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Page No: 28
1. The graphs of y = p(x) are given in the following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.
(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.
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Page No: 33
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
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.
(ii) 4s2 – 4s + 1
= (2s-1)2
The value of 4s2 – 4s + 1 is zero when 2s – 1 = 0, i.e., s = 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
= 6x2 – 7x – 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
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
= t2 – 0.t – 15
= (t – √15) (t + √15)
The value of t2 – 15 is zero when t – √15 = 0 or t + √15 = 0, i.e., when t = √15 or t = -√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.
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
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 x2 – x +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
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1. 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
Quotient = x-3 and remainder 7x – 9
Quotient = x2 + x – 3 and remainder 8
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:
(i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12
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
(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
and – √(5/3).
Answer
p(x) = 3x4 + 6x3 – 2x2 – 10x – 5
Since the two zeroes are √(5/3) and – √(5/3).
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,
∴ 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
(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 ) × x + 1
x3 + 1 = x3 + 1
Thus, the division algorithm is satisfied.
Page No: 36
Read also – Real number solution
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1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i) 2x3 + x2 – 5x + 2; 1/2, 1, -2
(ii) x3 – 4x2 + 5x – 2; 2, 1, 1
Answer
(i) p(x) = 2x3 + x2 – 5x + 2
Now for zeroes, putting the given value in x.
p(1/2) = 2(1/2)3 + (1/2)2 – 5(1/2)+ 2
= (2×1/8) + 1/4 – 5/2 + 2
= 1/4 + 1/4 – 5/2 + 2
= 1/2 – 5/2 + 2 = 0
p(1) = 2(1)3 + (1)2 – 5(1)+ 2
= (2×1) + 1 – 5 + 2
= 2 + 1 – 5 + 2 = 0
p(-2) = 2(-2)3 + (-2)2 – 5(-2)+ 2
= (2 × -8) + 4 + 10 + 2
= -16 + 16 = 0
Thus, 1/2, 1 and -2 are the zeroes of the given polynomial.
Comparing the given polynomial with ax3 + bx2 + cx + d, we get a=2, b=1, c=-5, d=2
Also, α=1/2, β=1 and γ=-2
Now,
-b/a = α+β+γ
⇒ 1/2 = 1/2 + 1 – 2
⇒ 1/2 = 1/2
c/a = αβ+βγ+γα
⇒ -5/2 = (1/2 × 1) + (1 × -2) + (-2 × 1/2)
⇒ -5/2 = 1/2 – 2 – 1
⇒ -5/2 = -5/2
-d/a = αβγ
⇒ -2/2 = (1/2 × 1 × -2)
⇒ -1 = 1
Thus, the relationship between zeroes and the coefficients are verified.
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(ii) p(x) = x3 – 4x2 + 5x – 2
Now for zeroes, putting the given value in x.
p(2) = 23 – 4(2)2 + 5(2)- 2
= 8 – 16 + 10 – 2
= 0
p(1) = 13 – 4(1)2 + 5(1)- 2
= 1 – 4 + 5 – 2
= 0
p(1) = 13 – 4(1)2 + 5(1)- 2
= 1 – 4 + 5 – 2
= 0
Thus, 2, 1 and 1 are the zeroes of the given polynomial.
Comparing the given polynomial with ax3 + bx2 + cx + d, we get a=1, b=-4, c=5, d=-2
Also, α=2, β=1 and γ=1
Now,
-b/a = α+β+γ
⇒ 4/1 = 2 + 1 + 1
⇒ 4 = 4
c/a = αβ+βγ+γα
⇒ 5/1 = (2 × 1) + (1 × 1) + (1 × 2)
⇒ 5 = 2 + 1 + 2
⇒ 5 = 5
-d/a = αβγ
⇒ 2/1 = (2 × 1 × 1)
⇒ 2 = 2
Thus, the relationship between zeroes and the coefficients are verified.
2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.
Answer
Let the polynomial be ax3 + bx2 + cx + d and the zeroes be α, β and γ
Then, α + β + γ = -(-2)/1 = 2 = -b/a
αβ + βγ + γα = -7 = -7/1 = c/a
αβγ = -14 = -14/1 = -d/a
∴ a = 1, b = -2, c = -7 and d = 14
So, one cubic polynomial which satisfy the given conditions will be x3 – 2x2 – 7x + 14
3. If the zeroes of the polynomial x3 – 3x2 + x + 1 are a–b, a, a+b, find a and b.
Answer
Since, (a – b), a, (a + b) are the zeroes of the polynomial x3 – 3x2 + x + 1.
Therefore, sum of the zeroes = (a – b) + a + (a + b) = -(-3)/1 = 3
⇒ 3a = 3 ⇒ a =1
∴ Sum of the products of is zeroes taken two at a time = a(a – b) + a(a + b) + (a + b) (a – b) =1/1 = 1
a2 – ab + a2 + ab + a2 – b2 = 1
⇒ 3a2 – b2 =1
Putting the value of a,
⇒ 3(1)2 – b2 = 1
⇒ 3 – b2 = 1
⇒ b2 = 2
⇒ b = ±√2
Hence, a = 1 and b = ±√2
4. If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2±√3, find other zeroes.
Answer
2+√3 and 2-√3 are two zeroes of the polynomial p(x) = x4 – 6x3 – 26x2 + 138x – 35.
Let x = 2±√3
So, x-2 = ±√3
On squaring, we get x2 – 4x + 4 = 3,
⇒ x2 – 4x + 1= 0
Now, dividing p(x) by x2 – 4x + 1
∴ p(x) = x4 – 6x3 – 26x2 + 138x – 35
= (x2 – 4x + 1) (x2 – 2x – 35)
= (x2 – 4x + 1) (x2 – 7x + 5x – 35)
= (x2 – 4x + 1) [x(x – 7) + 5 (x – 7)]
= (x2 – 4x + 1) (x + 5) (x – 7)
∴ (x + 5) and (x – 7) are other factors of p(x).
∴ – 5 and 7 are other zeroes of the given polynomial.
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Disclaimer – These questions are taken from Ncert official website.
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