WORKSHEET 1 - Polynomials

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:no zeroes.

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 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:One

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 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:Three

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 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:Two

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 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:Four

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 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:Three.

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 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.

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 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 a and b
  Given a + b =  = 
  a b = -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 a and b
  Given a + b =  = 
  a b =  = 
  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 a + b
  Given a + b = 0 = 
  a b =  = 
  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 a + b
  Given a + b =  = 1
  a b =  = 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 a + b
  Given a + b = = 
  a b == .
  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 a + b = = 4
  a b = = 1
  If a = 1, b = -4 and c = 1
  \ The quadratic polynomial is x² – 4x +1.

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 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

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 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
  
  
  \ 1^st is a factor of the second polynomial.
  (ii) x² + 3x+1, 3x⁴ + 5x³ – 7x² + 2x + 2
  
  
  Hence 1^st polynomial is a factor of the second polynomial
  (iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1
  
  No it is not a solution
  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 are , -, -1, -1.

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   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).
    
    
    Thus g(x) =x² – x + 1.

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   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 – 2x +2, g(x) = x² – x –1, q(x) = 2x + 4, r(x) = 6.