# NCERT Exemplar: Polynomials Notes | Study Mathematics (Maths) Class 10 - Class 10

## Class 10: NCERT Exemplar: Polynomials Notes | Study Mathematics (Maths) Class 10 - Class 10

The document NCERT Exemplar: Polynomials Notes | Study Mathematics (Maths) Class 10 - 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

Exercise 2.1

Choose the correct answer from the given four options in the following questions:
1. If one of the zeroes of the quadratic polynomial (k–1) x2 + k x + 1 is –3, then the value of is
(a) 4/3
(b) -4/3
(c) 2/3
(d) -2/3
Ans: (a)
Explanation:
According to the question,
-3 is one of the zeros of quadratic polynomial (k-1)x2+kx+1
Substituting -3 in the given polynomial,
(k - 1)(-3)² + k(-3) + 1 = 0
(k - 1)9 + k(-3) + 1 = 0
9k - 9 -3k + 1 = 0
6k - 8 = 0
k = 8/6
Therefore, k =4 /3
Hence, option (A) is the correct answer.

2. A quadratic polynomial, whose zeroes are –3 and 4, is
(a) x2 – x + 12
(b) x2 + x + 12
(c) (x2/2) - (x/2) - 6
(d) 2x2 + 2x –24
Ans: (c)
Explanation:
Sum of zeroes, α+ β= -3+4 =1
Product of Zeroes, αβ = -3× 4 = -12
x²- (sum of zeroes)x+(product of zeroes)
= x²- (α+ β)x+(αβ)
= x² – (1)x + (-12)
= x² – x -12
divide by 2, we get
= x²/2 – x/2 - 12/2
= x²/2 – x/2 -6
Hence, option (c) is the correct answer.

3. If the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and –3, then
(a) a = –7, b = –1
(b) a = 5, b = –1
(c) a = 2, b = – 6
(d) a = 0, b = – 6
Ans: (d)
Explanation:
According to the question,
x² + (a + 1)x + b
Given that, the zeroes of the polynomial = 2 and -3,
When x = 2
2² + (a + 1)(2) + b = 0
4 + 2a + 2 + b = 0
6 + 2a + b = 0
2a + b = -6 —– (1)
When x = -3,
(-3)² + (a + 1)(-3) + b = 0
9 – 3a - 3 + b = 0
6 – 3a + b = 0
-3a + b = -6 —– (2)
Subtracting equation (2) from (1)
2a + b – (-3a + b) = -6-(-6)
2a + b + 3a - b = -6 + 6
5a = 0
a = 0
Substituting the value of ‘a’ in equation (1), we get,
2a + b = -6
2(0) +b = -6
b = -6
Hence, option (d) is the correct answer.

4. The number of polynomials having zeroes as –2 and 5 is
(a) 1
(b) 2
(c) 3
(d) more than 3
Ans: (d)
Explanation:
According to the question,
The zeroes of the polynomials = -2 and 5
We know that the polynomial is of the form,
p(x) = ax2 + bx + c.
Sum of the zeroes = – (coefficient of x) ÷ coefficient of x2 i.e.
Sum of the zeroes = – b/a
– 2 + 5 = – b/a
3 = – b/a
b = – 3 and a = 1
Product of the zeroes = constant term ÷ coefficient of x2 i.e.
Product of zeroes = c/a
(- 2)5 = c/a
– 10 = c
Substituting the values of a, b and c in the polynomial p(x) = ax2 + bx + c.
We get, x2 – 3x – 10
Therefore, we can conclude that x can take any value.
Hence, option (d) is the correct answer.

5. Given that one of the zeroes of the cubic polynomial ax3 + bx2 + cx + d is zero, the product of the other two zeroes is
(a) (–c/a)
(b c/a
(c) 0
(d) (–b/a)
Ans: (b)
Explanation:
According to the question,
We have the polynomial,
ax3 + bx2 + cx + d
We know that,
Sum of product of roots of a cubic equation is given by c/a
It is given that one root = 0
Now, let the other roots be α, β
So, we get,
αβ + β(0) + (0)α = c/a
αβ = c/a
Hence the product of other two roots is c/a
Hence, option (b) is the correct answer

Exercise 2.2

1. Answer the following and justify:
(i) Can x2 – 1 be the quotient on division of x6 + 2x3 + – 1 by a polynomial in of degree 5?
Solution:
No, x2 – 1 cannot be the quotient on division of x6 + 2x3 + x – 1 by a polynomial in x of degree 5.
Justification:
When a degree 6 polynomial is divided by degree 5 polynomial,
The quotient will be of degree 1.
Assume that (x2 – 1) divides the degree 6 polynomial with and the quotient obtained is degree 5 polynomial (1)
According to our assumption,
(degree 6 polynomial) = (x2 – 1)(degree 5 polynomial) + r(x) [ Since, (a = bq + r)]
= (degree 7 polynomial) + r(x) [ Since, (x2 term × x5 term = x7 term)]
= (degree 7 polynomial)
From the above equation, it is clear that, our assumption is contradicted.
x2 – 1 cannot be the quotient on division of x6 + 2x3 + x – 1 by a polynomial in x of degree 5
Hence Proved.

(ii) What will the quotient and remainder be on division of ax2 + bx + c by px3 + qx2 + rx + s, p ≠ 0?
Solution:
Degree of the polynomial px3 + qx2 + rx + s is 3
Degree of the polynomial ax2 + bx + c is 2
Here, degree of px3 + qx2 + rx + s is greater than degree of the ax2 + bx + c
Therefore, the quotient would be zero,
And the remainder would be the dividend = ax2 + bx + c.

(iii) If on division of a polynomial (x) by a polynomial (x), the quotient is zero, what is the relation between the degrees of (x) and (x)?
Solution:
We know that,
p(x)= g(x) × q(x)+r(x)
According to the question,
q(x) = 0
When q(x)=0, then r(x) is also = 0
So, now when we divide p(x) by g(x),
Then p(x) should be equal to zero
Hence, the relation between the degrees of p (x) and g (x) is the degree p(x)<degree g(x)

(iv) If on division of a non-zero polynomial (x) by a polynomial (x), the remainder is zero, what is the relation between the degrees of (x) and (x)?
Solution:
In order to divide p(x) by g(x)
We know that,
Degree of p(x) > degree of g(x)
or
Degree of p(x)= degree of g(x)
Therefore, we can say that,
The relation between the degrees of (x) and (x) is degree of p(x) > degree of g(x)

(v) Can the quadratic polynomial x2 + kx + have equal zeroes for some odd integer > 1?
Solution:
A Quadratic Equation will have equal roots if it satisfies the condition:
b² – 4ac = 0
Given equation is x² + kx + k = 0
a = 1, b = k, x = k
Substituting in the equation we get,
k² – 4 ( 1 ) ( k ) = 0
k² – 4k = 0
k ( k – 4 ) = 0
k = 0 , k = 4
But in the question, it is given that k is greater than 1.
Hence the value of k is 4 if the equation has common roots.
Hence if the value of k = 4, then the equation ( x² + kx + k ) will have equal roots.

Exercise 2.3

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

2. 3x2 + 4– 4
Solution:
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

3. 5t2 + 12+ 7
Solution:
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

4. t3 – 2t2 – 15t
Solution:
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

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

Solution:
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

Exercise 2.4

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
Solution:
(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

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.
Solution:
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

3. Given that 2 is a zero of the cubic polynomial 6x32 x2 – 10– 42 , find its other two zeroes.
Solution:
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 document NCERT Exemplar: Polynomials Notes | Study Mathematics (Maths) Class 10 - Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
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