1
Which of these is a polynomial?
💡 Hint:
Every exponent must be a whole number.
Explanation:
In 3x² − 2x + 1 the exponents are 2, 1 and 0. The others have −1 or 1/2.
Class 9 · Maths · Chapter 2 · बिहार बोर्ड (BSEB)CBSE · NCERT 2026-27
Polynomials
Polynomials
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- 1 A polynomial in one variable and its degree — Exercise 2.1
- 2 The value of a polynomial and its zeroes — Exercise 2.2
- 3 Division and the Remainder Theorem — Exercise 2.3
- 4 The Factor Theorem and factorisation — Exercise 2.4
- 5 The first four identities — Exercise 2.5
- 6 Identities for three terms and cubes — Exercise 2.5
- Chapter winner — every lesson at mastery ★
🎬 Shorts
This chapter has no Short on the channel yet, so ten empty cards are not shown. The notes, diagrams and checks are complete.
📄 PDF download — notes + answer key
A polynomial in one variable and its degree — Exercise 2.1
एक चर का बहुपद और घात — प्रश्नावली 2.1 · NCERT 2.2 · Exercise 2.1
The exponent must be a whole numberNotes
A polynomial in one variable x is p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀, with aₙ ≠ 0. Here n is a whole number. a₀, a₁, … are the coefficients. The terms run from aₙxⁿ down to a₀.
One term is a monomial, two terms a binomial, three terms a trinomial. Degree 1 is linear, of the form ax + b with a ≠ 0. Degree 2 is quadratic. Degree 3 is cubic. 5x² + 3x + π is quadratic, because π may be a coefficient.
x + 1/x = x + x⁻¹ is not a polynomial. Neither is √x + 4 = x1/2 + 4. A non-zero constant has degree 0. The degree of the zero polynomial is not defined.
| Name | Degree | Example |
|---|---|---|
| Constant | 0 | 7 |
| Linear | 1 | 4x + 7 |
| Quadratic | 2 | x² − 3x + 2 |
| Cubic | 3 | 2x³ − x |
Worked exampleExample
Question: Find the degree of 2 − y² − y³ + 2y⁸. Also write the degree of the constant 5.
Step 1. Write the terms in descending degree: 2y⁸ − y³ − y² + 2.
Step 2. The highest power of the variable is 8. So the degree is 8.
Step 3. 5 is a non-zero constant. It has no variable, which means the coefficient of y⁰. The degree is 0.
Caution: If the polynomial is only 0, do not write a degree. It is not defined.
10-second revision
- Degree = the highest power
- A non-zero constant has degree 0
- 1/x and √x are not polynomials
Board tip · BSEBBoard tip
Before division, and when the degree is asked, arrange the terms in descending degree.
Board tip · CBSEBoard tip
A coefficient may be irrational. πx² + 2 is a polynomial. The degree belongs to the variable, not to the coefficient.
Check your understandingall correct = mastery ★
2
The degree of the zero polynomial is 0.
Explanation:
False. A non-zero constant has degree 0. The degree of the zero polynomial is not defined.
3
The degree of 4x + 7 is ______.
💡 Hint:
A linear polynomial.
Explanation:
The highest power is 1.
4
x² + 3x + 2 is —
💡 Hint:
Three terms, highest power 2.
Explanation:
There are three terms, so it is a trinomial. The degree is 2, so it is quadratic.
5
Say why 5x² + 3x + π is a polynomial and why √x + 1 is not. Write the degree where it exists.
Model answer (board format)
In 5x² + 3x + π the exponents are 2, 1 and 0, and π is a coefficient. It is a quadratic polynomial of degree 2. In √x + 1 the power of x is 1/2, which is not a whole number, so it is not a polynomial.
Check your answer:
The value of a polynomial and its zeroes — Exercise 2.2
बहुपद का मान और शून्यक — प्रश्नावली 2.2 · NCERT 2.3 · Exercise 2.2
p(c) = 0 is the zeroNotes
The number obtained by putting c in place of x in p(x) is p(c). If p(c) = 0, then c is a zero of the polynomial. The same c is a root of the equation p(x) = 0.
The linear polynomial ax + b has exactly one zero, x = −b/a. A non-zero constant has no zero, because 7 = 0 never holds. Every real number is a zero of the zero polynomial. That is the convention.
A polynomial can have more than one zero. The zeroes of x² − 2x are 0 and 2, because x(x − 2) = 0.
Worked exampleExample
Question: Find one zero of p(x) = 2x + 1 and check it.
Step 1. For a zero put p(x) = 0: 2x + 1 = 0.
Step 2. 2x = −1, so x = −1/2.
Formula: The zero of ax + b is −b/a. Here a = 2 and b = 1, so −b/a = −1/2.
Check: p(−1/2) = 2×(−1/2) + 1 = −1 + 1 = 0. So −1/2 is a zero.
10-second revision
- A zero means p(c) = 0
- A linear has one zero, −b/a
- A non-zero constant has no zero
Board tip · BSEBBoard tip
The zero of x + 2 is −2, not 2. The sign flips.
Board tip · CBSEBoard tip
Value and zero are different. p(1) may be 4. A zero is only where the value is 0.
Check your understandingall correct = mastery ★
1
The zero of p(x) = x − 3 is —
💡 Hint:
x − 3 = 0.
Explanation:
p(3) = 0. The zero is 3.
2
The constant polynomial 4 has no zero.
Explanation:
True. 4 is never 0. The zero polynomial is a different case.
3
If p(x) = 5x − 4x² + 3, then p(0) = ______.
💡 Hint:
Putting x = 0 leaves only the constant.
Explanation:
p(0) = 0 − 0 + 3 = 3.
4
Check that −2 is a zero of p(x) = x + 2 and that 2 is not. Write both values.
Model answer (board format)
p(−2) = −2 + 2 = 0, so −2 is a zero. p(2) = 2 + 2 = 4 ≠ 0, so 2 is not a zero.
Check your answer:
Division and the Remainder Theorem — Exercise 2.3
भाग और शेषफल प्रमेय — प्रश्नावली 2.3 · NCERT 2.4 · Exercise 2.3
The remainder is p(a)Notes
Polynomials are divided in the same spirit as numbers. In 15 = (6 × 2) + 3 the remainder is smaller than the divisor. For polynomials, p(x) = g(x) q(x) + r(x), and the degree of r is less than the degree of g.
If the divisor is a monomial, divide each term. Dividing 2x³ + x² + x by x gives 2x² + x + 1. But in 3x² + x + 1 the constant 1 does not divide by x into a polynomial term. The remainder stays 1 and x is not a factor.
Before dividing, write descending powers. Write x + 3x² − 1 as 3x² + x − 1. If the degree is at least 1 and the divisor is x − a, the remainder is p(a).
Worked exampleExample
Question: Find the remainder when p(x) = x³ + 1 is divided by x + 1. Also show the long division.
Formula: x + 1 = x − (−1), so a = −1 and the remainder is p(−1).
Substitution: p(−1) = (−1)³ + 1 = −1 + 1 = 0.
Long division: The dividend is x³ + 0x² + 0x + 1. The first term is x². (x + 1)x² = x³ + x². Subtraction leaves −x². The next term is −x. (x + 1)(−x) = −x² − x. Subtraction leaves x. The next term is +1. (x + 1)(1) = x + 1. Subtraction leaves 0.
Answer: The quotient is x² − x + 1 and the remainder is 0. The two methods agree.
10-second revision
- Remainder = p(a), divisor x − a
- x + 1 means a = −1
- If the remainder is 0, the divisor is a factor
Board tip · BSEBBoard tip
For x − 1 and x + 1 the value of a is different. Use p(1) for x − 1 and p(−1) for x + 1.
Board tip · CBSEBoard tip
Write a missing power with coefficient 0, such as x² and x in x³ + 1.
Check your understandingall correct = mastery ★
1
The remainder when p(x) = x² − 4 is divided by x − 2 is —
💡 Hint:
Find p(2).
Explanation:
p(2) = 4 − 4 = 0.
2
The remainder on division by x + 1 is p(1).
Explanation:
False. x + 1 = x − (−1), so the remainder is p(−1).
3
If p(x) = x³ − 1 and the divisor is x − 1, the remainder is ______.
💡 Hint:
p(1).
Explanation:
p(1) = 1 − 1 = 0.
4
Before division, x + 3x² − 1 is written as —
💡 Hint:
Descending degree.
Explanation:
The standard form is 3x² + x − 1.
5
Dividing 2x³ + x² + x by x gives the quotient ______.
💡 Hint:
Cancel x in each term.
Explanation:
2x² + x + 1.
6
Find the remainder when p(x) = x⁴ + x³ − 2x² + x + 1 is divided by x − 1. Write the formula and the substitution.
Model answer (board format)
Formula: divisor x − a with a = 1, remainder p(1). p(1) = 1 + 1 − 2 + 1 + 1 = 2. The remainder is 2.
Check your answer:
The Factor Theorem and factorisation — Exercise 2.4
गुणनखंड प्रमेय और गुणनखंडन — प्रश्नावली 2.4 · NCERT 2.5 · Exercise 2.4
p(a) = 0 means a factorNotes
By the Remainder Theorem, p(x) = (x − a) q(x) + p(a). If p(a) = 0, then p(x) = (x − a) q(x). (x − a) is a factor if and only if p(a) = 0.
Split the middle term of ax² + bx + c into two numbers whose product is a×c and whose sum is b. Then take out the common bracket.
For a cubic, try factors of the constant. If p(1) = 0, then (x − 1) is a factor. Divide, leave a quadratic, and split that too.
Worked exampleExample
Question: Factorise 6x² + 17x + 5 by splitting the middle term.
Step 1. a×c = 6×5 = 30 and b = 17. The two numbers are 15 and 2, because 15×2 = 30 and 15+2 = 17.
Step 2. 6x² + 15x + 2x + 5.
Step 3. 3x(2x + 5) + 1(2x + 5) = (3x + 1)(2x + 5).
Check: (3x + 1)(2x + 5) = 6x² + 15x + 2x + 5 = 6x² + 17x + 5.
10-second revision
- p(a) = 0 means (x − a) is a factor
- Middle term: product ac, sum b
- 6x² + 17x + 5 = (3x + 1)(2x + 5)
Board tip · BSEBBoard tip
If the question asks about the factor x + 2, put a = −2 and look at p(−2).
Board tip · CBSEBoard tip
The signs of the split numbers follow the sum. If the sum is negative, both can be negative.
Check your understandingall correct = mastery ★
1
x + 2 is a factor of x³ + 3x² + 5x + 6, because —
💡 Hint:
x + 2 = x − (−2).
Explanation:
p(−2) = −8 + 12 − 10 + 6 = 0. So x + 2 is a factor.
2
y² − 5y + 6 = (y − 2)(y − 3).
Explanation:
True. −2 and −3 multiply to 6 and add to −5.
3
If x − 1 is a factor of p(x), then p(1) = ______.
💡 Hint:
Factor Theorem.
Explanation:
p(1) = 0.
4
One factor of x² + 5x + 6 is —
💡 Hint:
2 and 3, product 6, sum 5.
Explanation:
(x + 2)(x + 3) = x² + 5x + 6.
5
For p(x) = x³ − 6x² + 11x − 6, find p(1) and say whether x − 1 is a factor.
Model answer (board format)
p(1) = 1 − 6 + 11 − 6 = 0. By the Factor Theorem, x − 1 is a factor.
Check your answer:
The first four identities — Exercise 2.5
पहली चार सर्वसमिकाएँ — प्रश्नावली 2.5 · NCERT 2.6 · Exercise 2.5 · (a+b)²
True for every valueNotes
An identity is an algebraic equality that is true for every value of the variables. The first four:
I. (x + y)² = x² + 2xy + y²
II. (x − y)² = x² − 2xy + y²
III. x² − y² = (x + y)(x − y)
IV. (x + a)(x + b) = x² + (a + b)x + ab
A square of side (a + b) splits into a², ab, ab and b², so (a + b)² = a² + 2ab + b². These work both for products and for factors.
Worked exampleExample
Question: Find 103 × 98 without multiplying directly.
Step 1. 103 = 100 + 3 and 98 = 100 + (−2).
Formula IV: (x + a)(x + b) = x² + (a + b)x + ab. Here x = 100, a = 3, b = −2.
Substitution: 100² + (3 + (−2))×100 + (3)(−2).
Step 2. 10000 + 1×100 + (−6) = 10000 + 100 − 6 = 10094.
Check: 103×100 − 103×2 = 10300 − 206 = 10094.
10-second revision
- (a+b)² = a² + 2ab + b²
- x² − y² = (x+y)(x−y)
- (x+a)(x+b) = x² + (a+b)x + ab
Board tip · BSEBBoard tip
See 49a² + 70ab + 25b² as (7a + 5b)². First spot the squares and the double product.
Board tip · CBSEBoard tip
For a product without direct multiplication, choose numbers near 100, then use Identity IV.
Check your understandingall correct = mastery ★
1
(x + 3)² equals —
💡 Hint:
Do not forget the middle 2xy.
Explanation:
x² + 2·x·3 + 3² = x² + 6x + 9.
2
(x − y)² = x² − y².
Explanation:
False. (x − y)² = x² − 2xy + y². The form x² − y² is a product of two terms.
3
x² − 9 = (x + 3)(______ ).
💡 Hint:
Difference of squares.
Explanation:
(x + 3)(x − 3).
4
(x − 3)(x + 5) equals —
💡 Hint:
a + b = −3 + 5 = 2 and ab = −15.
Explanation:
x² + (−3 + 5)x + (−3)(5) = x² + 2x − 15.
5
The figure of (a + b)² has two parts labelled ab.
Explanation:
True. The two rectangles together make 2ab.
6
Factorise 49a² + 70ab + 25b² by an identity. Write the steps.
Model answer (board format)
49a² = (7a)², 25b² = (5b)² and 70ab = 2(7a)(5b). So (7a + 5b)² = (7a + 5b)(7a + 5b).
Check your answer:
Identities for three terms and cubes — Exercise 2.5
तीन पद और घन की सर्वसमिकाएँ — प्रश्नावली 2.5 · NCERT 2.6 · Exercise 2.5 · cubes
From V to VIIINotes
V. (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx. It is obtained from (a + b)² by treating (x + y) as one term.
VI. (x + y)³ = x³ + y³ + 3xy(x + y)
VII. (x − y)³ = x³ − y³ − 3xy(x − y)
VIII. x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx)
If x + y + z = 0, then x³ + y³ + z³ = 3xyz. The expanded form is the right-hand side. The left-hand side is the compact form.
Worked exampleExample
Question: Expand (2x + 1)³.
Formula VI: (x + y)³ = x³ + y³ + 3xy(x + y). Here the first term is 2x and the second is 1.
Substitution: (2x)³ + 1³ + 3(2x)(1)(2x + 1).
Step 1. 8x³ + 1 + 6x(2x + 1).
Step 2. 6x(2x + 1) = 12x² + 6x.
Answer: 8x³ + 12x² + 6x + 1.
10-second revision
- (x+y+z)² has three squares and three double products
- (x+y)³ = x³ + y³ + 3xy(x+y)
- x+y+z = 0 implies x³+y³+z³ = 3xyz
Board tip · BSEBBoard tip
In an expansion of three squares, write all of 2xy, 2yz and 2zx. Leaving one out loses marks.
Board tip · CBSEBoard tip
If you see x + y + z = 0, the cubes are directly 3xyz. Open the full Identity VIII only when the sum is not zero.
Check your understandingall correct = mastery ★
1
If x + y + z = 0, then x³ + y³ + z³ equals —
💡 Hint:
In VIII the first factor is zero.
Explanation:
x³ + y³ + z³ − 3xyz = 0, so the sum is 3xyz.
2
(x − y)³ = x³ − y³ − 3xy(x − y).
Explanation:
True. This is Identity VII.
3
In (x + y + z)² the term in xy is ______xy.
💡 Hint:
Twice each pair.
Explanation:
2xy. Likewise 2yz and 2zx.
4
Given 3 + (−2) + (−1) = 0. Find x³ + y³ + z³ and check with 3xyz.
Model answer (board format)
The sum is 0, so x³ + y³ + z³ = 3xyz = 3×3×(−2)×(−1) = 18. Direct cubes: 27 + (−8) + (−1) = 18. Both agree.
Check your answer:
❓ Full question bank — with answers and explanations — 59 questions
Pick a type. The 30 lesson checks are separate — each lesson has as many as its topic needs. All correct earns mastery ★.
BSEB modelpractice · not annual60 tried · 0 correctMultiple choiceMCQ120 tried · 0 correctTrue / falseTrue / false60 tried · 0 correctFill in the blanksFill60 tried · 0 correctMatchMatch20 tried · 0 correctAssertion–reasonAssertion–reason40 tried · 0 correctBalancingBalancing40 tried · 0 correctIdentifyIdentify20 tried · 0 correctShort and longVSA, SA, LA110 tried · 0 correctCBSE-stylenot a PYQ60 tried · 0 correct
No question is marked as a verified past paper. The BSEB set is a model for practice. CBSE items are CBSE-style, not a copy of any year’s paper.
Multiple choice
0/12Pick one option. A wrong try brings a hint.
1
This is not a polynomial —
💡 Hint:
The exponent −1 is not a whole number.
Explanation:
x + 1/x is not a polynomial. πx + 2 and 5 are polynomials.
2
The degree of 2y⁸ − y³ + 2 is —
💡 Hint:
Look at the highest power.
Explanation:
The degree is 8.
3
A linear polynomial has one zero. The zero of 3x − 6 is —
💡 Hint:
−b/a = 6/3.
Explanation:
3x = 6, so x = 2.
4
One zero of p(x) = x² − 5x + 6 is —
💡 Hint:
(x − 2)(x − 3).
Explanation:
p(2) = 4 − 10 + 6 = 0.
5
On division by x − 1 the remainder is —
💡 Hint:
a = 1.
Explanation:
The remainder is p(1).
6
Dividing x³ + 1 by x + 1 leaves remainder —
💡 Hint:
p(−1).
Explanation:
(−1)³ + 1 = 0.
7
One factor of 6x² + 17x + 5 is —
💡 Hint:
15 and 2.
Explanation:
(3x + 1)(2x + 5).
8
The middle term of (x + 4)² is —
💡 Hint:
2xy with y = 4.
Explanation:
2·x·4 = 8x.
9
x² − 16 equals —
💡 Hint:
Difference of squares.
Explanation:
(x + 4)(x − 4).
10
In (x + y + z)² the coefficient of xy is —
💡 Hint:
Each pair is doubled.
Explanation:
2xy.
11
If x + y + z = 0, then x³ + y³ + z³ is —
💡 Hint:
The special case of VIII.
Explanation:
3xyz.
12
What is correct about the zero polynomial?
💡 Hint:
The constant 0 is a separate case.
Explanation:
The degree is not defined. Every real number is a zero of it.
True or false
0/61
5 is a polynomial and its degree is 0.
Explanation:
True. A non-zero constant has degree 0.
2
Every linear polynomial has two zeroes.
Explanation:
False. A linear polynomial has exactly one zero.
3
If the remainder is 0, the divisor is a factor.
Explanation:
True. That is one direction of the Factor Theorem.
4
(x − y)² = x² − 2xy + y².
Explanation:
True. This is Identity II.
5
(x + y)² = x² + y².
Explanation:
False. The middle term 2xy is also there.
6
(x + y + z) is a factor of x³ + y³ + z³ − 3xyz.
Explanation:
True. This is Identity VIII.
Fill in the blanks
0/61
The degree of a quadratic polynomial is ______.
💡 Hint:
The highest power.
Explanation:
An example is x² + 1.
2
The zero of ax + b is ______.
💡 Hint:
The sign is minus.
Explanation:
a ≠ 0.
3
The remainder on division by x + 3 is p(______).
💡 Hint:
x − (−3).
Explanation:
a = −3.
4
x² + 7x + 12 = (x + 3)(x + ______ ).
💡 Hint:
Product 12.
Explanation:
3 + 4 = 7 and 3×4 = 12.
5
(a + b)² = a² + ______ + b².
💡 Hint:
The two ab parts of the figure.
Explanation:
Two rectangles.
6
If x + y + z = 0, then x³ + y³ + z³ = ______.
💡 Hint:
The product of the three.
Explanation:
The special case of Identity VIII.
Match
0/21
Match the polynomial with its name.
Column B: A. Quadratic · B. Cubic · C. Constant · D. Linear
1. 7
2. 4x + 1
3. x² + 1
4. x³ − x
Explanation:
7 is constant, 4x+1 linear, x²+1 quadratic, x³−x cubic.
2
Match the identity with its left side.
Column B: A. (x+y)(x−y) · B. x³+y³+3xy(x+y) · C. 3xyz · D. x²+2xy+y²
1. (x+y)²
2. x²−y²
3. (x+y)³
4. Cubes when x+y+z=0
Explanation:
The square has 2xy, the difference has two brackets, the cube has 3xy(x+y), and a zero sum gives 3xyz.
Assertion–reason
0/4Check both statements, then see whether the reason explains the assertion.
1
Assertion (A): x − 2 is a factor of p(x) = x² − 4.
Reason (R): p(2) = 0.
Explanation:
Both are true and R explains A.
2
Assertion (A): The degree of 5 is 0.
Reason (R): The degree of the zero polynomial is also 0.
Explanation:
A is true. R is false. The degree of the zero polynomial is not defined.
3
Assertion (A): (x + 1)² = x² + 2x + 1.
Reason (R): (x + 1)² = x² + 1.
Explanation:
A is true. R is false, because 2x is missing.
4
Assertion (A): x + 1/x is a polynomial.
Reason (R): In a polynomial every power of the variable is a whole number.
Explanation:
A is false. R is true. The power of 1/x is −1.
Coefficient practice
0/4These equations are not results of this maths chapter. This is only practice in filling coefficients. A blank coefficient means 1.
1
This is coefficient practice, not a result of the polynomials chapter. H2 + O2 makes water. Balance by filling coefficients (blank = 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
2
This is coefficient practice, not a result of the polynomials chapter. N2 + H2 makes NH3. Balance by filling coefficients (blank = 1).
N2 + H2 → NH3
Explanation: N2 + 3H2 → 2NH3
3
This is coefficient practice, not a result of the polynomials chapter. C + O2 makes CO2. Balance by filling coefficients (blank = 1).
C + O2 → CO2
Explanation: C + O2 → CO2
4
This is coefficient practice, not a result of the polynomials chapter. H2 + Cl2 makes HCl. Balance by filling coefficients (blank = 1).
H2 + Cl2 → HCl
Explanation: H2 + Cl2 → 2HCl
Classify
0/21
Place each expression as a polynomial or not.
3x² + 1
x + 1/x
7
√x + 2
Explanation:
3x²+1 and 7 are polynomials. In 1/x and √x the power is not a whole number.
2
Choose the degree class.
4x + 7
x² + 1
2x³ − x
0
Explanation:
4x+7 is linear, x²+1 quadratic, 2x³−x cubic. The degree of the zero polynomial is not defined.
Very short answer
0/41
Define the degree of a polynomial in one line.
Model answer (board format)
The degree is the highest power of the variable, when the leading coefficient is not zero.
Check your answer:
2
Find the zeroes of p(x) = x² − 2x.
Model answer (board format)
x(x − 2) = 0, so the zeroes are 0 and 2.
Check your answer:
3
State the Remainder Theorem.
Model answer (board format)
If p(x) of degree at least 1 is divided by x − a, the remainder is p(a).
Check your answer:
4
Write the expansion of (x + y)².
Model answer (board format)
x² + 2xy + y².
Check your answer:
Short answer
0/41
Find the remainder when p(x) = 2x³ − x² + 4 is divided by x − 1.
Model answer (board format)
a = 1. p(1) = 2 − 1 + 4 = 5. The remainder is 5.
Check your answer:
2
Factorise y² − 5y + 6.
Model answer (board format)
−2 and −3 multiply to 6 and add to −5. y² − 2y − 3y + 6 = y(y − 2) − 3(y − 2) = (y − 3)(y − 2).
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3
Find 105 × 106 by an identity.
Model answer (board format)
(100 + 5)(100 + 6) = 100² + (5 + 6)×100 + 5×6 = 10000 + 1100 + 30 = 11130.
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4
Write the expansion of (3a + 4b + 5c)² without multiplying everything out by hand.
Model answer (board format)
Three squares and three double products: 9a² + 16b² + 25c² + 24ab + 40bc + 30ca.
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Long answer
0/31
Write the idea of the Remainder Theorem. Then divide x³ + 1 by x + 1 and find both the remainder and the quotient.
Model answer (board format)
p(x) = (x − a)q(x) + r and r = p(a). Here a = −1 and p(−1) = 0. Long division gives quotient x² − x + 1 and remainder 0. So x + 1 is a factor.
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2
Explain the figure of (a+b)² in four parts. Then write the factors of 49a² + 70ab + 25b².
Model answer (board format)
A square of side a+b: an a by a square, a b by b square and two a by b rectangles. The sum is a² + 2ab + b². Here 49a² = (7a)², 25b² = (5b)² and 70ab = 2(7a)(5b). So the factor form is (7a + 5b)².
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3
Write Identity VIII. Then show that for 3, −2 and −1 the sum of cubes equals 3xyz.
Model answer (board format)
x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx). Here 3 + (−2) + (−1) = 0, so x³ + y³ + z³ = 3xyz = 18. Directly, 27 − 8 − 1 = 18.
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BSEB model questions · practice
0/6This model set is for practice. It is not a question from any year annual examination.
1
Which of these has degree 3?
💡 Hint:
A cubic.
Explanation:
The degree of 2x³ − 5 is 3.
2
The zero of p(x) = x + 5 is —
💡 Hint:
x + 5 = 0.
Explanation:
The zero is −5.
3
(x − 5)² = x² − ______ x + 25.
💡 Hint:
10.
Explanation:
2×5 = 10.
4
Before division the terms are written in descending degree.
Explanation:
True. x + 3x² − 1 is written as 3x² + x − 1.
5
State the Factor Theorem and check x − 1 for x³ − 1.
Model answer (board format)
If p(a) = 0 then (x − a) is a factor, and the converse too. For p(x) = x³ − 1, p(1) = 0, so x − 1 is a factor.
Check your answer:
6
Expand (2x + 1)³. Then say how the condition x + y + z = 0 shortens Identity VIII.
Model answer (board format)
(2x)³ + 1³ + 3(2x)(1)(2x + 1) = 8x³ + 1 + 12x² + 6x = 8x³ + 12x² + 6x + 1. If x + y + z = 0 the right side is 0, so x³ + y³ + z³ = 3xyz.
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CBSE-style questions
0/6These are competency-based practice questions. They are not copies of a past paper.
1
A student wrote (x + 3)² = x² + 9. The missing term is —
💡 Hint:
2xy.
Explanation:
2·x·3 = 6x was left out. The full form is x² + 6x + 9.
2
p(x) = 2x² − 3x + 1 and a student finds p(1) for a zero. p(1) = 0. The conclusion is —
💡 Hint:
p(1) = 2 − 3 + 1.
Explanation:
p(1) = 0, so 1 is a zero and x − 1 is a factor.
3
Assertion (A): x² + 5x + 6 = (x + 2)(x + 3).
Reason (R): The product of 2 and 3 is 6 and their sum is 5.
Explanation:
Both are true and R explains A.
4
Assertion (A): The degree of the zero polynomial is 0.
Reason (R): A non-zero constant polynomial has degree 0.
Explanation:
A is false. R is true.
5
Amit evaluated p(1) while dividing x³ + 1 by x + 1. The remainder came out wrong. Which value should he use, and what is the correct remainder?
Model answer (board format)
x + 1 = x − (−1), so p(−1) = −1 + 1 = 0. The remainder is 0. p(1) = 2 is not the remainder for this divisor.
Check your answer:
6
A shop wants the square of 104 without direct multiplication. Find the value using an identity and the idea of the figure.
Model answer (board format)
104 = 100 + 4. (a+b)² = a² + 2ab + b², the four parts of the figure. 100² + 2·100·4 + 4² = 10000 + 800 + 16 = 10816.
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🏛️ Board exam corner — Bihar Board (BSEB)— CBSE
Switch board with BSEB | CBSE above. The lessons follow the same NCERT chapter.
BSEB model · practiceBoard tip
This page has no verified annual-exam question, because no source page has been added. The model set below is practice in the board pattern.
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CBSE-style · competency-based
These are case and assertion-reason practice items. Do not treat them as past CBSE questions.
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🧠 What you learned + equation sheet
What you learned
- In a polynomial in one variable every exponent is a whole number. A negative or fractional power does not make a polynomial.
- The degree is the highest power of the variable. A non-zero constant has degree 0. The degree of the zero polynomial is not defined.
- If p(c) = 0, then c is a zero. A linear polynomial has exactly one zero. A non-zero constant has none.
- The remainder on division by x − a is p(a). Before dividing, write the terms in descending degree.
- If p(a) = 0, then (x − a) is a factor. The middle term of a quadratic is split into two numbers.
- (a+b)² = a² + 2ab + b². If x + y + z = 0, then x³ + y³ + z³ = 3xyz.
| What | Keep this |
|---|---|
| Degree | highest power; 0 if the constant is not zero |
| Zero | p(c) = 0 |
| Remainder | division by x − a leaves p(a) |
| Factor | p(a) = 0 means (x − a) is a factor |
| (a+b)² | a² + 2ab + b² |
| Three cubes | x + y + z = 0 implies x³ + y³ + z³ = 3xyz |
The notes are original writing. The textbook was used only for activity order and numbers. “Verified” will be used only on a question that has a source page.