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
Answer
(i) The number of zeroes is 0 as the graph does not cut the xaxis at any point.
(ii) The number of zeroes is 1 as the graph intersects the xaxis at only 1 point.
(iii) The number of zeroes is 3 as the graph intersects the xaxis at 3 points.
(iv) The number of zeroes is 2 as the graph intersects the xaxis at 2 points.
(v) The number of zeroes is 4 as the graph intersects the xaxis at 4 points.
(vi) The number of zeroes is 3 as the graph intersects the xaxis 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) x^{2} – 2x – 8
(ii) 4s^{2} – 4s 1
(iii) 6x^{2} – 3 – 7x
(iv) 4u^{2} 8u
(v) t^{2} – 15
(vi) 3x^{2} – x – 4
Answer
(i) x^{2} – 2x – 8
= (x  4) (x 2)
The value of x^{2} – 2x – 8 is zero when x  4 = 0 or x 2 = 0, i.e., when x = 4 or x = 2
Therefore, the zeroes of x^{2} – 2x – 8 are 4 and 2.
Sum of zeroes = 4 (2) = 2 = (2)/1 = (Coefficient of x)/Coefficient of x^{2}
Product of zeroes = 4 × (2) = 8 = 8/1 = Constant term/Coefficient of x^{2}
(ii) 4s^{2} – 4s + 1= (2s1)^{2}
The value of 4s^{2}  4s +1 is zero when 2s  1 = 0, i.e., s = 1/2
Therefore, the zeroes of 4s^{2}  4s + 1 are 1/2 and 1/2.
Sum of zeroes = 1/2 1/2 = 1 = (4)/4 = (Coefficient of s)/Coefficient of s^{2}
Product of zeroes = 1/2 × 1/2 = 1/4 = Constant term/Coefficient of s^{2}.
(iii) 6x^{2} – 3 – 7x
= 6x^{2 }– 7x – 3
= (3x 1) (2x  3)
The value of 6x^{2}  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 6x^{2}  3  7x are 1/3 and 3/2.
Sum of zeroes = 1/3 3/2 = 7/6 = (7)/6 = (Coefficient of x)/Coefficient of x^{2}
Product of zeroes = 1/3 × 3/2 = 1/2 = 3/6 = Constant term/Coefficient of x^{2}.
(iv) 4u^{2} 8u
= 4u^{2} 8u 0
= 4u(u 2)
The value of 4u^{2} 8u is zero when 4u = 0 or u 2 = 0, i.e., u = 0 or u =  2
Therefore, the zeroes of 4u^{2} 8u are 0 and  2.
Sum of zeroes = 0 (2) = 2 = (8)/4 = (Coefficient of u)/Coefficient of u^{2}
Product of zeroes = 0 × (2) = 0 = 0/4 = Constant term/Coefficient of u^{2}.
(v) t^{2} – 15
= t^{2 } 0.t  15
= (t  √15) (t √15)
The value of t^{2}  15 is zero when t  √15 = 0 or t √15 = 0, i.e., when t = √15 or t = √15
Therefore, the zeroes of t^{2}  15 are √15 and √15.Sum of zeroes = √15 √15 = 0 = 0/1 = (Coefficient of t)/Coefficient of t^{2}
Product of zeroes = (√15) (√15) = 15 = 15/1 = Constant term/Coefficient of u^{2}.
(vi) 3x^{2} – x – 4
= (3x  4) (x 1)
The value of 3x^{2} – 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 3x^{2} – x – 4 are 4/3 and 1.
Sum of zeroes = 4/3 (1) = 1/3 = (1)/3 = (Coefficient of x)/Coefficient of x^{2}
Product of zeroes = 4/3 × (1) = 4/3 = Constant term/Coefficient of x^{2}.
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 ax^{2} 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 4x^{2}  x 4.
(ii) √2 , 1/3
Let the polynomial be ax^{2} 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 3x^{2} 3√2x 1.
(iii) 0, √5
Let the polynomial be ax^{2} 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 x^{2} √5.
(iv) 1, 1
Let the polynomial be ax^{2} 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 x^{2}  x 1.
(v) 1/4 ,1/4
Let the polynomial be ax^{2} 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 4x^{2} x 1.
(vi) 4,1
Let the polynomial be ax^{2} 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 x^{2}  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) = x^{3} – 3x^{2} 5x – 3, g(x) = x^{2} – 2
Quotient = x3 and remainder 7x  9
(ii) p(x) = x^{4} – 3x^{2} 4x 5, g(x) = x^{2} 1 – x
Quotient = x^{2} x  3 and remainder 8
(iii) p(x) = x^{4} – 5x 6, g(x) = 2 – x^{2}
Quotient = x^{2} 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) t^{2} – 3, 2t^{4} 3t^{3} – 2t^{2} – 9t – 12
t^{2} – 3 exactly divides 2t^{4} 3t^{3} – 2t^{2} – 9t – 12 leaving no remainder. Hence, it is a factor of 2t^{4} 3t^{3} – 2t^{2} – 9t – 12.
(ii) x^{2} 3x 1, 3x^{4} 5x^{3} – 7x^{2} 2x 2
x^{2} 3x 1 exactly divides 3x^{4} 5x^{3} – 7x^{2} 2x 2 leaving no remainder. Hence, it is factor of 3x^{4} 5x^{3} – 7x^{2} 2x 2.
(iii) x^{3} – 3x 1, x^{5} – 4x^{3} x^{2} 3x 1
x^{2} 3x 1 exactly divides 3x^{4} 5x^{3} – 7x^{2} 2x 2 leaving no remainder. Hence, it is factor of 3x^{4} 5x^{3} – 7x^{2} 2x 2.
(iii) x^{3} – 3x 1, x^{5} – 4x^{3} x^{2} 3x 1
x^{3} – 3x 1 didn't divides exactly x^{5} – 4x^{3} x^{2} 3x 1 and leaves 2 as remainder. Hence, it not a factor of x^{5} – 4x^{3} x^{2} 3x 1.
3. Obtain all other zeroes of 3x^{4} 6x^{3} – 2x^{2} – 10x – 5, if two of its zeroes are √(5/3) and  √(5/3).
Answer
p(x) = 3x^{4} 6x^{3} – 2x^{2} – 10x – 5
Since the two zeroes are √(5/3) and  √(5/3).
We factorize x^{2} + 2x + 1
= (x + 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 x^{3}  3x^{2} 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 = x^{3}  3x^{2} x 2
Quotient = x  2
Remainder = 2x 4
Divisor = g(x)
We know that,
Dividend = Quotient × Divisor Remainder
⇒ x^{3}  3x^{2} x 2 = (x  2) × g(x) (2x 4)⇒ x^{3}  3x^{2} x 2  (2x 4) = (x  2) × g(x)
⇒ x^{3}  3x^{2} 3x  2 = (x  2) × g(x)
⇒ g(x) = (x^{3}  3x^{2} 3x  2)/(x  2)
∴ g(x) = (x^{2}  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 6x^{2} 2x 2 by 2
Here, p(x) = 6x^{2} 2x 2
g(x) = 2
q(x) = 3x^{2} 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, 6x^{2} 2x 2 = 2x (3x^{2} x 1)
Hence, division algorithm is satisfied.
(ii) Let us assume the division of x^{3} x by x^{2},
Here, p(x) = x^{3} x
g(x) = x^{2}
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)
x^{3} x = (x^{2} ) × x x
x^{3} x = x^{3} x
Thus, the division algorithm is satisfied.
(iii) Let us assume the division of x^{3} 1 by x^{2}.
Here, p(x) = x^{3} 1
g(x) = x^{2}
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)
x^{3} 1 = (x^{2} ) × x 1
x^{3} 1 = x^{3} 1
Thus, the division algorithm is satisfied.
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