NCERT Exemplar: Polynomials - 2

# NCERT Exemplar: Polynomials - 2 | Mathematics (Maths) Class 10 PDF Download

### Exercise 2.3

Find the zeroes of the following polynomials by factorisation method.
Q.1. 4x2 – 3– 1

4x2 – 3– 1
Splitting the middle term, we get,
4x2- 4x + 1x - 1
Taking the common factors out, we get,
4x(x - 1) +1(x - 1)
On grouping, we get,
(4x + 1)(x - 1)
So, the zeroes are,
4x + 1 = 0⇒ 4x = -1 ⇒x = (-1/4)
(x-1) = 0 ⇒ x = 1
Therefore, zeroes are (-1/4) and 1
Verification:
Sum of the zeroes = – (coefficient of x) ÷ coefficient of x2
α + β = – b/a
1 – 1/4 = – (- 3)/4 = ¾
Product of the zeroes = constant term ÷ coefficient of x2
α β = c/a
1(- 1/4) = – ¼
– 1/4 = – 1/4

Q.2. 3x2 + 4– 4

3x2 + 4x – 4
Splitting the middle term, we get,
3x2 + 6x – 2x – 4
Taking the common factors out, we get,
3x(x + 2) -2(x + 2)
On grouping, we get,
(x + 2)(3x - 2)
So, the zeroes are,
x + 2 =0 ⇒ x = -2
3x - 2 = 0 ⇒ 3x = 2 ⇒ x = 2/3
Therefore, zeroes are (2/3) and -2
Verification:
Sum of the zeroes = – (coefficient of x) ÷ coefficient of x2
α + β = – b/a
– 2 + (2/3) = – (4)/3
= – 4/3 = – 4/3
Product of the zeroes = constant term ÷ coefficient of x2
α β = c/a
Product of the zeroes = (- 2) (2/3) = – 4/3

Q.3. 5t2 + 12+ 7

5t2 + 12t + 7
Splitting the middle term, we get,
5t2 + 5t + 7t + 7
Taking the common factors out, we get,
5t (t + 1) +7(t + 1)
On grouping, we get,
(t + 1)(5t + 7)
So, the zeroes are,
t + 1 = 0 ⇒ y = -1
5t+7=0 ⇒ 5t=-7⇒t=-7/5
Therefore, zeroes are (-7/5) and -1
Verification:
Sum of the zeroes = – (coefficient of x) ÷ coefficient of x2
α + β = – b/a
(- 1) + (- 7/5) = – (12)/5
= – 12/5 = – 12/5
Product of the zeroes = constant term ÷ coefficient of x2
α β = c/a
(- 1)(- 7/5) =  7/5
7/5 =  7/5

Q.4. t3 – 2t2 – 15t

t3 – 2t2 – 15t
Taking t common, we get,
t ( t2 -2t -15)
Splitting the middle term of the equation t2 -2t -15, we get,
t( t2 -5t + 3t -15)
Taking the common factors out, we get,
t (t (t-5) +3(t-5)
On grouping, we get,
t (t+3)(t-5)
So, the zeroes are,
t=0
t+3=0 ⇒ t= -3
t -5=0 ⇒ t=5
Therefore, zeroes are 0, 5 and -3
Verification:
Sum of the zeroes = – (coefficient of x2) ÷ coefficient of x3
α + β + γ = – b/a
(0) + (- 3) + (5) = – (- 2)/1
= 2 = 2
Sum of the products of two zeroes at a time = coefficient of x ÷ coefficient of x3
αβ + βγ + αγ = c/a
(0)(- 3) + (- 3) (5) + (0) (5) = – 15/1
= – 15 = – 15
Product of all the zeroes = – (constant term) ÷ coefficient of x3
αβγ = – d/a
(0)(- 3)(5) = 0

Q.5. 2x2 +(7/2)+3/4

2x2 +(7/2)+3/4
The equation can also be written as,
8x2+14x+3
Splitting the middle term, we get,
8x2+12x+2x+3
Taking the common factors out, we get,
4x (2x+3) +1(2x+3)
On grouping, we get,
(4x+1)(2x+3)
So, the zeroes are,
4x+1=0 ⇒ x = -1/4
2x+3=0 ⇒ x = -3/2
Therefore, zeroes are -1/4 and -3/2
Verification:
Sum of the zeroes = – (coefficient of x) ÷ coefficient of x2
α + β = – b/a
(- 3/2) + (- 1/4) = – (7)/4
= – 7/4 = – 7/4
Product of the zeroes = constant term ÷ coefficient of x2
α β = c/a
(- 3/2)(- 1/4) = (3/4)/2
3/8 = 3/8

Q.6. t3 - 2t2 - 15t

Let t³ - 2t² - 15t = 0
Taking out common term,
t(t² - 2t - 15) = 0
t = 0
On factoring,
t² - 2t - 15 = t² - 5t + 3t - 15
= t(t - 5) + 3(t - 5)
= (t + 3)(t - 5)
Now, t + 3 = 0
t = -3
Also, t - 5 = 0
t = 5

Therefore, the zeros of the polynomial are 0, -3 and 5.
We know that, if α and β are the zeroes of a polynomial ax² + bx + c, then
Sum of the roots is α + β = -coefficient of x/coefficient of x² = -b/a
Product of the roots is αβ = constant term/coefficient of x² = c/a
From the given polynomial,
coefficient of x = -(1+2√2)
Coefficient of x² = 2
Constant term = √2
Sum of the roots:
LHS: α + β
= 1/2 + √2
= (1 + 2√2)/2
RHS: -coefficient of x/coefficient of x²
= -[-(1+2√2)/2]
= (1 + 2√2)/2
LHS = RHS
Product of the roots
LHS: αβ
= (1/2)(√2)
= √2/2
RHS: constant term/coefficient of x²
= √2/2
LHS = RHS
Therefore, the zeros of the polynomial are ½ and √2. The relation between the coefficients is -b/a
= (1+2√2)/2 and c/a = √2/2.

Q.7. 2x2 - (1 + 2√2)s + √2

Given, the polynomial is 2s² - (1+ 2√2)s + √2.
We have to find the relation between the coefficients and zeros of the polynomial
Let 2s² - (1+ 2√2)s + √2 = 0
On factoring,
2s² - s - 2√2s + √2 = 0
2s² - 2√2s - s + √2 = 0
2s(s - √2 ) - 1(s - √2) = 0
(2s - 1)(s - √2) = 0
Now, 2s - 1 = 0
2s = 1
s = 1/2
Also, s - √2 = 0
s = √2
Therefore, the zeros of the polynomial are 1/2 and √2.

Q.8. v2 + 4√3v - 15

Given, the polynomial is v² + 4√3v - 15.

We have to find the relation between the coefficients and zeros of the polynomial

Let v² + 4√3v - 15 = 0

On factoring,

v² + 5√3v - √3v - 15 = 0

v(v + 5√3) - √3(v + 5√3) = 0

(v - √3)(v + 5√3) = 0

Now, v - √3 = 0

v = √3

Also, v + 5√3 = 0

v = -5√3

Therefore, the zeros of the polynomial are -5√3 and √3.

We know that, if α and β are the zeroes of a polynomial ax² + bx + c, then

Sum of the roots is α + β = -coefficient of x/coefficient of x² = -b/a

Product of the roots is αβ = constant term/coefficient of x² = c/a

From the given polynomial,

coefficient of x = 4√3

Coefficient of x² = 1

Constant term = -15

Sum of the roots:

LHS: α + β

= √3 - 5√3

= -4√3

RHS: -coefficient of x/coefficient of x²

= -4√3/1

= -4√3

LHS = RHS

Product of the roots

LHS: αβ

= (-5√3)(√3)

= -15

RHS: constant term/coefficient of x²

= -15/1

= -15

LHS = RHS

Q.9.

Given, the polynomial is y² + (3√5/2)y - 5.
We have to find the relation between the coefficients and zeros of the polynomial
The polynomial can be rewritten as (1/2)[2y² + 3√5y - 10].
Let 2y² + 3√5y - 10 = 0
On factoring,
2y² + 4√5y - √5y - 10 = 0
2y(y + 2√5) - √5(y - 2√5) = 0
(y + 2√5)(2y - √5) = 0
Now, y + 2√5 = 0
y = -2√5
Also, 2y - √5 = 0
2y = √5
x = √5/2
Therefore, the zeros of the polynomial are √5/2 and -2√5.
We know that, if α and β are the zeroes of a polynomial ax² + bx + c, then
Sum of the roots is α + β = -coefficient of x/coefficient of x² = -b/a
Product of the roots is αβ = constant term/coefficient of x² = c/a
From the given polynomial,
coefficient of x = 3√5
Coefficient of x² = 2
Constant term = -10
Sum of the roots:
LHS: α + β
= √5/2 - 2√5
= (√5 - 4√5)/2
= -3√5/2
RHS: -coefficient of x/coefficient of x²
= -3√5/2
LHS = RHS
Product of the roots
LHS: αβ
= (√5/2)(-2√5)
= -10/2
= -5
RHS: constant term/coefficient of x²
= -10/2
= -5
LHS = RHS

Q.10. 7y2-11/3y - 2/3

Given, the polynomial is 7y² - (11/3)y - (2/3).

We have to find the relation between the coefficients and zeros of the polynomial

The polynomial can be rewritten as (1/3)[21y² - 11y - 2]

Let (1/3)[21y² - 11y - 2] = 0

21y² - 11y - 2 = 0

On factoring,

21y² - 14y + 3y - 2 = 0

7y(3y - 2) + (3y - 2) = 0

(7y + 1)(3y - 2) = 0

Now, 7y + 1 = 0

7y = -1

y = -1/7

Also, 3y - 2 = 0

3y = 2

y = 2/3

Therefore, the zeros of the polynomial are 2/3 and -1/7.

We know that, if α and β are the zeroes of a polynomial ax² + bx + c, then

Sum of the roots is α + β = -coefficient of x/coefficient of x² = -b/a

Product of the roots is αβ = constant term/coefficient of x² = c/a

From the given polynomial,

coefficient of x = -11

Coefficient of x² = 21

Constant term = -2

Sum of the roots:

LHS: α + β

= -1/7 + 2/3

= (-3+14)/21

= 11/21

RHS: -coefficient of x/coefficient of x²

= -(-11)/21

= 11/21

LHS = RHS

Product of the roots

LHS: αβ

= (-1/7)(2/3)

= -2/21

RHS: constant term/coefficient of x²

= -2/21

LHS = RHS

### Exercise 2.4

Q.1. For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also find the zeroes of these polynomials by factorisation.
(i) (–8/3), 4/3
(ii) 21/8, 5/16
(iii) -2√3, -9
(iv) (-3/(2√5)), -½

(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

Q.2. Given that the zeroes of the cubic polynomial x3 – 6x2 + 3x + 10 are of the form a, a + b, a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.

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

Q.3. Given that 2 is a zero of the cubic polynomial 6x32 x2 – 10– 42 , find its other two zeroes.

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

Q.4. Find k so that x2 + 2x + k is a factor of 2x4 + x3 – 14x+ 5x + 6. Also find all the zeroes of the two polynomials.

Given, p(x) = 2x⁴ + x³ - 14 x² + 5x + 6.

g(x) = x² + 2x + k

We have to find the zeros of the polynomial.

The division algorithm states that given any polynomial p(x) and any non-zero

polynomial g(x), there are polynomials q(x) and r(x) such that

p(x) = g(x) q(x) + r(x), where r(x) = 0 or degree r(x) < degree g(x).

Let r(x) = 0

So, p(x) = g(x) q(x)

By using long division,

So, q(x) = 2x² - 3x - 8 - 2k

r(x) = (21+7k)x + (2k² + 8k + 6)

By comparing the coefficients of (21+7k)x and 2k² + 8k + 6

2k² + 8k + 6 = 0

2(k² + 4k + 3) = 0

k² + 3k + k + 3 = 0

k(k + 3) + (k + 3) = 0

(k + 1)(k + 3) = 0

Now, k + 1 = 0

k = -1

Also, k + 3 = 0

k = -3

So, k = -1, -3.

When k = -1,

21 + 7k = 0

= 21 + 7(-1)

= 21 - 7

= 14

21 + 7k is not equal to zero.

So, k = -1 is neglected.

When k = -3,

21 + 7k = 0

= 21 + 7(-3)

= 21 - 21

= 0

Therefore, the value of k is -3.

Now, g(x) = x² + 2x - 3

x² + 2x - 3 = 0

x² - x + 3x - 3 = 0

x(x - 1) + 3(x - 1) = 0

(x + 3)(x - 1) = 0

Now, x + 3 = 0

x = -3

Also, x - 1 = 0

x = 1

Now, q(x) = 2x² - 3x - 8 - 2k

= 2x² - 3x - 8 - 2(-3)

= 2x² - 3x - 8 + 6

= 2x² - 3x - 2

On factoring,

2x² - 3x - 2 = 0

2x² - 4x + x - 2 = 0

2x(x - 2) + (x - 2) = 0

(2x + 1)(x - 2) = 0

Now, 2x + 1 = 0

2x = -1

x = -1/2

Also, x - 2 = 0

x = 2

We know that g(x) and q(x) are the factors of p(x).

So, the zeros of g(x) and q(x) will be the zeros of p(x).

Therefore, the zeros of p(x) = -3, -1/2, 1 and 2.

Q.5. Given that x – √5 is a factor of the cubic polynomial x3 – 3√5x2 + 13x – 3√5, find all the zeroes of the polynomial.

Give, the cubic polynomial p(x) = x³ - 3√5x² + 13x - 3√5

g(x) = x - √5

We have to find all the zeros of the polynomial.

By using long division,

The quotient is g(x) = x² - 2√5x + 3
On factoring,
x² - 2√5x + 3 = 0
x= [-b ± √b² - 4ac]/2a
Here, a = 1, b = -2√5 and c = 3
x = [2√5 ± √(-2√5)² - 4 (1) (3)]/ 2(1)
x = [2√5 ± √20 - 12]/ 2
x = [2√5 ± √8]/ 2
x = [2√5 ± 2√2]/ 2
Taking out common terms,
x = 2[√5 ± √2]/ 2
x = √5 ± √2
Therefore, the zeros are √5, √5 ± √2.

The document NCERT Exemplar: Polynomials - 2 | Mathematics (Maths) Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10

## Mathematics (Maths) Class 10

126 videos|477 docs|105 tests

## FAQs on NCERT Exemplar: Polynomials - 2 - Mathematics (Maths) Class 10

 1. What is a polynomial?
Ans. A polynomial is an algebraic expression consisting of variables, coefficients, and exponents, combined using addition, subtraction, multiplication, and non-negative integer exponents.
 2. How do you classify polynomials based on the number of terms?
Ans. Polynomials can be classified based on the number of terms they contain. A polynomial with one term is called a monomial, a polynomial with two terms is called a binomial, and a polynomial with three terms is called a trinomial.
 3. How do you find the degree of a polynomial?
Ans. The degree of a polynomial is the highest power of the variable in the polynomial. To find the degree, identify the term with the highest exponent and that exponent will be the degree of the polynomial.
 4. What are the different operations that can be performed on polynomials?
Ans. The different operations that can be performed on polynomials are addition, subtraction, multiplication, and division. Addition and subtraction involve combining like terms, while multiplication involves multiplying each term of one polynomial with each term of the other polynomial. Division of polynomials can be done using long division or synthetic division.
 5. Can a polynomial have a negative degree?
Ans. No, a polynomial cannot have a negative degree. The degree of a polynomial is always a non-negative integer. If a polynomial has a term with a negative exponent, it is not considered a polynomial, but a rational function.

## Mathematics (Maths) Class 10

126 videos|477 docs|105 tests

### Up next

 Explore Courses for Class 10 exam

### Top Courses for Class 10

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;