Class 10 Mathematics Chapter 2

Chapter 2 - Polynomials

Excercise 2.1

Solution 1

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

At all these points where the graph intersects x axis  the value of the polynomial y = p(x) will be zero.

Excercise 2.2

Solution 1

So, the zeroes of x² - 2x - 8 are 4 and -2.











Solution 2

(i)    Let the required polynomial be  ax² + bx + c, and let its zeroes  and    

        If a = 4k, then b = -k, c = -4k  Therefore, the quadratic polynomial is k(4 x 2 - x - 4), where k is a real number .

(ii)     Let the polynomial be  ax² + bx + c, and let its zeroes be  and  

   (iii)    Let the polynomial be  ax² + bx + c, and let its zeroes be  and               (iv)    Let the polynomial be  ax² + bx + c, and let its zeroes be  and        Therefore, the quadratic polynomial is k(x² - x + 1),where k is a real number .

(v)    Let the polynomial be ax² + bx + c, and its zeroes be  and   Therefore, the quadratic polynomial is k(4x² + x + 1),where k is a real number .

(vi)    Let the polynomial be ax² + bx + c.
   Therefore, the quadratic polynomial is k(x² - 4x + 1),where k is a real number .

Polynomials Page/Excercise 2.3

Solution 1

Quotient = x - 3
Remainder = 7x - 9




Quotient = x2 + x - 3
Remainder = 8




Quotient = -x2 - 2
Remainder = -5x + 10

 After performing division, one can check his/her answer obtained by the division algorithm which is as below:

Dividend = Divisor x Quotient + Remainder
Also, remember that the quotient obtained is a polynomial only.

Solution 2

The polynomial 2t4 + 3t3 - 2t2 - 9t - 12  can be divided by the polynomial t2 - 3 = t2 + 0.t - 3 as follows:


Since the remainder is 0, t² - 3 is a factor of 2t4 + 3t3 - 2t2 - 9t - 12 .

(ii) The polynomial 3x4 + 5x3 - 7x2 + 2x + 2 can be divided by the polynomial x2 + 3x + 1 as follows:



Since the remainder is 0, x² + 3x + 1 is a factor of 3x4 + 5x3 - 7x2 + 2x + 2

(iii) The polynomial x5 - 4x3 + x2 + 3x + 1 can be divided by the  polynomial x3 - 3x + 1 as follows:



Since the remainder is not equal to 0, x3 - 3x + 1 is not a factor of x5 - 4x3 + x2 + 3x + 1.

Solution 3

Let p(x) = 3x4 + 6x3 - 2x2 - 10x -5



Now, x² + 2x + 1 = (x + 1)2
So, the two zeroes of  x² + 2x + 1 are -1 and -1.



Solution 4

Divided, p(x) = x3 - 3x2 + x + 2
Quotient = (x - 2)
Remainder = (-2x + 4)
Let g(x) be the divisor.

According to the division algorithm,

Dividend = Divisor x Quotient + Remainder





 When a polynomial is divided by any other non-zero polynomial, then it satisfies the division algorithm which is as below:
Dividend = Divisor x Quotient + Remainder


Divisor x Quotient = Dividend - Remainder

So, from this relation, the divisor can be obtained by dividing the result of (Dividend - Remainder) by the quotient.

Solution 5
According to the division algorithm, if p(x) and g(x) are two polynomials with g(x)  0, then we can find polynomials q(x) and r(x) such that

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

(i)    Degree of quotient will be equal to degree of dividend when divisor is constant.
Let us consider the division of 18x2 + 3x + 9  by 3.
Here, p(x) = 18x2 + 3x + 9  and g(x) = 3
q(x) = 6x2 + x + 3  and r(x) = 0
Here, degree of p(x) and q(x) is the same which is 2.


Checking:
p(x) = g(x) x q(x) + r(x)



Thus, the division algorithm is satisfied.

(ii)    Let us consider the division of 2x4 + 2x by 2x3,
Here, p(x) = 2x4 + 2x and g(x) = 2x3
q(x) = x and r(x) = 2x
Clearly, the degree of q(x) and r(x) is the same which is 1.

Checking,
p(x) = g(x) x q(x) + r(x)
2x4 + 2x =  (2x3 ) x x  + 2x
2x4 + 2x = 2x4 + 2x
Thus, the division algorithm is satisfied.

(iii)    Degree of remainder will be 0 when remainder obtained on division is a constant.
Let us consider the division of 10x3 + 3 by 5x2.
Here, p(x) = 10x3 + 3 and g(x) = 5x2
q(x) = 2x and r(x) = 3
Clearly, the degree of r(x) is 0.

Checking:
p(x) = g(x) x q(x) + r(x)
10x3 + 3 = (5x2 ) x 2x  +  3
10x3 + 3 = 10x3 + 3
Thus, the division algorithm is satisfied.

Concept insight: In order to answer such type of questions, one should remember the division algorithm. Also, remember the condition on the remainder polynomial r(x). The polynomial r(x) is either 0 or its degree is strictly less than g(x). The answer may not be unique in all the cases because there can be multiple polynomials which satisfies  the given conditions.

Polynomials Page/Excercise 2.4

Solution 1

On comparing the given polynomial with the polynomial ax3 + bx2 + cx + d, we obtain a = 2, b = 1, c = -5, d = 2



Thus, the relationship between the zeroes and the coefficients is verified.



On comparing the given polynomial with the polynomial ax3 + bx2 + cx + d, we obtain a = 1, b = -4, c = 5, d = -2.



Thus, the relationship between the zeroes and the coefficients is verified.

Concept insight: The zero of a polynomial is that value of the variable which makes the polynomial 0. Remember that there are three  relationships between the zeroes of a cubic polynomial and its coefficients which involve the sum of zeroes, product of all zeroes and the product of zeroes taken two at a time.

Solution 2

Let the polynomial be ax3  + bx2 + cx + d and its zeroes be   .
According to the given information:



If a = k, then b = -2k, c = -7k, d = 14k

Thus, the required cubic polynomial is k(x3 - 2x2 - 7x + 14), where k is any real number.
The simplest polynomial will be obtained by taking k = 1.

Solution 3

Let p(x) = x3 - 3x2 + x + 1
The zeroes of the polynomial p(x) are given as a - b, a, a + b.
Comparing the given polynomial with dx3 + ex2 + fx + g, we can observe that
d = 1, e = -3, f = 1, g = 1





Therefore, the zeroes of p(x) are 1-b, 1, 1+b .



Concept insight: When the polynomial and its zeroes are given then, remember to apply relationships between the zeroes and coefficients .
These relations  involve the sum of the zeroes, product of zeroes and the product of zeroes taken two at a time.

Solution 4

Therefore, x2 - 4x + 1 is a factor of the given polynomial. For finding the remaining zeroes of the given polynomial, we will carry out the division of



Hence, 7 and -5 are also zeroes of the given polynomial.

Solution 5

By division algorithm,
Dividend = Divisor x Quotient + Remainder
Divisor x Quotient = Dividend - Remainder





It will be perfectly divisible by x2 - 2x + k.




10 - a - 8 x 5 + 25 = 0
10 - a - 40 + 25 = 0
- 5 - a = 0
a = -5

Thus, k = 5 and a = -5.

Concept insight: When a polynomial is divided by any non-zero polynomial, then it satisfies the division algorithm which is as below:
Dividend = Divisor x Quotient + Remainder

Divisor x Quotient = Dividend - Remainder

So, from this relation, it can be said that the result (Dividend - Remainder) will be completely divisible by the divisor.