polynomials Archives – Mathemafia

Exercise 2.1 (NCERT) – Polynomials, Class 10 – All Solutions

Sunil Kumar — Thu, 12 Jun 2025 12:20:08 +0000

Exercise 2.1 (NCERT) - Polynomials, Class 10

Chapter 2	Exercise 2.1	Class 10

Question 1

The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

The number of zeros of a graph is the number of times the graph intersects the x-axis.

The answers are written on the graphs below in red circle.

The post Exercise 2.1 (NCERT) – Polynomials, Class 10 – All Solutions appeared first on Mathemafia.

Quiz – Polynomials

Sunil Kumar — Sat, 07 Jun 2025 11:04:11 +0000

Quiz – Polynomials

Topic: Polynomials, Mathematics.

Target Audience: Middle to high school

Total Questions available

This quiz is about one of the most sort after topics of Mathematics – Polynomials.

Good Luck!!!

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For class 10 students, complete solution of Chapter Polynomials, NCERT is available on mathemafia.com. Click the following link to view the solutions and watch the video tutorials.

Class 10, Ex: 2.2 – Polynomials – All Solutions

The post Quiz – Polynomials appeared first on Mathemafia.

Exercise 2.2 (NCERT) – Polynomials, Class 10 – All Solutions

Sunil Kumar — Wed, 05 Mar 2025 08:52:17 +0000

Exercise 2.2 (NCERT) - Polynomials, Class 10

Chapter 2	Exercise 2.2	Class 10

Below are the Quick links for all questions of Exercise 2.2, Real Numbers, Class 10.

Click a link to view the solution of corresponding question.

Relationships of Zeros and Coefficients

Splitting the Middle Term

Ex: 2.2, Question 1
Ex: 2.2, Question 2

Relationships of Zeros and Coefficients

This exercise is about very basic concepts of Zeros of Polynomials. The zero of a polynomial, P(x), is the value of x for which the value of P(x) is 0. For a quadratic polynomial ax²+bx+c there can be exactly 2 real zeros, say α and β. The relationships between the zeros of the quadratic polynomial and the coefficients of the polynomial are:

𝑃(𝑥)=𝑎𝑥2+𝑏𝑥+𝑐𝑆=Sum of zeros=α+β=−𝑏𝑎𝑃=Product of zeros=α×β=𝑐𝑎

Now if we already know the Sum (S) and Product (P) of a polynomial, the we can directly get the polynomial using the formula written below.

𝑆=Sum of zeros𝑃=Product of zeros𝑡ℎ𝑒𝑛 𝑃(𝑥)=𝑥2−𝑆𝑥+𝑃

Splitting the Middle Term

If you are able to understand the image displayed below then you will solve all questions really easily.

Before we start solving the questions of this exercise, it is very important to understand the concept used to solve these questions – Splitting the Middle Term of a quadratic polynomial. You are suggested to watch the video embedded below completely to understand 🧠 the concept. The tutorial is slightly lengthy so please have some patience in case you want to dive deep into it. 🤿 👀 ❤️ 💃

How it works?

The coefficient of x, b, is split in two parts, say b1 and b2, such that

b1 + b2 = b and

b1 x b2 = a x c

To find the required pair of b1 and b2 you must use prime factorisation of b.

Do you love Mathematics? If yes, then watch this video to dive deep into the underlying concept.

Below are the solutions of every question of exercise 2.2 of chapter 1, Polynomials, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.

Question 1

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) \( \begin{aligned} x^2-2x-8 \end{aligned}\)

𝑥2−2𝑥−8=𝑥2−4𝑥+2𝑥−8=𝑥(𝑥−4)+2(𝑥−4)=(𝑥−4)(𝑥+2)

𝑃𝑢𝑡𝑡𝑖𝑛𝑔 (𝑥−4)(𝑥+2)=0

⇒ 𝑥−4=0 ⇒ 𝑥=4 𝑜𝑟 𝑥+2=0𝑜𝑟 𝑥=−2

Answer: 4 & -2 are the zeros of the polynomial.

Verification

Comparing the polynomial with ax²+ bx + c

𝑎=1,𝑏=−2,𝑐=−8Sum of roots=−𝑏𝑎=−(−2)1=2Sum of roots=α+β=4+(−2)=2𝐻𝑒𝑛𝑐𝑒 α+β=−𝑏𝑎

Product of roots=𝑐𝑎=−81=−8Product of roots=α×β=4×(−2)=−8𝐻𝑒𝑛𝑐𝑒 α×β=𝑐𝑎

(ii) \( \begin{aligned} 4s^2-4s + 1 \end{aligned}\)

4𝑠2−4𝑠+1=4𝑠2−2𝑠−2𝑠+1=2𝑠(2𝑠−1)−1(2𝑠−1)=(2𝑠−1)(2𝑠−1)

𝑃𝑢𝑡𝑡𝑖𝑛𝑔 (2𝑠−1)(2𝑠−1)=0

⇒2𝑠−1=0 𝑜𝑟 2𝑠−1=0⇒𝑠=12 𝑜𝑟 𝑠=12

Answer: 1/2 & 1/2 are the zeros of the polynomial.

Verification

Comparing the polynomial with ax²+ bx + c

𝑎=4,𝑏=−4,𝑐=1Sum of roots=−𝑏𝑎=−(−4)4=1Sum of roots=α+β=12+12=1𝐻𝑒𝑛𝑐𝑒 α+β=−𝑏𝑎

Product of roots=𝑐𝑎=14Product of roots=α×β=12×12=14𝐻𝑒𝑛𝑐𝑒 α×β=𝑐𝑎

(iii) \( \begin{aligned} 6x^2-3-7x \end{aligned}\)

6𝑥2−3−7𝑥=6𝑥2−7𝑥−3=6𝑥2−9𝑥+2𝑥−3=3𝑥(2𝑥−3)+1(2𝑥−3)=(2𝑥−3)(3𝑥+1)

𝑃𝑢𝑡𝑡𝑖𝑛𝑔 (2𝑥−3)(3𝑥+1)=0

⇒2𝑥−3=0 ⇒𝑥=32 𝑜𝑟 3𝑥+1=0𝑜𝑟 𝑥=−13

Answer: 3/2 & -1/3 are the zeros of the polynomial.

Verification

Comparing the polynomial with ax²+ bx + c

𝑎=6,𝑏=−7,𝑐=−3Sum of roots=−𝑏𝑎=−(−7)6=76Sum of roots=α+β=32–13=9−26=76𝐻𝑒𝑛𝑐𝑒 α+β=−𝑏𝑎

Product of roots=𝑐𝑎=−36=−12Product of roots=α×β=32×−13=−12𝐻𝑒𝑛𝑐𝑒 α×β=𝑐𝑎

(iv) \( \begin{aligned} 4u^2+ 8u \end{aligned}\)

4𝑢2+8𝑢=4𝑢(𝑢+2)

𝑃𝑢𝑡𝑡𝑖𝑛𝑔 4𝑢(𝑢+2)=0

⇒4𝑢=0 ⇒𝑢=0 𝑜𝑟 𝑢+2=0𝑜𝑟 𝑢=−2

Answer: 0 & -2 are the zeros of the polynomial.

Verification

Comparing the polynomial with ax²+ bx + c

𝑎=4,𝑏=8,𝑐=0Sum of roots=−𝑏𝑎=−84=−2Sum of roots=α+β=0+(−2)=−2𝐻𝑒𝑛𝑐𝑒 α+β=−𝑏𝑎

Product of roots=𝑐𝑎=04=0Product of roots=α×β=0×(−2)=0𝐻𝑒𝑛𝑐𝑒 α×β=𝑐𝑎

(v) \( \begin{aligned} t^2-15 \end{aligned}\)

𝑡2−15=(𝑡+15‾‾‾√)(𝑡−15‾‾‾√)

𝑃𝑢𝑡𝑡𝑖𝑛𝑔 (𝑡+15‾‾‾√)(𝑡−15‾‾‾√)=0

⇒𝑡+15‾‾‾√=0 ⇒𝑡=−15‾‾‾√ 𝑜𝑟 𝑡−15‾‾‾√=0𝑜𝑟 𝑡=15‾‾‾√

Answer: √15 & -√15 are the zeros of the polynomial.

Verification

Comparing the polynomial with ax²+ bx + c

𝑎=1,𝑏=0,𝑐=−15Sum of roots=−𝑏𝑎=01=0Sum of roots=α+β=15‾‾‾√–15‾‾‾√=0𝐻𝑒𝑛𝑐𝑒 α+β=−𝑏𝑎

Product of roots=𝑐𝑎=−151=−15Product of roots=α×β=15‾‾‾√×(−15‾‾‾√)=−15𝐻𝑒𝑛𝑐𝑒 α×β=𝑐𝑎

(vi) \( \begin{aligned} 3x^2-x-4 \end{aligned}\)

3𝑥2−𝑥−4=3𝑥2−4𝑥+3𝑥−4=𝑥(3𝑥−4)+1(3𝑥−4)=(3𝑥−4)(𝑥+1)

𝑃𝑢𝑡𝑡𝑖𝑛𝑔 (3𝑥−4)(𝑥+1)=0

⇒3𝑥−4=0 ⇒𝑥=43 𝑜𝑟 𝑥+1=0𝑜𝑟 𝑥=−1

Answer: 4/3 & -1 are the zeros of the polynomial.

Verification

Comparing the polynomial with ax²+ bx + c

𝑎=3,𝑏=−1,𝑐=−4Sum of roots=−𝑏𝑎=−(−1)3=13Sum of roots=α+β=43−1=13𝐻𝑒𝑛𝑐𝑒 α+β=−𝑏𝑎

Product of roots=𝑐𝑎=−43Product of roots=α×β=43×(−1)=−43𝐻𝑒𝑛𝑐𝑒 α×β=𝑐𝑎

Question 2

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) 1/4, -1

𝑆=Sum of zeros=14𝑃=Product of zeros=−1

Now P(x) = x²– Sx + P

𝑃(𝑥)=𝑥2−𝑆𝑥+𝑃=𝑥2−(14)𝑥+(−1)=𝑥2−𝑥4−1

We can stop solving the question further and the polynomial written above is the correct answer. However, if we multiply this polynomial by any non-zero real number, the resulting polynomial will also have the same zeros.

Multiply the P(x) by 4 to remove 4 from the denominator.

4(𝑥2−𝑥4−1)=4𝑥2−𝑥−4

(ii) √2 , 1/3

𝑆=Sum of zeros=2‾√𝑃=Product of zeros=13

Now P(x) = x²– Sx + P

𝑃(𝑥)=𝑥2−𝑆𝑥+𝑃=𝑥2−2‾√𝑥+13

Multiply the P(x) by 3 to remove 3 from the denominator.

3(𝑥2−2‾√𝑥+13)=3𝑥2−32‾√𝑥+1

(iii) 0, √5

𝑆=Sum of zeros=0𝑃=Product of zeros=5‾√

Now P(x) = x²– Sx + P

𝑃(𝑥)=𝑥2−𝑆𝑥+𝑃=𝑥2−0𝑥+5‾√=𝑥2+5‾√

(iv) 1, 1

𝑆=Sum of zeros=1𝑃=Product of zeros=1

Now P(x) = x²– Sx + P

𝑃(𝑥)=𝑥2−𝑆𝑥+𝑃=𝑥2−1𝑥+1=𝑥2−𝑥+1

(v) -1/4, 1/4

𝑆=Sum of zeros=−14𝑃=Product of zeros=14

Now P(x) = x²– Sx + P

𝑃(𝑥)=𝑥2−𝑆𝑥+𝑃=𝑥2−(−14)𝑥+14=𝑥2+14𝑥+14

Multiply the P(x) by 4 to remove 4 from the denominator.

4(𝑥2+14𝑥+14)=4𝑥2+𝑥+1

(vi) 4, 1

𝑆=Sum of zeros=4𝑃=Product of zeros=1

Now P(x) = x²– Sx + P

𝑃(𝑥)=𝑥2−𝑆𝑥+𝑃=𝑥2−4𝑥+1

\( \begin{aligned} \end{aligned}\)

The post Exercise 2.2 (NCERT) – Polynomials, Class 10 – All Solutions appeared first on Mathemafia.

Rational Root Theorem – Proof

Sunil Kumar — Sat, 15 Feb 2025 14:54:11 +0000

Rational Root Theorem

Statement: Given a polynomial

𝑃(𝑥)=𝑎𝑛𝑥𝑛+𝑎𝑛−1𝑥𝑛−1+𝑎𝑛−2𝑥𝑛−2+…….+𝑎1𝑥+𝑎0

with integral coefficients where a_n ≠ 0, a₀ ≠ 0. if P(x) has a rational root p/q where p and q are relatively prime integers, then p is a divisor of a₀ and q is a divisor of a_n.

Proof

𝑃(𝑥)=𝑎𝑛𝑥𝑛+𝑎𝑛−1𝑥𝑛−1+𝑎𝑛−2𝑥𝑛−2+…….+𝑎1𝑥+𝑎0𝑤ℎ𝑒𝑟𝑒 𝑎0,…,𝑎𝑛∈ℤ 𝑎𝑛𝑑 𝑎0,𝑎𝑛≠0.

Let p/q be a rational root of this equation where p and q are coprime integers and q ≠ 0.

⇒𝑃(𝑝𝑞)=0⇒𝑎𝑛(𝑝𝑞)𝑛+𝑎𝑛−1(𝑝𝑞)𝑛−1+𝑎𝑛−2(𝑝𝑞)𝑛−2+…….+𝑎1(𝑝𝑞)+𝑎0=0

Multiply both sides by qⁿ

𝑞𝑛×[𝑎𝑛(𝑝𝑞)𝑛+𝑎𝑛−1(𝑝𝑞)𝑛−1+𝑎𝑛−2(𝑝𝑞)𝑛−2+…….+𝑎2(𝑝𝑞)2+𝑎1(𝑝𝑞)+𝑎0]=𝑞𝑛×0𝑎𝑛(𝑝𝑞)𝑛𝑞𝑛+𝑎𝑛−1(𝑝𝑞)𝑛−1𝑞𝑛+𝑎𝑛−2(𝑝𝑞)𝑛−2𝑞𝑛+…….+𝑎2(𝑝𝑞)2𝑞𝑛+𝑎1(𝑝𝑞)𝑞𝑛+𝑎0𝑞𝑛=0𝑎𝑛𝑝𝑛+𝑎𝑛−1𝑝𝑛−1𝑞+𝑎𝑛−2𝑝𝑛−2𝑞2+…….+𝑎2𝑝2𝑞𝑛−2+𝑎1𝑝𝑞𝑛−1+𝑎0𝑞𝑛=0 …(1)

Imp: We multiplied by qⁿto get rid of terms in the denominator.

Video tutorial: Proof of Rational Root Theorem

Proof of Part 1: p divides a₀

Let’s write equation (1) again

𝑎𝑛𝑝𝑛+𝑎𝑛−1𝑝𝑛−1𝑞+𝑎𝑛−2𝑝𝑛−2𝑞2+…….+𝑎2𝑝2𝑞𝑛−2+𝑎1𝑝𝑞𝑛−1+𝑎0𝑞𝑛=0 …(1)

Take a₀qⁿ to the other side of the equation.

𝑎𝑛𝑝𝑛+𝑎𝑛−1𝑝𝑛−1𝑞+𝑎𝑛−2𝑝𝑛−2𝑞2+…….+𝑎2𝑝2𝑞𝑛−2+𝑎1𝑝𝑞𝑛−1= –𝑎0𝑞𝑛

Now p is common in every term on LHS. This implies:

𝑝[𝑎𝑛𝑝𝑛−1+𝑎𝑛−1𝑝𝑛−2𝑞+𝑎𝑛−2𝑝𝑛−3𝑞2+…….+𝑎2𝑝𝑞𝑛−2+𝑎1𝑞𝑛−1]= –𝑎0𝑞𝑛

Now p, q and a₀, a₁,….a_n,all are integers. Hence LHS is an integer.

⇒𝑝|𝐿𝐻𝑆⇒𝑝|𝑅𝐻𝑆⇒𝑝|𝑎0𝑞𝑛

As we know that p and q are coprime

⇒𝑝 does not divide 𝑞⇒𝑝 does not divide 𝑞𝑛

But because 𝑝|𝑎0𝑞𝑛⇒𝑝|𝑎0

Proof of Part 2: q divides a_n

Let’s go back to equation (1)

𝑎𝑛𝑝𝑛+𝑎𝑛−1𝑝𝑛−1𝑞+𝑎𝑛−2𝑝𝑛−2𝑞2+…….+𝑎2𝑝2𝑞𝑛−2+𝑎1𝑝𝑞𝑛−1+𝑎0𝑞𝑛=0 …(1)

Take a_npⁿ to the other side of the equation.

𝑎𝑛−1𝑝𝑛−1𝑞+𝑎𝑛−2𝑝𝑛−2𝑞2+…….+𝑎2𝑝2𝑞𝑛−2+𝑎1𝑝𝑞𝑛−1+𝑎0𝑞𝑛=−𝑎𝑛𝑝𝑛

Now q is common in every term on LHS. This implies:

𝑞[𝑎𝑛−1𝑝𝑛−1+𝑎𝑛−2𝑝𝑛−2𝑞+…….+𝑎2𝑝2𝑞𝑛−3+𝑎1𝑝𝑞𝑛−2+𝑎0𝑞𝑛−1]=−𝑎𝑛𝑝𝑛

Now p, q and a₀, a₁,….a_n,all are integers. Hence LHS is an integer.

⇒𝑞|𝐿𝐻𝑆⇒𝑞|𝑅𝐻𝑆⇒𝑞|𝑎𝑛𝑝𝑛

As we know that p and q are coprime

⇒𝑞 does not divide 𝑝⇒𝑞 does not divide 𝑝𝑛

But because 𝑞|𝑎𝑛𝑝𝑛⇒𝑞|𝑎𝑛

Hope you find this proof useful.

Let’s now use this theorem to find the roots of a polynomial of degree 3.

Example of Rational Root Theorem

𝑃(𝑥)=2𝑥3–3𝑥2–2𝑥+3

Find the roots of P(x) using Rational Root Theorem.

This equation is of the form

𝑃(𝑥)=𝑎3𝑥3+𝑎2𝑥2+𝑎1𝑥+𝑎0

⇒𝑎3=2 𝑎𝑛𝑑 𝑎0=3

As per rational root theorem, if this polynomial has any rational root, say p/q, then

⇒ 𝑝 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 𝑎0 , 𝑞 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 𝑎3𝑝 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 3 , 𝑞 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 2

Here are the possible values of p/q:

𝑝𝑞⇒𝑝𝑞=±1, ±3=±1, ±2=±11,±31,±12,±32=±1,±3,±12,±32

Now, if P(x) has any rational root, it has to be from the possible values of p/q written above. If P(x) = 0 for any of these values, that will be a rational root of the polynomial. The three values for which P(x) = 0 are 1, -1 and 3/2.

𝑃(𝑥)𝑃(1)𝑃(−1)𝑃(32)=2𝑥3–3𝑥2–2𝑥+3=2−3−2+3=0=−2−3+2+3=0=2×278−3×94−2×32+3=274–274−3+3=0

Rational Root Theorem is a very important theorem of Polynomials and is intended for the students learning higher mathematics and for those who are willing to dive deep into the core mathematics. For lower classes, like class 9 and class 10, this theorem comes handy when we have to find the roots of polynomials with degree 3 or more.

\(\begin{aligned} \end{aligned}\)

The post Rational Root Theorem – Proof appeared first on Mathemafia.

Class 10 – Chapter 2 – Polynomials – Last Minute Formula Revision

Sunil Kumar — Sun, 02 Feb 2025 09:06:02 +0000

⏰ Last minute Formula Revision 🔥 Polynomials, Chapter 2, Class 10

This short video covers all the formulas of chapter 2 (Polynomials) of class 10. It is recommended to watch this before your exam to quickly recall all the formulas of this chapter.

The tutorial is available in Hinglish + 9 different languages: English, French, German, Hindi, Spanish, Indonesian, Italian, Japanese, Portuguese

The post Class 10 – Chapter 2 – Polynomials – Last Minute Formula Revision appeared first on Mathemafia.

Class 10 | Chapter 2 | All Solutions | Polynomials

Sunil Syal — Fri, 17 Dec 2021 07:37:47 +0000

This playlist covers all the questions, concepts and examples of chapter 2 (Polynomials) of class 10, NCERT book. Click a thumbnail to watch the tutorial.

Watch complete chapter on YouTube

Example 1 – Chapter 2 – Class 10, Maths | Polynomials | NCERT Maths | CBSE

Example 2 – Chapter 2 – Class 10, Maths | Polynomials | NCERT Maths | CBSE

Example 3 – Chapter 2 – Class 10, Maths | Polynomials | NCERT Maths | CBSE

Example 4 – Chapter 2 – Class 10, Maths | Polynomials | NCERT Maths | CBSE

Example 5* – Chapter 2 – Class 10, Maths | Polynomials | NCERT Maths | CBSE

The post Class 10 | Chapter 2 | All Solutions | Polynomials appeared first on Mathemafia.

Class 9 Chapter 2 | All Solutions | Polynomials

Sunil Syal — Thu, 16 Dec 2021 15:32:46 +0000

Here you will find complete solution of all questions, concepts, theorems and examples of class 9 chapter 2 (Polynomials) of class 9, NCERT textbook.

There are few very important concepts that students must learn to get a good understanding of polynomials:

All above mentioned tutorials are included in the list displayed below for class 9 chapter 2 (Polynomials). You are strongly advised to watch all these videos thoroughly and do not miss any of these if you really want to get good hold on polynomials.

Click a thumbnail to watch the tutorial.

Watch complete chapter on YouTube

Introduction to Polynomials | Introduction to Algebra | Class 9, Chapter 2, NCERT Maths

Class 9 | Chapter 2 | Polynomials Example 1 | Finding Degree | NCERT Maths

Question 1 Exercise 2.1 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 2 Exercise 2.1 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 3 Exercise 2.1 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 4 Exercise 2.1 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 5 Exercise 2.1 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Class 9 | Chapter 2 | Polynomials Example 2 explained | NCERT Maths

Class 9 | Chapter 2 | Polynomials Example 3 explained | NCERT Maths

Class 9 | Chapter 2 | Polynomials Example 4 explained | NCERT Maths

Class 9 | Chapter 2 | Polynomials Example 5 explained | NCERT Maths

Question 1 Exercise 2.2 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 2 Exercise 2.2 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 3 Exercise 2.2 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 4 Exercise 2.2 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

FACTOR THEOREM Proof Explained | Class 9 | Chapter 2 | Polynomials, 2.5 Factorisation of Polynomials

Example 6 Chapter 2 Class 9 | Polynomials, Factor Theorem Questions | NCERT Maths

Example 7 Chapter 2 Class 9 | Polynomials, Factor Theorem Questions | NCERT Maths

Factorisation of Polynomials by SPLITTING THE MIDDLE TERM Explained| Class 9 |Chapter 2| NCERT Maths

Example 8 Chapter 2 Class 9 | Polynomials | 1St Solution | Splitting Middle Term Questions | Maths

Example 8 Chapter 2 Class 9 | Polynomials | 2nd Solution | Factor Theorem Questions |NCERT Maths

Example 9 Chapter 2 Class 9 | Polynomials | Factorise using Factor Theorem Questions |NCERT Maths

Example 10 Chapter 2 Class 9 | Polynomials | Factorise cubic or degree 3 polynomials | NCERT Maths

Question 1 Exercise 2.3 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 2 Exercise 2.3 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 3 Exercise 2.3 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 4 Exercise 2.3 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 5 Exercise 2.3 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

EASY TIPS & TRICKS 🔥 TO LEARN ALGEBRAIC IDENTITIES | Class 9 | Chapter 2 | Polynomials | NCERT Maths

Example 11 Chapter 2 Class 9 | Polynomials | Product of polynomials using algebraic identities

Example 12 Chapter 2 Class 9 | Polynomials| Example 17 |Product of big numbers without multiplying

Example 13 Chapter 2 Class 9 | Polynomials | Factorise polynomials using algebra identities | NCERT

Example 14 Chapter 2 Class 9 | Polynomials | Expand polynomials using algebraic identities | NCERT

Example 15 Chapter 2 Class 9 | Polynomials | How to expand polynomials using algebraic formulae |

Example 16 Chapter 2 Class 9 | Polynomials | Factorise polynomials using algebraic identities

Example 17 Chapter 2 Class 9 | Polynomials | How to expand polynomials using algebraic formulae

Example 18 Chapter 2 Class 9 | Polynomials | Example 23 | Cubes of big numbers without multiplying

Example 19 Chapter 2 Class 9 | Polynomials |Factorise cubic polynomials using algebraic identities

Example 20 Chapter 2 Class 9 | Polynomials | Factorise cubic polynomials using algebraic identities

Question 1 Exercise 2.4 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 2 Exercise 2.4 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 3 Exercise 2.4 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 4 Exercise 2.4 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 5 Exercise 2.4 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 6 Exercise 2.4 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 7 Exercise 2.4 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 8 Exercise 2.4 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 9 Exercise 2.4 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 10 Exercise 2.4 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Question 11 Exercise 2.4 Class 9 | Chapter 2 (Polynomials) | NCERT Maths | Solution

Class 9 Chapter 3 | Coordinate Geometry

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Class 10 | Chapter 2 | Polynomials | Exercise 2.1

Sunil Syal — Sun, 20 Jun 2021 04:41:12 +0000

Question 1: The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

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Class 10 – Mathematics – NCERT Solutions

Real Numbers
- Exercise 1.1
- Exercise 1.2
Polynomials
- Exercise 2.1
- Exercise 2.2
Pair of Linear Equation in Two Variables

Quadratic Equations
Arithmetic Progressions
- Exercise 5.1
- Exercise 5.2
Triangles
- Exercise 6.1
- Exercise 6.2
- Exercise 6.3

Coordinate Geometry
- Exercise 7.1
- Exercise 7.2
Introduction to Trigonometry
Some Applications of Trigonometry
- Exercise 9.1
Circles
- Exercise 10.1
- Exercise 10.2

Areas Related to Circles
- Exercise 11.1
Surface Areas and Volumes
- Exercise 12.1
- Exercise 12.2
Statistics
- Exercise 13.1
- Exercise 13.2
- Exercise 13.3
Probability
- Exercise 14.1

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Calculators
- Quadratic Equation Calculator – Solve any Equation
Super Concepts

Theorems
Quizzes

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Class 10 | Chapter 2 | Polynomials | Theorems & Concepts

Sunil Syal — Sun, 20 Jun 2021 03:44:55 +0000

Concept: In this tutorial, you will understand what is a Linear, Quadratic and Cubic polynomial and also what is a zero of a polynomial and its geometric meaning. Concepts are explained with examples and graphically.

Concept: Relationship between zeroes and coefficients of Linear Polynomials, Ch 2 Class 10 Maths Concepts

Concept: Relationship between zeroes and coefficients of Quadratic Polynomials, Ch 2 Class 10 Maths Concepts

Concept: Relationship between zeroes and coefficients of Cubic Polynomials, Ch 2 Class 10 Maths Concepts

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Class 10 – Mathematics – NCERT Solutions

Real Numbers
- Exercise 1.1
- Exercise 1.2
Polynomials
- Exercise 2.1
- Exercise 2.2
Pair of Linear Equation in Two Variables

Quadratic Equations
Arithmetic Progressions
- Exercise 5.1
- Exercise 5.2
Triangles
- Exercise 6.1
- Exercise 6.2
- Exercise 6.3

Coordinate Geometry
- Exercise 7.1
- Exercise 7.2
Introduction to Trigonometry
Some Applications of Trigonometry
- Exercise 9.1
Circles
- Exercise 10.1
- Exercise 10.2

Areas Related to Circles
- Exercise 11.1
Surface Areas and Volumes
- Exercise 12.1
- Exercise 12.2
Statistics
- Exercise 13.1
- Exercise 13.2
- Exercise 13.3
Probability
- Exercise 14.1

Super Cool Stuff

Calculators
- Quadratic Equation Calculator – Solve any Equation
Super Concepts

Theorems
Quizzes

The post Class 10 | Chapter 2 | Polynomials | Theorems & Concepts appeared first on Mathemafia.

Class 10 | Chapter 2 | Polynomials | Exercise 2.3

Sunil Syal — Sun, 20 Jun 2021 03:19:37 +0000

Question 1: Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following :
(i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2
(ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2

Question 2: Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: (i) t2 – 3 , 2t4 + 3t3 – 2t2 – 9t – 12
(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2 (iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1

Question 3: Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are 5/3 and – 5/3⋅

Question 4: On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Find g(x).

Question 5: Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x) (ii) deg q(x) = deg r(x) (iii) deg r(x) = 0

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Class 10 – Mathematics – NCERT Solutions

Real Numbers
- Exercise 1.1
- Exercise 1.2
Polynomials
- Exercise 2.1
- Exercise 2.2
Pair of Linear Equation in Two Variables

Quadratic Equations
Arithmetic Progressions
- Exercise 5.1
- Exercise 5.2
Triangles
- Exercise 6.1
- Exercise 6.2
- Exercise 6.3

Coordinate Geometry
- Exercise 7.1
- Exercise 7.2
Introduction to Trigonometry
Some Applications of Trigonometry
- Exercise 9.1
Circles
- Exercise 10.1
- Exercise 10.2

Areas Related to Circles
- Exercise 11.1
Surface Areas and Volumes
- Exercise 12.1
- Exercise 12.2
Statistics
- Exercise 13.1
- Exercise 13.2
- Exercise 13.3
Probability
- Exercise 14.1

Super Cool Stuff

Calculators
- Quadratic Equation Calculator – Solve any Equation
Super Concepts

Theorems
Quizzes

The post Class 10 | Chapter 2 | Polynomials | Exercise 2.3 appeared first on Mathemafia.

polynomials Archives – Mathemafia

Exercise 2.1 (NCERT) – Polynomials, Class 10 – All Solutions

Exercise 2.1 (NCERT) - Polynomials, Class 10

Question 1

Quiz – Polynomials

Quiz – Polynomials

Total Questions available

Complexity Meter

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Exercise 2.2 (NCERT) – Polynomials, Class 10 – All Solutions

Exercise 2.2 (NCERT) - Polynomials, Class 10

Relationships of Zeros and Coefficients

Splitting the Middle Term

Question 1

Question 2

Rational Root Theorem – Proof

Rational Root Theorem

𝑃(𝑥)=2𝑥3–3𝑥2–2𝑥+3

Find the roots of P(x) using Rational Root Theorem.

Class 10 – Chapter 2 – Polynomials – Last Minute Formula Revision

⏰ Last minute Formula Revision 🔥 Polynomials, Chapter 2, Class 10

Class 10 | Chapter 2 | All Solutions | Polynomials

Class 9 Chapter 2 | All Solutions | Polynomials

Class 10 | Chapter 2 | Polynomials | Exercise 2.1

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Class 10 – Mathematics – NCERT Solutions

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Class 10 | Chapter 2 | Polynomials | Theorems & Concepts

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Class 10 – Mathematics – NCERT Solutions

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Class 10 | Chapter 2 | Polynomials | Exercise 2.3

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Class 10 – Mathematics – NCERT Solutions

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