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Page No: 28

 Exercises 2.1

 1. The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case

Polynomials,NCERT,Solutions,10th,Class,Cbse

Answer

(i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.
(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.
(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.
(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.
(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.
(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

Page No: 33

 Exercise 2.2

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
 (i) x2 – 2x – 8
 (ii) 4s2 – 4s 1
 (iii) 6x2 – 3 – 7x
 (iv) 4u2 8u
  (v) t2 – 15
 (vi) 3x2 x – 4

Answer
(i) x2 – 2x – 8
= (x - 4) (x  2)
The value of x2 – 2x – 8 is zero when x - 4 = 0 or x 2 = 0, i.e., when x = 4 or x = -2
Therefore, the zeroes of x2 – 2x – 8 are 4 and -2.

Sum of zeroes = 4 (-2) = 2 = -(-2)/1 = -(Coefficient of x)/Coefficient of x2

Product of zeroes = 4 × (-2) = -8 = -8/1 = Constant term/Coefficient of x2

(ii) 4s2 – 4s + 1= (2s-1)2
The value of 4s2 - 4s +1 is zero when 2s - 1 = 0, i.e., s = 1/2

Therefore, the zeroes of 4s2 - 4s + 1 are 1/2 and 1/2.

Sum of zeroes = 1/2 1/2 = 1 = -(-4)/4 = -(Coefficient of s)/Coefficient of s2
Product of zeroes = 1/2 × 1/2 = 1/4 = Constant term/Coefficient of s2.

(iii) 6x2 – 3 – 7x
6x– 7– 3
= (3x  1) (2x - 3)
The value of 6x2 - 3 - 7x is zero when 3x + 1 = 0 or 2x - 3 = 0, i.e., x = -1/3 or x = 3/2

Therefore, the zeroes of 6x2 - 3 - 7x are -1/3 and 3/2.

Sum of zeroes = -1/3 3/2 = 7/6 = -(-7)/6 = -(Coefficient of x)/Coefficient of x2
Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of x2.

(iv) 4u2   8u
4u2  8u  0
= 4u(u 2)
The value of 4u2 8u is zero when 4u = 0 or u 2 = 0, i.e., u = 0 or u = - 2
Therefore, the zeroes of 4u2 8u are 0 and - 2.
Sum of zeroes = 0 (-2) = -2 = -(8)/4 = -(Coefficient of u)/Coefficient of u2
Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of u2.

(v) t2 – 15
t- 0.t - 15
= (- √15) (t  √15)
The value of t2 - 15 is zero when t - √15 = 0 or t  √15 = 0, i.e., when t = √15 or = -√15
Therefore, the zeroes of t2 - 15 are √15 and -√15.Sum of zeroes = √15   -√15 = 0 = -0/1 = -(Coefficient of t)/Coefficient of t2
Product of zeroes = (√15) (-√15) = -15 = -15/1 = Constant term/Coefficient of u2.

(vi) 3x2 x – 4
= (3x - 4) (x  1)
The value of 3x2 x – 4 is zero when 3x - 4 = 0 and x  1 = 0,i.e., when x = 4/3 or x = -1
Therefore, the zeroes of 3x2 x – 4 are 4/3 and -1.
Sum of zeroes = 4/3  (-1) = 1/3 = -(-1)/3 = -(Coefficient of x)/Coefficient of x2
Product of zeroes = 4/3 × (-1) = -4/3 = Constant term/Coefficient of x2.

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) 1/4 , -1

(ii) √2 , 1/3 

(iii) 0, √5

(iv) 1,1 

(v) -1/4 ,1/4 

(vi) 4,1

Answer

(i) 1/4 , -1
Let the polynomial be ax2  bx  c, and its zeroes be α and ß
α  ß = 1/4 = -b/a
αß = -1 = -4/4 = c/a
If a = 4, then b = -1, c = -4
Therefore, the quadratic polynomial is 4x2 - x -4.

(ii) √2 , 1/3
Let the polynomial be ax2   bx   c, and its zeroes be α and ß
α  ß = √2 = 3√2/3 = -b/a
αß = 1/3 = c/a
If a = 3, then b = -3√2, c = 1
Therefore, the quadratic polynomial is 3x2 -3√2x  1.

(iii) 0, √5
Let the polynomial be ax2   bx   c, and its zeroes be α and ß
α  ß = 0 = 0/1 = -b/a
αß = √5 = √5/1 = c/a
If a = 1, then b = 0, c = √5
Therefore, the quadratic polynomial is x2  √5.

(iv) 1, 1
Let the polynomial be ax2   bx   c, and its zeroes be α and ß
α  ß = 1 = 1/1 = -b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -1, c = 1
Therefore, the quadratic polynomial is x2x  1.

(v) -1/4 ,1/4
Let the polynomial be ax2   bx   c, and its zeroes be α and ß
α  ß = -1/4 = -b/a
αß = 1/4 = c/a
If a = 4, then b = 1, c = 1
Therefore, the quadratic polynomial is 4x2   x  1.

(vi) 4,1
Let the polynomial be ax2   bx   c, and its zeroes be α and ß
α  ß = 4 = 4/1 = -b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -4, c = 1
Therefore, the quadratic polynomial is x2 - 4x  1.

Page No: 36

Exercise 2.3

1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

Answer

(i) p(x) = x3 – 3x2  5x – 3, g(x) = x2 – 2

Polynomials,NCERT,Solutions,10th,Class,Cbse

Quotient = x-3 and remainder 7x - 9

(ii) p(x) = x4 – 3x2  4x 5, g(x) = x2  1 – x

Polynomials,NCERT,Solutions,10th,Class,Cbse

Quotient = x2  - 3 and remainder 8

(iii) p(x) = x4 – 5x  6, g(x) = 2 – x2

Polynomials,NCERT,Solutions,10th,Class,Cbse

Quotient = -x2 -2 and remainder -5x 10

2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

Answer

(i) t2 – 3,  2t4 3t3 – 2t2 – 9t – 12

Polynomials,NCERT,Solutions,10th,Class,Cbse

t2 – 3 exactly divides  2t4  3t3 – 2t2 – 9t – 12 leaving no remainder. Hence, it is a factor of  2t4  3t3 – 2t2 – 9t – 12.

(ii) x2 3x 1, 3x4 5x3 – 7x2 2x 2

Polynomials,NCERT,Solutions,10th,Class,Cbse

x2  3x  1 exactly divides 3x4  5x3 – 7x2  2x  2 leaving no remainder. Hence, it is factor of 3x4  5x3 – 7x2  2x  2.

(iii) x3 – 3x 1, x5 – 4x3 x2 3x 1

Polynomials,NCERT,Solutions,10th,Class,Cbse

x2  3x  1 exactly divides 3x4  5x3 – 7x2  2x  2 leaving no remainder. Hence, it is factor of 3x4  5x3 – 7x2  2x  2.

(iii) x3 – 3x 1, x5 – 4x3 x2 3x 1

Polynomials,NCERT,Solutions,10th,Class,Cbse

x3 – 3x  1 didn't divides exactly x5 – 4x3   x2  3x  1 and leaves 2 as remainder. Hence, it not a factor of x5 – 4x3   x2  3x  1.

3. Obtain all other zeroes of 3x4 6x3 – 2x2 – 10x – 5, if two of its zeroes are √(5/3) and - √(5/3).

Answer

p(x) = 3x4  6x3 – 2x2 – 10x – 5
Since the two zeroes are √(5/3) and - √(5/3).

Polynomials, Class 10, Maths NCERT Solutions
Polynomials, Class 10, Maths NCERT Solutions
Polynomials, Class 10, Maths NCERT Solutions  

We factorize x2 + 2x + 1

= ( + 1)2

Therefore, its zero is given by x  + 1 = 0

x = -1

As it has the term (x + 1)2 , therefore, there will be 2 zeroes at x = - 1.
Hence, the zeroes of the given polynomial are √(5/3) and - √(5/3), - 1 and - 1.

4.  On dividing x3 - 3x2  x 2 by a polynomial g(x), the quotient and remainder were x - 2 and  -2x  4, respectively. Find g(x).

Answer
Here in the given question,

Dividend = x3 - 3x2  x 2

Quotient = x - 2

Remainder = -2x  4

Divisor = g(x)

We know that,

Dividend = Quotient × Divisor  Remainder

⇒ x3 - 3x2  x 2 = (x - 2) × g(x)  (-2x  4)⇒ x3 - 3x2  x 2 - (-2x  4) = (x - 2) × g(x)
⇒ x3 - 3x2  3x - 2 = (x - 2) × g(x)
g(x) =  (x3 - 3x2  3x - 2)/(x - 2)

Polynomials,NCERT,Solutions,10th,Class,Cbse

∴ g(x) = (x2 - x + 1)

5.Give examples of polynomial p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
 (i) deg p(x) = deg q(x)
 (ii) deg q(x) = deg r(x)
 (iii) deg r(x) = 0


Answer
(i) Let us assume the division of 6x2 2x 2 by 2
Here, p(x) = 6x2 2x 2
g(x) = 2
q(x) = 3x2 x 1
r(x) = 0
Degree of p(x) and q(x) is same i.e. 2.
Checking for division algorithm,
p(x) = g(x) × q(x) r(x)
Or, 6x2 2x 2 = 2x (3x2 1)
Hence, division algorithm is satisfied.

(ii) Let us assume the division of x3 x by x2,
Here, p(x) = x3 x
g(x) = x2
q(x) = x and r(x) = x
Clearly, the degree of q(x) and r(x) is the same i.e., 1.
Checking for division algorithm,
p(x) = g(x) × q(x) r(x)
x3 x = (x2 ) × x x
x3 x = x3 x
Thus, the division algorithm is satisfied.

(iii) Let us assume the division of x3 1 by x2.
Here, p(x) = x3 1
g(x) = x2
q(x) = x and r(x) = 1
Clearly, the degree of r(x) is 0.
Checking for division algorithm,
p(x) = g(x) × q(x) r(x)
x3  1 = (x2 ) × 1
x3 1 = x3 1
Thus, the division algorithm is satisfied.

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FAQs on Polynomials, Class 10, Maths NCERT Solutions

1. What are polynomials?
Ans. Polynomials are expressions containing variables and coefficients, which involve only the operations of addition, subtraction, multiplication, and non-negative integer exponents. For example, 3x^2 + 2x - 5 is a polynomial of degree 2.
2. What is the degree of a polynomial?
Ans. The degree of a polynomial is the highest power of the variable present in the polynomial. For example, the degree of the polynomial 3x^2 + 2x - 5 is 2.
3. What is the remainder theorem?
Ans. The remainder theorem states that if a polynomial f(x) is divided by x-a, then the remainder is f(a). In other words, if we divide f(x) by (x-a), the remainder we get is equal to f(a).
4. What is the factor theorem?
Ans. The factor theorem states that if a polynomial f(x) has a factor (x-a), then f(a) = 0. In other words, if we substitute x=a in f(x) and the value of f(a) is zero, then (x-a) is a factor of f(x).
5. How to find the zeroes of a polynomial?
Ans. To find the zeroes of a polynomial, we need to solve the equation f(x) = 0, where f(x) is the polynomial. There are various methods to find the zeroes of a polynomial, such as factorization, using the remainder theorem, using the factor theorem, or by using graphical methods.
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