)=(x+2)(x²−5x+4)

=x³−3x²−6x+8.

y-intercept occurs at x=0: p(0)=8.

Therefore the y-intercept is (0,8).

Challenge 5

Level 5 Ungraded Challenge

If α and β are the zeroes of x²−4x+2, form a monic quadratic polynomial whose zeroes are 1/(α−1) and 1/(β−1).

View Solution

S=α+β=4, P=αβ=2.

New sum

1α−1+1β−1=(β−1)+(α−1)(α−1)(β−1)

=S−2P−S+1=2−1=−2.

New product

1(α−1)(β−1)=1P−S+1=1−1=−1.

Required polynomial=x²−(−2)x−1

=x²+2x−1.

Board Examination Strategy

How to Score Full Marks in Polynomials

Write in descending powers

This reduces sign and missing-term errors, especially in long division.

Identify a, b, c with signs

In 2x²−7x+3, b is −7, not 7.

Show factorisation

Do not jump from a quadratic directly to its roots when method marks are available.

State both zeroes clearly

After factorisation, write the two equations and the two resulting x-values.

Verify sum and product separately

Show α+β and −b/a, then αβ and c/a.

Write formulas before substitution

It makes the method explicit and protects method marks.

Define transformed zeroes

Show the new sum and new product before forming the new polynomial.

Verify long division when asked

Write Dividend = Divisor × Quotient + Remainder.

Label graphs

Axes and x-intercepts should be clear; state the number of zeroes in words.

Respect the question wording

If a monic polynomial is requested, use leading coefficient 1; otherwise non-zero scalar multiples may also be valid.

Last-Minute Recall

60-Second Polynomials Revision

Polynomial: powers of the variable are non-negative integers.

Degree: highest power with non-zero coefficient; zero polynomial degree is not defined at this level.

Zero: α is a zero if p(α)=0.

Graph: real zeroes are x-coordinates where y=p(x) meets or touches the x-axis.

Linear ax+b: zero = −b/a.

Quadratic ax²+bx+c: α+β=−b/a and αβ=c/a.

Form from zeroes: k(x−α)(x−β), k≠0; monic form x²−(α+β)x+αβ.

Transformed zeroes: find the new sum and new product first.

Division: p(x)=g(x)q(x)+r(x), with r=0 or deg r<deg g.

Shortcut: if only a symmetric expression in the roots is needed, use sum/product relations instead of solving the roots.

SK Tuitions • Learn the method, then practise the reasoning.