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Class 9 Maths Chapter 2 HOTS Questions - Polynomials

Q.1. If  p(x) = x2 – 2√2 x + 1, then find  p (2√2) .

Sol. Since, p(x) = x– 2 √2 x + 1 , replacing value of x=2√2
Class 9 Maths Chapter 2 HOTS Questions - Polynomials
= 4 (2) – 4 (2) + 1

= 8 – 8 + 1 

= 1

Q.2. Factorize 3x2-14x - 5  2x^2 + 7x + 3using the middle term splitting method.

Sol. a=3,b=14,c=5, We need two numbers that multiply to ac=3×(5)=15ac = 3 \times (-5) = -15 and add up to b=14.= -14

The numbers are -15 and 1 because 15×1=15-15 \times 1 = -15 and 15+1=14.-15 + 1 = -14

Rewrite the middle term: 3x215x+x5

 - 15x + x - 5Factor by grouping: (3x215x)+(x5)(3x^2 - 15x) + (x - 5)   ⇒ 3x(x5)+1(x5)3x(x - 5) + 1(x - 5)  ⇒ (3x+1)(x5)(3x + 1)(x - 5)(3x+1)(x−5)

So, the factorized form of 3x214x53x^2 - 14x - 5  is (3x+1)(x5)(3x + 1)(x - 5).

Q.3. If  a + b + c = 9, and ab + bc + ca = 26, find a+ b2 + c2

Sol.nGiven: a + b + c = 9, ab + bc + ca = 26

Squaring, we get

(a+b+c)2=(9)2

a2+b2+c2+2(ab+bc+ca)=81

 a2+b2+c2+2×26=81

a2+b2+c2=8152  =29

Q.4. Factorise 4x2 – 12x + 9 

Sol: To factorize this expression, we need to find two numbers α and β such that α + β = –12 and αβ = 36

4x2 – 6x – 6x + 9

2x(x – 3) – 3(x – 3)

(2x – 3)(x – 3)

Q.5. If a, b, c are all non-zero and a + b + c = 0, prove that Class 9 Maths Chapter 2 HOTS Questions - Polynomials

Sol. ∵ a + b + c = 0  ⇒  a3 + b3 + c3 – 3abc = 0
or a+ b+ c3 = 3abc 

dividing the whole equation by abc, we get 

Class 9 Maths Chapter 2 HOTS Questions - Polynomials

Q.6. If Class 9 Maths Chapter 2 HOTS Questions - Polynomialsthen find the value of Class 9 Maths Chapter 2 HOTS Questions - Polynomials

Sol: 
Class 9 Maths Chapter 2 HOTS Questions - Polynomials

Q.7. Factorize 12x2-29x +15  using the middle term splitting method.

Sol. a=12,b=29,c=+15, We need two numbers that multiply to ac=12×(15)= 180 and add up to b=−29.

The numbers are -20 and -9 because 20×9=180-20 \times -9 = 180 and 20+9=29-20 + -9 = -29.

Rewrite the middle term: 12x2−20x−9x+15

Factor by grouping: (12x2−20x)+(−9x+15) ⇒ 4x(3x5)3(3x5)4x(3x - 5) - 3(3x - 5) ⇒ (4x−3)(3x−5)
So, the factorized form of 12x2−29x+15 is (4x3)(3x5)(4x - 3)(3x - 5).

Q.8. Factorise: (a – b)+ (b – c)3 + (c – a)3

Sol: Put a – b = x, b – c = y and c – a = z so that
                       x + y + z = 0 ⇒ x3 + y3 + z3 = 3xyz
                       ⇒ (a – b)3 + (b – c)3 + (c – a)3 = 3(a – b) (b – c) (c – a)

Q.10.  Factorise 10y2 – 28y + 14

Sol: To factorize this expression, we need to find two numbers α and β such that α + β = –28 and αβ = 140

10y2 – 14y – 10y + 14

2y(5y – 7) – 2(5y – 7)

(2y – 2)(5y – 7)

2(y – 1)(5y – 7)
Q.11 Compute the value of 9x2 + 4y2 if xy = 6 and 3x + 2y = 12. 

Sol: Consider the equation 3x + 2y = 12
Now, square both sides:
(3x + 2y)2 = 122
=> 9x2 + 12xy + 4y2 = 144
=>9x2 + 4y2 = 144 – 12xy
From the questions, xy = 6
So,
9x2 + 4y2 = 144 – 72
Thus, the value of 9x2 + 4y= 72

Q.12.Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: 

(i) Area: 25a2–35a+12

(ii) Area: 35y2+13y–12

Sol:(i) Area: 25a2–35a+12

Using the splitting the middle term method,

We have to find a number whose sum = -35 and product =25×12 = 300

We get -15 and -20 as the numbers [-15+-20=-35 and -15×-20 = 300]

25a2–35a+12 = 25a2–15a−20a+12

= 5a(5a–3)–4(5a–3)

= (5a–4)(5a–3)

Possible expression for length  = 5a–4

Possible expression for breadth  = 5a –3

(ii) Area: 35y2+13y–12

Using the splitting the middle term method,

We have to find a number whose sum = 13 and product = 35×-12 = 420

We get -15 and 28 as the numbers [-15+28 = 13 and -15×28=420]

35y2+13y–12 = 35y2–15y+28y–12

= 5y(7y–3)+4(7y–3)

= (5y+4)(7y–3)

Possible expression for length  = (5y+4)

Possible expression for breadth  = (7y–3)

The document Class 9 Maths Chapter 2 HOTS Questions - Polynomials is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on Class 9 Maths Chapter 2 HOTS Questions - Polynomials

1. What are polynomials and how are they classified?
Ans.Polynomials are algebraic expressions that consist of variables raised to whole number powers and coefficients. They can be classified based on the number of terms: a monomial (one term), binomial (two terms), trinomial (three terms), or polynomial (four or more terms). Additionally, they can be classified by their degree, which is the highest power of the variable.
2. How do you add and subtract polynomials?
Ans.To add or subtract polynomials, combine like terms. Like terms are terms that have the same variable raised to the same power. For example, to add \(3x^2 + 4x^2\), you add the coefficients to get \(7x^2\). For subtraction, such as \(5x^3 - 2x^3\), subtract the coefficients to get \(3x^3\).
3. What is the process of multiplying polynomials?
Ans.Multiplying polynomials involves using the distributive property, also known as the FOIL method for binomials. For example, to multiply \((x + 2)(x + 3)\), you multiply each term in the first polynomial by each term in the second: \(x^2 + 3x + 2x + 6 = x^2 + 5x + 6\).
4. How do you factor polynomials?
Ans.Factoring polynomials involves rewriting the polynomial as a product of its factors. Common methods include taking out the greatest common factor (GCF), using the difference of squares, and applying the quadratic formula for trinomials. For example, to factor \(x^2 - 9\), recognize it as a difference of squares: \((x - 3)(x + 3)\).
5. What is the significance of the degree of a polynomial?
Ans.The degree of a polynomial indicates the highest exponent of the variable and helps determine the polynomial's behavior and properties. For example, a polynomial of degree 2 is a quadratic, which has a parabolic graph, while a polynomial of degree 3 is a cubic, which may have one or two turns. The degree also influences the number of possible roots and the end behavior of the graph.
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