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Question-1

The graphs of y = p(x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p(x).

 


Solution:
The graph does not intersects at x -axis.
The no. of zeroes of p(x) is 0

Question-2

The graphs of y = p(x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p(x).
 

 


Solution:
The graph intersects x - axis at one point. 
∴ The no. of zeroes of p(x) is One

Question-3

The graphs of y = p(x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p(x).
 

 

Solution:
The graph intersects x- axis at three points
∴ The no. of zeroes of p(x) is three

Question-4

The graphs of y = p(x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p(x).
 

 

Solution:
The graph intersects x - axis at two points
∴ The no. of zeroes of p(x) is Two

Question-5

The graphs of y = p(x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p(x).
 

 

Solution:
The graph intersects x -axis at four points
∴ The no. of zeroes of p(x) is four

Question-6

The graphs of y = p(x) are given in the figure below for some polynomials p(x). Find the number of zeroes of p(x).

 

Solution:
The graph intersects x -axis at three points
∴ The no. of zeroes of p(x) is Three.

Question-7

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.

Solution:
(i) x2 – 2x – 8 = x2 – 4x + 2x – 8
                      = x(x – 4) + 2(x – 4)
                      = (x – 4)(x + 2)
Therefore the zeroes of the polynomial x2 – 2x – 8 are {4, -2}.
Relationship between the zeroes and the coefficients of the polynomial:
Sum of the zeroes =  
Also sum of the zeroes of the polynomial = 4 – 2 = 2.
Product of the zeroes = =
Also product of the zeroes = 4 x –2 =  8
Hence verified.

(ii) 4s2 – 4s + 1 = 4s2 – 2s – 2s + 1
                      = 2s(2s – 1) – 1(2s – 1)
                      = (2s – 1)(2s – 1)
                      = 2(s  2(s  
Therefore the zeroes of the polynomial are
Relationship between the zeroes and the coefficients of the polynomial:
Sum of the zeroes =   =  = 1
Also sum of the zeroes = = 1
Product of the zeroes =
Also product of the zeroes =
Hence verified.

(iii) 6x2 – 3 – 7x = 6x2 – 7x – 3
                      = 6x2 – 9x + 2x – 3
                      = 3x(2x – 3) + 1(2x – 3)
                      = (2x –3)(3x + 1)
                      = 2(x  3(x +
                      = 6(x  (x +
The zeroes of the polynomials are {

Relationship between the zeroes and the coefficients of the polynomial:

Sum of the zeroes = - =   =
Also sum of the zeroes =
Product of the zeroes =  
Also product of the zeroes =
Hence verified.

(iv) 4u2 + 8u = 4u(u + 2)
                  = 4[u – 0][u –( 2)]
The zeroes of the polynomials are {0,  2}
Relationship between the zeroes and the coefficients of the polynomial:
Sum of the zeroes =   
Also sum of the zeroes =
Product of the zeroes = =
Also product of the zeroes =
Hence verified.

(v) t2 – 15  = (t +
The zeroes of the polynomials are {
Relationship between the zeroes and the coefficients of the polynomial:
Sum of the zeroes =   
Also sum of the zeroes =
Product of the zeroes = =
Also product of the zeroes =
Hence verified.  

(vi) 3x2 – x – 4 = 3x2 – 4x + 3x – 4
                      = x(3x – 4) + 1(3x – 4)
                      = (3x – 4)(x + 1)
The zeroes of the polynomials are {
Relationship between the zeroes and the coefficients of the polynomial:
Sum of the zeroes =   =  
Also sum of the zeroes =
Product of the zeroes = 
Also product of the zeroes =
Hence verified.

Question-8

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) (ii)  (iii) 0, (iv) 1, 1 (v) -, (vi) 4, 1

Solution:
(i)
Let the quadratic polynomial be ax2 + bx + c, and its zeroes be α and β
Given α + β = =
α β = -1 =
If a = 4 , b = -1 and c = -4
The quadratic polynomial is 4x2 - x - 4.

 (ii) 
Let the quadratic polynomial be ax2 + bx + c, and its zeroes be α and β
Given α + β = =
α β = =
If a = 1, b = - and c =
The quadratic polynomial x2 - x + (or) 3x2 - 3x + 1 .

(iii) 0,
Let the quadratic polynomial be ax2 + bx + c and the zeroes be α + β
Given α + β = 0 =
α β = =
If a = 1, b = 0 and c =
The quadratic polynomial is x2 +
 
(iv) 1, 1
Let the quadratic polynomial be ax2 + bx + c , and its zeroes be α + β
Given α + β = = 1
α β = = 1
If a = 1 , b = -1 and c = 1
The quadratic polynomial is x2 - x + 1.

(v),
Let the quadratic polynomial be ax2 + bx + c , and its zeroes be α + β
Given α + β = =

α β == .
If a = 4, b = 1 and c = 1
The quadratic polynomial is 4x2 + x + 1.

(vi) 4, 1
Let the quadratic polynomial be ax2 + bx + c
Given α + β = = 4
α β = = 1
If a = 1, b = -4 and c = 1
The quadratic polynomial is x2 – 4x +1.

Question-9

Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
(i) p(x) = x3 – 3x2 + 5x - 3, g(x) = x2 – 2
(ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2

Solution:
(i) p(x) = x3 – 3x2 + 5x - 3, g(x) = x2 – 2



Quotient is (x - 3)

Remainder is 7x – 9

(ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
Rearrange g(x) as x2 - x + 1


The Quotient is x2 + x – 3
Remainder is 8

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

Quotient is –x2 – 2
Remainder is –5x + 10

Question-10

Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.
(i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12
(ii) x2 + 3x +1, 3x4 + 5x3 – 7x2 + 2x + 2
(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
 

Solution:
(i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12


 t2 – 3 is a factor of the polynomial 2t4 + 3t3 – 2t2 – 9t – 12 .
 
(ii) x2 + 3x+1, 3x4 + 5x3 – 7x2 + 2x + 2


Hence the polynomial
x2 + 3x+1 is a factor of the second polynomial 3x4 + 5x3 – 7x2 + 2x + 2

(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1

Hence
x3 – 3x + 1 is not a factor of  x5 – 4x3 + x2 + 3x + 1.

Question-11

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

Solution:
Two zeros are and -.
Since the two zeros are and-
(x -) (x +) = x2 is a factor of the polynomial (i.e.,) (3x2 – 5)
Divide the given polynomial by 3x2 – 5.

(3x4 + 6x3 – 2x2 – 10x – 5) = (3x2 – 5) (x2 + 2x + 1)
                                           = (3x2 – 5) (x + 1)2
Its zeroes are given by x = -1, x = -1
The zeroes of the given polynomial 3x4 + 6x3 – 2x2 – 10x – 5 are , -, -1, -1.

Question-12

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).

Solution:
By the division algorithm,
Dividend = Divisor × Quotient + Remainder.
In this problem f(x)= x3 – 3x2 + x + 2, since (-2x + 4 ) is the remainder, subtract (-2x + 4) from f(x)
then divide the result by(x - 2).
f(x) - (-2x + 4)= x3 – 3x2 + x + 2 + 2x - 4
                     =x3 – 3x2 + 3x - 2 


Thus g(x) =x2 – x + 1.

Question-13

Give examples of polynomials p(x), g(x), q(x), 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.

Solution:
If p(x) denotes the dividend, g(x) denotes the divisor, q(x) denotes the quotient, r(x) denotes the remainder.
p(x) = 3x2 – 6x + 12, g(x) = 3, q(x) = x2 - 2x + 4, r(x) = 0.
p(x) = x3 + x2 + x + 1, g(x) = x2 – 1, q(x) = x + 1, r(x) = 2x + 2.
p(x) = 2x3 + 2x2 – 6x + 2, g(x) = x2 – x – 1, q(x) = 2x + 4, r(x) = 6.




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