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State Board · Class 9 · Mathematics Part 1

Lesson 3: Polynomials

5 concepts · 40 practice questions with answers · Free to read

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Polynomials and their types by number of terms

From daily life

A shopkeeper's bill for x notebooks costing ₹ 12 each plus a ₹ 5 bag is 12x + 5. Such neat expressions, where x only appears as a whole-number power, are called polynomials.

Simple definition

A polynomial (बहुपदी) is an algebraic expression in which the powers of the variables are whole numbers (0, 1, 2, ...).

Rule 1: Is it a polynomial? — 3x2 + 5x − 7 is a polynomial. Numbers like 6, −7, 0 are constant polynomials (स्थिर बहुपदी) because 6 = 6x0. √x + 2 is not one, as the power of x is 12. 4x + 2 = 4x−1 + 2 is not one, as −1 is not a whole number. A polynomial in x is named p(x), in m is q(m), in y is r(y).

  • Easy : Easy: 5x3 − 2 is a polynomial.
  • Medium : Medium: x2 + 3x is not, since 3x = 3x−1.
  • Exam-type : Exam: Which is a polynomial: y2 − √y, 7y5 + y − 1, 1y + 4? Only 7y5 + y − 1.

Rule 2: Types by number of terms — Monomial (एकपदी): one term, e.g. 7y3, 5. Binomial (द्विपदी): two terms, e.g. a2 − 9. Trinomial (त्रिपदी): three terms, e.g. m2 + 3m + 2.

A monomial has one term such as 7p, a binomial two terms such as q² + 4 and a trinomial three terms such as z³ − 2z + 9Polynomials by number of termsMonomial – 1 term7pBinomial – 2 termsq²+ 4Trinomial – 3 termsz³− 2z+ 9

In the figure : In the figure count the terms: 1 term is a monomial, 2 a binomial and 3 a trinomial.

  • Easy : Easy: 4x is a monomial.
  • Medium : Medium: x3 − 2x is a binomial.
  • Exam-type : Exam: Classify 2p2 − p + 8 and 9. Trinomial; monomial (a constant).
Remember

Whole-number powers only. Terms: 1 = mono, 2 = bi, 3 = tri.

Common mistake

Calling 1x + 1 a polynomial. Its power −1 is not a whole number.

Shortcut

Memory clue: "घात पूर्ण = बहुपदी".

Spot the difference

ExpressionPolynomial?
3x2 + 5x − 7Yes
√x + 2No
4x + 2No
6Yes (constant)

Practice questions

Easy
  1. 1What is a polynomial?
  2. 2Is 0 a polynomial?
  3. 3Which has two terms?
Medium
  1. 4Is 8x2 − 3x a binomial?
  2. 5Is 2x3 + 4 a polynomial?
  3. 6Is √m + 5 a polynomial?
Challenging
  1. 7Name the type: y2 − 5y + 6.
  2. 8Is 1x + x2 a polynomial? Why?
Show answers (try them yourself first)
  1. 1An expression whose variable powers are whole numbers.
  2. 2Yes, a constant polynomial.
  3. 3A binomial.
  4. 4Yes.
  5. 5Yes.
  6. 6No, the power is 12.
  7. 7Trinomial.
  8. 8No, because 1x = x−1 and −1 is not a whole number.

Degree and forms of a polynomial

From daily life

A ball thrown up follows a curve, a rolling cart follows a straight line. The shape depends on how big the highest power is. That power is called the degree.

Simple definition

The degree (घात) of a polynomial in one variable is its highest power of the variable.

Rule 1: Finding the degree — In one variable: 5x4 − x2 + 8 has degree 4. A non-zero constant such as 10 has degree 0; the degree of the zero polynomial is not defined. In more than one variable, take the highest sum of powers in a term: 5m2n4 + 2mn3 − 7 has term degrees 6, 4, 0, so the degree is 6.

  • Easy : Easy: Degree of 3x5 + x is 5.
  • Medium : Medium: Degree of 9 is 0.
  • Exam-type : Exam: Degree of 4a3b2 − a2b5 + 1: terms 5 and 7, so 7.

Rule 2: Linear, quadratic, cubic — Degree 1: linear (ax + b). Degree 2: quadratic (ax2 + bx + c). Degree 3: cubic (ax3 + bx2 + cx + d). Here a ≠ 0.

Graphs of a linear, a quadratic and a cubic polynomial: a straight line, a U-shaped curve and an S-shaped curveDegree decides the shape of the graphDegree 1linear: 2x + 1Degree 2quadratic: x²Degree 3cubic: x³

In the figure : In the figure degree 1 gives a straight line, degree 2 a U-shaped curve and degree 3 an S-shaped curve.

  • Easy : Easy: 2x + 1 is linear.
  • Medium : Medium: x2 − 4 is quadratic.
  • Exam-type : Exam: Name 5x3 − x + 2. Cubic.

Rule 3: Forms of a polynomial — Take 2 + x3 − 5x. Standard form (प्रमाण रूप): descending powers, x3 − 5x + 2. Index form (घात रूप): add missing terms with coefficient 0, x3 + 0x2 − 5x + 2. Coefficient form (सहगुणक रूप): (1, 0, −5, 2).

  • Easy : Easy: Standard form of 3 + x is x + 3.
  • Medium : Medium: Coefficient form of x2 − 4 is (1, 0, −4).
  • Exam-type : Exam: Write 6 − x4 + 2x in the three forms: −x4 + 2x + 6; −x4 + 0x3 + 0x2 + 2x + 6; (−1, 0, 0, 2, 6).
Remember

Degree = highest power (sum of powers in many variables). Linear 1, quadratic 2, cubic 3.

Common mistake

Taking the degree as the number of terms. 4x5 + 1 has 2 terms but degree 5.

Shortcut

Memory clue: "घात = सर्वांत मोठा वर चढलेला अंक".

Spot the difference

DegreeNameForm
1Linearax + b
2Quadraticax2 + bx + c
3Cubicax3 + bx2 + cx + d

Practice questions

Easy
  1. 1What is the degree?
  2. 2Degree of 7x6 + 2?
  3. 3Degree of a non-zero constant?
Medium
  1. 4Which degree is cubic?
  2. 5Standard form of 5 − 2x + x2?
  3. 6Coefficient form of x3 + 4?
Challenging
  1. 7Degree of 3m2n3 + mn5 − 2?
  2. 8Why is the degree of the zero polynomial not defined?
Show answers (try them yourself first)
  1. 1The highest power of the variable.
  2. 26.
  3. 30.
  4. 43.
  5. 5x2 − 2x + 5.
  6. 6(1, 0, 0, 4).
  7. 7Terms 5 and 6, so 6.
  8. 8It has no non-zero term, so there is no highest power.

Operations on polynomials and division

From daily life

Dividing 17 sweets among 5 children gives 3 each and 2 left: 17 = 5 × 3 + 2. Polynomials divide the same way.

Simple definition

Dividend = Divisor × Quotient + Remainder (भाज्य = भाजक × भागाकार + बाकी).

Rule 1: Add, subtract, multiply — Like terms are added or subtracted by their coefficients. (4a2 + 3a − 5) + (2a2 − 7a + 1) = 6a2 − 4a − 4. (5x2 + 2x) − (3x2 − 4) = 2x2 + 2x + 4. (x + 3)(x2 − 2x + 1) = x3 − 2x2 + x + 3x2 − 6x + 3 = x3 + x2 − 5x + 3, degree 3.

  • Easy : Easy: 3m + 4m = 7m.
  • Medium : Medium: (x + 2)(x + 5) = x2 + 7x + 10.
  • Exam-type : Exam: Subtract (2y2 − y + 3) from (5y2 + 2y − 1): 3y2 + 3y − 4.

Rule 2: Division — Example: (x2 + 5x + 6) ÷ (x + 2) = x + 3 with remainder 0. Example: (2x2 + 3x + 1) ÷ (x − 1) gives quotient 2x + 5 and remainder 6, so 2x2 + 3x + 1 = (x − 1)(2x + 5) + 6. Check: (x − 1)(2x + 5) = 2x2 + 3x − 5, plus 6 gives 2x2 + 3x + 1.

x² + 3x + 5 divided by x + 1 gives quotient x + 2 and remainder 3Dividend = Divisor × Quotient + Remainderx² + 3x + 5dividend=(x + 1)divisor×(x + 2)quotient+3remainderCheck: (x + 1)(x + 2) = x² + 3x + 2, and 2 + 3 = 5 ✓

In the figure : In the figure division of polynomials works like division of numbers: dividend = divisor × quotient + remainder.

  • Easy : Easy: 17 = 5 × 3 + 2.
  • Medium : Medium: Remainder 0 means exact division.
  • Exam-type : Exam: (x3 − 3x2 + 5) ÷ (x − 2): quotient x2 − x − 2, remainder 1.

Rule 3: Euclid's division lemma — s(x) = p(x) × q(x) + r(x), where r(x) = 0 or the degree of r(x) is less than the degree of the divisor p(x).

Live figure

Operations on polynomials and division

In the app you can move this figure with your finger.

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  • Easy : Easy: The remainder is smaller in degree than the divisor.
  • Medium : Medium: A linear divisor leaves a constant remainder.
  • Exam-type : Exam: Why must deg r(x) < deg p(x)?
Remember

Dividend = divisor × quotient + remainder. The remainder is smaller in degree.

Common mistake

Forgetting to add the remainder back when checking.

Shortcut

Memory clue: "भाज्य = भाजक × भागाकार + बाकी".

Spot the difference

PartExample (2x2 + 3x + 1) ÷ (x − 1)
Dividend2x2 + 3x + 1
Divisorx − 1
Quotient2x + 5
Remainder6

Practice questions

Easy
  1. 1Write the division relation.
  2. 2Add 3x + 2 and 4x − 5.
  3. 3Multiply (x + 1)(x + 4).
Medium
  1. 4Divide x2 + 5x + 6 by x + 3.
  2. 5What is the degree of (x + 3)(x2 − 2x + 1)?
  3. 6Quotient 2x + 5, divisor x − 1, remainder 6. Find the dividend.
Challenging
  1. 7Subtract (3x2 − 4) from (5x2 + 2x).
  2. 8State Euclid's division lemma for polynomials.
Show answers (try them yourself first)
  1. 1Dividend = divisor × quotient + remainder.
  2. 27x − 3.
  3. 3x2 + 5x + 4.
  4. 4x + 2.
  5. 53.
  6. 62x2 + 3x + 1.
  7. 72x2 + 2x + 4.
  8. 8s(x) = p(x)q(x) + r(x), with r(x) = 0 or deg r(x) < deg p(x).

Synthetic division and value of a polynomial

From daily life

Long division is like writing out every step of a big sum. A shortcut that keeps only the numbers is much quicker. That shortcut is synthetic division.

Simple definition

Synthetic division (संश्लेषी भागाकार) divides by (x − a) or (x + a) using only coefficients.

Rule 1: Steps — Write the dividend's coefficient form. Put the opposite of a on the left (for x − 2 use 2; for x + 3 use −3). Bring down the first coefficient; multiply by that number, add to the next coefficient, and repeat. The last number is the remainder.

Synthetic division table: coefficients 1, 2, −5, 1 with divisor number 1 give 1, 3, −2 and remainder −1(x³ + 2x² − 5x + 1) ÷ (x − 1) by synthetic divisiondivisor1number12−51coeffs13−2× 1 →13−2−1add ↓Quotient: x² + 3x − 2Remainder: −1Rule: bring down, multiply by 1, add, repeat.

In the figure : In the figure synthetic division needs only the coefficients, so it is quick and neat.

  • Easy : Easy: Divisor x − 4 uses the number 4.
  • Medium : Medium: (x2 + 5x + 6) ÷ (x + 2): coefficients 1, 5, 6; number −2; 1, 3, 0. Quotient x + 3, remainder 0.
  • Exam-type : Exam: (2x3 − 5x2 + 3x − 4) ÷ (x − 2): number 2; 2, then −5 + 4 = −1, then 3 − 2 = 1, then −4 + 2 = −2. Quotient 2x2 − x + 1, remainder −2.

Rule 2: Another division and the value of p(x) — (x3 + 4x2 − x + 6) ÷ (x + 3): number −3; 1, then 4 − 3 = 1, then −1 − 3 = −4, then 6 + 12 = 18. Quotient x2 + x − 4, remainder 18. The value of a polynomial (किंमत) at x = a is p(a): for p(x) = x2 − 4x + 7, p(3) = 9 − 12 + 7 = 4 and p(−1) = 1 + 4 + 7 = 12. For p(x) = 3x3 − 2x + 1, p(2) = 24 − 4 + 1 = 21 and p(0) = 1.

  • Easy : Easy: p(x) = x + 5 gives p(1) = 6.
  • Medium : Medium: p(x) = 2x2 − 3: p(2) = 8 − 3 = 5.
  • Exam-type : Exam: p(x) = x3 − x2 + 2: p(−1) = −1 − 1 + 2 = 0.
Remember

Opposite sign number on the left. Multiply, then add. Last = remainder.

Common mistake

Using +2 for x + 2. Use −2.

Shortcut

Memory clue: "x − a → a; x + a → −a".

Spot the difference

DivisorNumber used
x − 22
x + 3−3
x − 55

Practice questions

Easy
  1. 1Which divisors suit synthetic division?
  2. 2Number used for x + 4?
  3. 3What is the last number in the last row?
Medium
  1. 4Find p(2) if p(x) = x2 + 3.
  2. 5Find p(−2) if p(x) = x2 − 3x.
  3. 6Divide x2 + 5x + 6 by x + 3 synthetically.
Challenging
  1. 7(x3 + 4x2 − x + 6) ÷ (x + 3): find quotient and remainder.
  2. 8Why does p(0) equal the constant term?
Show answers (try them yourself first)
  1. 1Those of the form x − a or x + a.
  2. 2−4.
  3. 3The remainder.
  4. 47.
  5. 54 + 6 = 10.
  6. 6Number −3; 1, 2, 0. Quotient x + 2, remainder 0.
  7. 7x2 + x − 4 and 18.
  8. 8Every term with x becomes 0.

Remainder theorem, factor theorem and factorization

From daily life

If 12 divided by 4 leaves no remainder, 4 is a factor of 12. A polynomial works the same way: if dividing by (x − a) leaves remainder 0, then (x − a) is a factor.

Simple definition

Remainder theorem (बाकीचे प्रमेय): dividing p(x) by (x + a) leaves p(−a). Factor theorem (विभाजक प्रमेय): (x − a) is a factor of p(x) if and only if p(a) = 0.

Rule 1: Remainder theorem — Instead of dividing, just substitute. 2x2 + 3x − 5 divided by (x + 2) leaves p(−2) = 8 − 6 − 5 = −3. x3 − 2x + 7 divided by (x − 3) leaves p(3) = 27 − 6 + 7 = 28.

  • Easy : Easy: Remainder of p(x) ÷ (x − 1) is p(1).
  • Medium : Medium: x2 + 1 ÷ (x + 2): p(−2) = 5.
  • Exam-type : Exam: Remainder of 4x3 − x + 2 on dividing by (x + 1): p(−1) = −4 + 1 + 2 = −1.

Rule 2: Factor theorem — For p(x) = x3 − 6x2 + 11x − 6: p(1) = 1 − 6 + 11 − 6 = 0, so (x − 1) is a factor; p(2) = 0 and p(3) = 0 too, so p(x) = (x − 1)(x − 2)(x − 3). But p(−1) = −1 − 6 − 11 − 6 = −24 ≠ 0, so (x + 1) is not a factor.

Putting x = 2 in x² − 5x + 6 gives 0, so (x − 2) is a factorFactor theorem in action✏️Divide p(x)by (x − 2)🔢Put x = 2 inx² − 5x + 6:4 − 10 + 6 = 0✅Remainder 0,so (x − 2) isa factor

In the figure : In the figure if p(a) = 0, then (x − a) is a factor of p(x).

  • Easy : Easy: p(x) = x − 4, p(4) = 0, factor (x − 4).
  • Medium : Medium: x2 − 5x + 6: p(2) = 0, so (x − 2) is a factor.
  • Exam-type : Exam: Is (x + 2) a factor of x3 + 8? p(−2) = −8 + 8 = 0, yes.

Rule 3: Factorization — x2 − 16 = (x + 4)(x − 4). 2x2 + 7x + 3 = (2x + 1)(x + 3). 6x2 − x − 2 = (3x − 2)(2x + 1). For x3 − 2x2 − 5x + 6: p(1) = 0, synthetic division by (x − 1) gives 1, −1, −6, remainder 0, so the quotient is x2 − x − 6 = (x − 3)(x + 2); thus the factors are (x − 1)(x − 3)(x + 2). A new variable helps: (y2 + y)2 − 8(y2 + y) + 12 = (t − 2)(t − 6) with t = y2 + y, which is (y + 2)(y − 1)(y + 3)(y − 2).

  • Easy : Easy: x2 − 9 = (x + 3)(x − 3).
  • Medium : Medium: x2 + 7x + 12 = (x + 3)(x + 4).
  • Exam-type : Exam: Factorize x3 − 7x + 6. p(1) = 0; quotient x2 + x − 6 = (x + 3)(x − 2); so (x − 1)(x + 3)(x − 2).
Remember

Remainder of ÷ (x + a) is p(−a). p(a) = 0 ⇔ (x − a) is a factor.

Common mistake

Using p(a) for the divisor x + a. Use p(−a).

Shortcut

Memory clue: "x चे मूल्य = शून्य करणारी किंमत टाका".

Spot the difference

TheoremStatement
Remainder(x + a) ÷ → p(−a)
Factorp(a) = 0 ⇔ (x − a) factor

Practice questions

Easy
  1. 1State the remainder theorem.
  2. 2State the factor theorem.
  3. 3Remainder of x2 + 1 ÷ (x − 3)?
Medium
  1. 4Is (x − 1) a factor of x2 + x − 2?
  2. 5Factorize x2 − 16.
  3. 6Factorize 2x2 + 7x + 3.
Challenging
  1. 7Find the remainder of x3 − 2x + 7 ÷ (x − 3).
  2. 8For what k is (x − 2) a factor of x2 + kx − 10?
Show answers (try them yourself first)
  1. 1Dividing p(x) by (x + a) leaves p(−a).
  2. 2(x − a) is a factor of p(x) iff p(a) = 0.
  3. 3p(3) = 10.
  4. 4p(1) = 0, yes.
  5. 5(x + 4)(x − 4).
  6. 6(2x + 1)(x + 3).
  7. 727 − 6 + 7 = 28.
  8. 8p(2) = 4 + 2k − 10 = 0, so k = 3.

Quick revision

  • A polynomial has whole-number powers of the variable; degree = highest power; the zero polynomial has no degree.
  • Dividend = Divisor × Quotient + Remainder; the remainder on dividing by (x + a) is p(−a).
  • (x − a) is a factor of p(x) if and only if p(a) = 0.