Class 9 Maths Chapter 2 Polynomials (বহুপদ) Question Answer Assamese Medium SEBA
Are you looking for the solutions for Class 9 Maths Chapter 2 (বহুপদ) for SEBA Board? Mathematics can be tough, but understanding the concepts in your own language makes it much easier.
In this post, we provide easy, step-by-step solutions in Assamese Medium for the entire chapter (Exercise 2.1 to 2.5). Whether you are preparing for your Unit Test or Final Exams, these notes will help you score full marks.
What is a Polynomial? (বহুপদ ৰাশি কি?)
Before solving the questions, let's understand the basics.
গণিতত, চলক (variable) আৰু ধ্ৰুৱক (constant) লগ লাগি যি গাণিতিক ৰাশি গঠন কৰে, তাক বীজগণিতীয় ৰাশি বোলে। যিবোৰ ৰাশিত চলকৰ সূচক (power) অঋণাত্মক অখণ্ড সংখ্যা (Whole number like 0, 1, 2, 3...) হয়, তেনে ৰাশিক বহুপদ ৰাশি (Polynomials) বোলা হয়।
| English Name | Assamese Name | Example |
|---|---|---|
| Monomial | একপদ ৰাশি | $$2x, 5y^2$$ |
| Binomial | দ্বিপদ ৰাশি | $$x + 1, y^2 - 4$$ |
| Trinomial | ত্ৰিপদ ৰাশি | $$x^2 + 2x + 1$$ |
Class 9 Maths Exercise 2.1 Solution (অনুশীলনী 2.1)
Q1. Which of the following expressions are polynomials? (তলৰ কোনবোৰ ৰাশি বহুপদ?)
- (i) $$4x^2 - 3x + 7$$
✔ উত্তৰ: হয়। (কাৰণ ইয়াত x-ৰ সূচকবোৰ 2 আৰু 1, যি অখণ্ড সংখ্যা।) - (ii) $$y^2 + \sqrt{2}$$
✔ উত্তৰ: হয়। - (iii) $$3\sqrt{t} + t\sqrt{2}$$
✘ উত্তৰ: নহয়। (কাৰণ ইয়াত প্ৰথম পদটোত t-ৰ সূচক $$1/2$$, যি অখণ্ড সংখ্যা নহয়।)
Q2. Write the coefficient of $$x^2$$. ($$x^2$$-ৰ সহগ লিখা)
- (i) $$2 + x^2 + x$$ → উত্তৰ: 1
- (ii) $$2 - x^2 + x^3$$ → উত্তৰ: -1
- (iii) $$\frac{\pi}{2}x^2 + x$$ → উত্তৰ: $$\frac{\pi}{2}$$
Zeros of a Polynomial (বহুপদৰ শূন্য) - Exercise 2.2
Q1. Find the value of the polynomial $$5x - 4x^2 + 3$$ at $$x = 0$$.
($$x = 0$$ ত $$5x - 4x^2 + 3$$ ৰ মান নিৰ্ণয় কৰা।)
Solution:
ধৰা হ'ল, $$p(x) = 5x - 4x^2 + 3$$
এতিয়া, $$x = 0$$ বহুৱাই পাওঁ,
$$p(0) = 5(0) - 4(0)^2 + 3$$
$$p(0) = 0 - 0 + 3$$
$$p(0) = 3$$ (উত্তৰ)
Remainder Theorem (ভাগশেষ উপপাদ্য) - Exercise 2.3
Rule: যদি এটা বহুপদ $$p(x)$$-ক $$(x - a)$$-ৰে হৰণ কৰা হয়, তেন্তে ভাগশেষ হ'ব $$p(a)$$।
Q1. Find the remainder when $$x^3 + 3x^2 + 3x + 1$$ is divided by $$x + 1$$.
Solution:
ধৰা হ'ল ভাজক, $$x + 1 = 0 \implies x = -1$$
এতিয়া, $$p(x)$$ ত $$x = -1$$ বহুৱাই পাওঁ:
$$p(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1$$
$$= -1 + 3(1) - 3 + 1$$
$$= 0$$
উত্তৰ: নিৰ্ণেয় ভাগশেষ 0।
Factorisation (উৎপাদক বিশ্লেষণ) - Exercise 2.4
Q4. Factorise: $$12x^2 - 7x + 1$$
This method is called Middle Term Splitting (মাজৰ পদ বিভাজন)।
$$= 12x^2 - (4 + 3)x + 1$$
$$= 12x^2 - 4x - 3x + 1$$
$$= 4x(3x - 1) - 1(3x - 1)$$
$$= (3x - 1)(4x - 1)$$ (উত্তৰ)
Algebraic Identities (বীজগণিতীয় অভেদ)
Memorize these for Exercise 2.5:
- 1. $$(x + y)^2 = x^2 + 2xy + y^2$$
- 2. $$(x - y)^2 = x^2 - 2xy + y^2$$
- 3. $$x^2 - y^2 = (x + y)(x - y)$$
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