THE GOAL | Class 10 Mathematics

01. What is a Polynomial?

A polynomial in one variable is an algebraic expression in which the powers of the variable are non-negative integers.

A general polynomial in x can be written as:

aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₂x² + a₁x + a₀

Here, a₀, a₁, a₂, ... are real numbers.

💡 Simple meaning: Polynomial mein variable ki powers 0, 1, 2, 3... jaise non-negative integers honi chahiye.

✏ Example

2x³ − 5x² + 7x − 4

This is a polynomial in x.

But 1 x + 2 is not a polynomial because 1 x = x⁻¹.

02. Terms and Coefficients

The separate parts of a polynomial connected by + or − are called its terms. The numerical multiplier of a variable term is called its coefficient.

✏ Example

For:
5x³ − 7x² + 4x − 9

Terms are:
5x³, −7x², 4x, −9

Coefficient of x² = −7
Coefficient of x = 4
Constant term = −9

💡 Yaad rakho: Sign bhi coefficient ka part hota hai. Isliye x² ka coefficient −7 hai, 7 nahi.

03. Zero Polynomial

A polynomial in which all coefficients are zero is called the zero polynomial.

Example:

p(x) = 0

💡 Zero polynomial mein koi non-zero coefficient nahi hota.

04. Degree of a Polynomial

The highest power of the variable having a non-zero coefficient is called the degree of the polynomial.

✏ Example

p(x) = 7x⁴ − 3x² + 8x − 5

Highest power of x = 4

Therefore, Degree = 4

💡 Simple trick: Sabse badi power dekho — wahi degree hai.

05. Types of Polynomials According to Degree

Degree	Name	Example
0	Constant	7
1	Linear	2x + 3
2	Quadratic	x² − 5x + 6
3	Cubic	x³ − 4x + 1

✏ Example

3x² + 7x − 2

Highest power = 2
Therefore, it is a quadratic polynomial.

06. Monomial, Binomial and Trinomial

A polynomial having:
1 term → Monomial
2 terms → Binomial
3 terms → Trinomial

✏ Examples

7x² → Monomial

x + 5 → Binomial

x² + 3x + 2 → Trinomial

💡 Terms count karo, degree nahi.

07. Zero of a Polynomial ⭐

A number k is called a zero of a polynomial p(x) if:

p(k) = 0

💡 Simple meaning: Jis value ko x ki jagah put karne par polynomial ki value 0 ho jaaye, wahi zero hai.

✏ Example

p(x) = x − 5

Put x = 5:
p(5) = 5 − 5
= 0

Therefore, 5 is a zero of p(x).

08. Finding the Zero of a Linear Polynomial

✏ Example

Find the zero of:
p(x) = 3x − 12

For zero:
3x − 12 = 0
3x = 12
x = 4

Therefore, zero = 4.

💡 Linear polynomial ka zero find karne ke liye simply p(x) = 0 put karke x solve karo.

09. Geometrical Meaning of Zeroes ⭐⭐⭐

The zeroes of a polynomial are the x-coordinates of the points where its graph intersects or touches the x-axis.

x = 2

x = 5

x-axis →

y-axis ↑

✏ Example

If a graph intersects the x-axis at x = 2 and x = 5,

then its zeroes are:
2 and 5

💡 Graph mein x-axis ko cut/touch karne wale points ke x-coordinates zeroes hote hain.

10. Number of Zeroes from a Graph

The number of zeroes of a polynomial is equal to the number of points where its graph intersects or touches the x-axis.

✏ Example

If a graph cuts the x-axis at three distinct points,

Number of zeroes = 3

⭐ Exam Point:

More x-axis intersections → More zeroes.
No intersection with x-axis → No real zero.

11. Remainder Theorem ⭐⭐⭐

If a polynomial p(x) is divided by (x − a), then the remainder is:

Remainder = p(a)

💡 Shortcut: Long division karne ki zarurat nahi. Divisor x − a hai to directly x = a put karo.

✏ Example

Find the remainder when:
p(x) = x³ + 2x² − x + 5
is divided by x − 2.

By Remainder Theorem:
Remainder = p(2)
= 2³ + 2(2²) − 2 + 5
= 8 + 8 − 2 + 5
= 19

12. Remainder with Divisor (x + a)

If the divisor is:

x + a

write it as:

x − (−a)

Therefore, remainder = p(−a).

✏ Example

Find the remainder when:
p(x) = x² + 3x + 5
is divided by x + 2.

Remainder = p(−2)
= (−2)² + 3(−2) + 5
= 4 − 6 + 5
= 3

💡 x + 2 mein value −2 put hogi.

13. Factor Theorem ⭐⭐⭐

If p(a) = 0, then (x − a) is a factor of p(x). Conversely, if (x − a) is a factor of p(x), then p(a) = 0.

(x − a) is a factor of p(x) ⇔ p(a) = 0

💡 Factor check karna hai? Bas p(a) calculate karo. Answer 0 aaya → factor hai.

✏ Example

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

p(2) = 8 − 12 + 4
= 0

Therefore, x − 2 is a factor of p(x).

14. Proof of Factor Theorem

Suppose p(x) is divided by (x − a). By the division algorithm:

p(x) = (x − a)q(x) + r

Since the divisor is linear, the remainder r is a constant. Putting x = a:

p(a) = (a − a)q(a) + r
p(a) = r

Therefore, if p(a) = 0,
r = 0

Hence,
(x − a) is a factor of p(x).

💡 Exam mein proof likhte waqt equations ka order maintain karna. Last line mein clearly conclusion likhna.

15. Factorisation of a Quadratic Polynomial ⭐⭐⭐

A quadratic polynomial has the form:

ax² + bx + c, a ≠ 0

It can often be factorised by splitting the middle term.

✏ Example

Factorise:
x² + 7x + 12

We need two numbers whose:
Product = 12
Sum = 7

Numbers are 3 and 4.

Therefore:
x² + 7x + 12
= (x + 3)(x + 4)

💡 Aise questions mein do numbers find karo: product = last term and sum = middle coefficient when the leading coefficient is 1.

16. Splitting the Middle Term ⭐⭐⭐

✏ Example

Factorise:
2x² + 7x + 3

First × last coefficient:
2 × 3 = 6

Two numbers with:
Product = 6
Sum = 7

are 6 and 1.

Therefore:
2x² + 6x + x + 3
= 2x(x + 3) + 1(x + 3)
= (2x + 1)(x + 3)

💡 Pehle middle term ko suitable two terms mein split karo, phir grouping se common factor nikalo.

17. Factorising a Cubic Polynomial ⭐⭐⭐

For a cubic polynomial, we can use the Factor Theorem to find one factor and then factorise the remaining quadratic.

✏ Example

Factorise:
p(x) = x³ − 6x² + 11x − 6

Try x = 1:
p(1) = 1 − 6 + 11 − 6
= 0

Therefore:
x − 1 is a factor.

Hence:
p(x) = (x − 1)(x² − 5x + 6)
= (x − 1)(x − 2)(x − 3)

💡 Cubic factorisation mein possible zeroes try karke Factor Theorem use karna very useful hai.

18. Algebraic Identity: (x + y + z)²

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

✏ Example

For x = 1, y = 2, z = 3:

(1 + 2 + 3)² = 36

Using identity:
1² + 2² + 3² + 2(1)(2) + 2(2)(3) + 2(3)(1)
= 1 + 4 + 9 + 4 + 12 + 6
= 36

19. Identity: (x + y)³

(x + y)³ = x³ + y³ + 3xy(x + y)

✏ Example

(2 + 3)³
= 2³ + 3³ + 3(2)(3)(5)
= 8 + 27 + 90
= 125

20. Identity: (x − y)³

(x − y)³ = x³ − y³ − 3xy(x − y)

✏ Example

(5 − 2)³
= 5³ − 2³ − 3(5)(2)(3)
= 125 − 8 − 90
= 27

21. Sum of Two Cubes

x³ + y³ = (x + y)(x² − xy + y²)

✏ Example

Factorise:
x³ + 8

= x³ + 2³
= (x + 2)(x² − 2x + 4)

💡 Sum of cubes mein first bracket (x + y) hota hai, second bracket mein middle sign negative hota hai.

22. Difference of Two Cubes

x³ − y³ = (x − y)(x² + xy + y²)

✏ Example

Factorise:
x³ − 27

= x³ − 3³
= (x − 3)(x² + 3x + 9)

💡 Difference of cubes: same sign first bracket, opposite sign middle term in second bracket.

23. Identity: x³ + y³ + z³ − 3xyz

x³ + y³ + z³ − 3xyz

= (x + y + z) (x² + y² + z² − xy − yz − zx)

✏ Example

For x = 1, y = 2, z = 3:

LHS:
1³ + 2³ + 3³ − 3(1)(2)(3)
= 1 + 8 + 27 − 18
= 18

24. Special Case: x + y + z = 0 ⭐

If:

x + y + z = 0

then:

x³ + y³ + z³ = 3xyz

✏ Example

Let:
x = 1, y = 2
Since x + y + z = 0:
1 + 2 + z = 0
z = −3

Therefore:
x³ + y³ + z³
= 1 + 8 − 27
= −18

3xyz = 3(1)(2)(−3)
= −18 ✓

25. Using Identities for Factorisation

✏ Example

Factorise:
8x³ + 27

= (2x)³ + 3³

Using:
a³ + b³ = (a + b)(a² − ab + b²)

Therefore:
= (2x + 3)(4x² − 6x + 9)

💡 Pehle expression ko identity ke pattern mein convert karo.

26. Finding an Unknown Constant Using Factor Theorem

✏ Example

Find k if x − 2 is a factor of:
p(x) = x³ + kx² − 4x + 4

Since x − 2 is a factor:
p(2) = 0

8 + 4k − 8 + 4 = 0
4 + 4k = 0
4k = −4
k = −1

💡 “x − a is a factor” dikhe → directly p(a) = 0 use karo.

27. Finding a Missing Factor

✏ Example

If x − 1 is a factor of:
x³ − 4x² + 5x − 2

First verify:
p(1) = 1 − 4 + 5 − 2 = 0

So x − 1 is a factor.

After factorisation:
x³ − 4x² + 5x − 2
= (x − 1)(x² − 3x + 2)
= (x − 1)(x − 1)(x − 2)

28. Important Exam Type — Check a Factor

Write p(x).

Identify a from factor (x − a).

Calculate p(a).

If p(a) = 0 → factor is correct.

If p(a) ≠ 0 → factor is not a factor.

✏ Quick Example

Is x + 1 a factor of p(x)?

Since x + 1 = x − (−1),
calculate p(−1).

If p(−1) = 0,
x + 1 is a factor.

29. Common Mistakes ⚠️

1. Degree ≠ number of terms.

2. Zero of p(x) means p(k) = 0, not p(0).

3. Divisor x + a → put x = −a.

4. Factor x − a → check p(a).

5. In x³ + y³ identity, the middle sign is negative.

6. In x³ − y³ identity, the middle sign is positive.

7. Don't confuse coefficient with constant term.

30. Formula Sheet ⭐⭐⭐

Remainder on division by (x − a) = p(a)

p(a) = 0 ⇔ (x − a) is a factor

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

(x + y)³ = x³ + y³ + 3xy(x + y)

(x − y)³ = x³ − y³ − 3xy(x − y)

x³ + y³ = (x + y)(x² − xy + y²)

x³ − y³ = (x − y)(x² + xy + y²)

x³ + y³ + z³ − 3xyz

= (x + y + z) (x² + y² + z² − xy − yz − zx)

31. One-Minute Revision

Polynomial → powers of variable are non-negative integers.

Degree → highest power with non-zero coefficient.

Zero → p(k) = 0.

Graph → x-axis intersections/touching points give zeroes.

Remainder → put the zero of the linear divisor.

Factor Theorem → p(a) = 0 means x − a is a factor.

Quadratic factorisation → split middle term when suitable.

Cubic factorisation → Factor Theorem + quadratic factorisation.

Identities → recognise the pattern before expanding.

32. Exam Practice Mindset 🎯

Question says “remainder”?
→ Think Remainder Theorem.

Question says “factor”?
→ Think Factor Theorem.

Question gives x − a?
→ Put x = a.

Question gives x + a?
→ Put x = −a.

Question gives cubic polynomial?
→ Try a suitable value → find one factor → factorise the remaining quadratic.

Question looks like a³ ± b³?
→ Use cube identities.

🏆 POLYNOMIALS — FINAL CHECK

Concept → Understand

Example → Observe

Formula → Remember

Practice → Master

💡 THE GOAL Tip: Formula ratne se pehle uska pattern samjho. Question ko identify karo — phir correct theorem/identity apply karo.

✍ POLYNOMIALS

Class 10 Mathematics • Chapter 2