Chapter 2: Polynomials

(i) Sum of the zeroes = – 8/3

Product of the zeroes = 4/3

P(x) = x² – (sum of the zeroes) + (product of the zeroes)

Then, P(x)= x² – (-8x)/3 + 4/3

P(x)= 3x² + 8x + 4

Using splitting the middle term method,

3x² + 8x + 4 = 0

3x² + (6x + 2x) + 4 = 0

3x² + 6x + 2x + 4 = 0

3x(x + 2) + 2(x + 2) = 0

(x + 2)(3x + 2) = 0

⇒ x = -2, -2/3

(ii) Sum of the zeroes = 21/8

Product of the zeroes = 5/16

P(x) = x² – (sum of the zeroes) + (product of the zeroes)

Then, P(x)= x² – 21x/8 + 5/16

P(x)= 16x² – 42x + 5

Using splitting the middle term method,

16x² – 42x + 5 = 0

16x² – (2x + 40x) + 5 = 0

16x² – 2x – 40x + 5 = 0

2x (8x – 1) – 5(8x – 1) = 0

(8x – 1)(2x – 5) = 0

⇒ x = 1/8, 5/2

(iii) Sum of the zeroes = – 2√3

Product of the zeroes = – 9

P(x) = x² – (sum of the zeroes) + (product of the zeroes)

Then, P(x) = x² – (-2√3x) – 9

Using splitting the middle term method,

x² + 2√3x – 9 = 0

x² + (3√3x – √3x) – 9 = 0

x(x + 3√3) – √3(x + 3√3) = 0

(x – √3)(x + 3√3) = 0

⇒ x =  √3, -3√3

(iv) Sum of the zeroes = -3/2√5x

Product of the zeroes = – ½

P(x) = x² – (sum of the zeroes) + (product of the zeroes)

Then, P(x)= x²  -(-3/2√5x) – ½

P(x)= 2√5x² + 3x – √5

Using splitting the middle term method,

2√5x² + 3x – √5 = 0

2√5x² + (5x – 2x) – √5 = 0

2√5x² – 5x + 2x – √5 = 0

√5x (2x + √5) – (2x + √5) = 0

(2x + √5)(√5x – 1) = 0

⇒ x = 1/√5, -√5/2

Given that a, a+b, a+2b are roots of given polynomial x³-6x²+3x+10

Sum of the roots ⇒ a+2b+a+a+b = -coefficient of x²/ coefficient of x³

⇒ 3a+3b = -(-6)/1 = 6

⇒ 3(a+b) = 6

⇒ a+b = 2 ——— (1) b = 2-a

Product of roots ⇒ (a+2b)(a+b)a = -constant/coefficient of x³

⇒ (a+b+b)(a+b)a = -10/1

Substituting the value of a+b=2 in it

⇒ (2+b)(2)a = -10

⇒ (2+b)2a = -10

⇒ (2+2-a)2a = -10

⇒ (4-a)2a = -10

⇒ 4a-a² = -5

⇒ a²-4a-5 = 0

⇒ a²-5a+a-5 = 0

⇒ (a-5)(a+1) = 0

a-5 = 0 or a+1 = 0

a = 5 a = -1

a = 5, -1 in (1) a+b = 2

When a = 5, 5+b=2 ⇒ b=-3

a = -1, -1+b=2 ⇒ b= 3

∴ If a=5 then b= -3

or

If a= -1 then b=3

Given, √2 is one of the zero of the cubic polynomial.

Then, (x-√2) is one of the factor of the given polynomial p(x) = 6x³+√2x²-10x- 4√2.

So, by dividing p(x) by x-√2

6x³+√2x²-10x-4√2= (x-√2) (6x² +7√2x + 4)

By splitting the middle term,

We get,

(x-√2) (6x² + 4√2x + 3√2x + 4)

= (x-√2) [ 2x(3x+2√2) + √2(3x+2√2)]

= (x-√2) (2x+√2)   (3x+2√2)

To get the zeroes of p(x),

Substitute p(x)= 0

(x-√2) (2x+√2)  (3x+2√2)= 0

x= √2 , x= -√2/2 ,x= -2√2/3

Hence, the other two zeroes of p(x) are -√2/2 and -2√2/3

The exemplars for Chapter 2 Polynomials are provided in PDF, which can be downloaded easily. Students can use these solved questions as a reference tool while preparing for the board exam to clear their doubts. These exemplar problems and solutions have been designed by subject experts with respect to the CBSE Syllabus and as per the exercise questions available in the NCERT book. Click here to get exemplars for all chapters.

Students of Class 10 can also go through the online learning materials such as notes, exemplar books, and question papers available in downloadable PDFs to prepare and score well in the board exam. They are also advised to solve sample papers and previous years’ question papers to get an idea of the type of questions asked from the chapter Polynomials and the marking scheme for the same.