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Chapter 2: Polynomials Exercise – 2.2 Question: 1 Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and coefficients in each of the following cases: (i) f(x) = 2x3+ x2– 5x + 2; 1/2, 1, – 2 (ii) g(x) = x3– 4x2 + 5x – 2; 2, 1, 1 Solution: (i) f(x) = 2x3 + x2 – 5x + 2; 1/2, 1, – 2 (a) By putting x = 1/2 in the above equation, we will get (b) By putting x = 1 in the above equation, we will get f(1) = 2(1)3 + (1)2 – 5(1) + 2 = 2 + 1 – 5 + 2 = 0 (c) By putting x = -2 in the above equation, we will get f(−2) = 2(−2)3 + (−2)2 – 5(−2) + 2 = -16 + 4 + 10 + 2 = – 16 + 16 = 0 Now, Sum of zeroes = α + β + γ = - b/a Product of the zeroes = αβ + βγ + αγ = c/a Hence, verified. (ii) g(x) = x3 – 4x2 + 5x – 2; 2, 1, 1 (a) By putting x = 2 in the given equation, we will get g(2) = (2)3 – 4(2)2 + 5(2) – 2 = 8 – 16 + 10 – 2 = 18 – 18 = 0 (b) By putting x = 1 in the given equation, we will get g(1) = (1)3 – 4(1)2 + 5(1) – 2 = 1 – 4 + 5 – 2 = 0 Now, Sum of zeroes= α + β + γ =-b/a ⇒ 2 + 1 + 1 = −(−4) 4 = 4 Product of the zeroes = αβ + βγ + αγ = c/a 2 × 1 + 1 × 1 + 1 × 2 = 5 2 + 1 + 2 = 5 5 = 5 αβγ = –(−2) 2 × 1 × 1 = 2 2 = 2 Hence, verified. Question: 2 Find a cubic polynomial with the sum, sum of the product of its zeroes is taken two at a time, and product of its zeroes as 3, – 1 and – 3 respectively. Solution: Any cubic polynomial is of the form ax3 + bx2 + cx + d: = x3 – (sum of the zeroes) x2 + (sum of the products of its zeroes) x – (product of the zeroes) = x3 – 3x2 + (−1)x + (−3) = k[x3 – 3x2 – x – 3] k is any non-zero real numbers. Question: 3 If the zeroes of the polynomial f(x) = 2x3 – 15x2 + 37x – 30, find them. Solution: Let, α = a – d, β = a and γ = a + d be the zeroes of the polynomial. f(x) = 2x3 – 15x2 + 37x – 30 And, a (a2 + d2) = 15 Question: 4 Find the condition that the zeroes of the polynomial f(x) = x3 + 3px2 + 3qx + r may be in A.P. Solution: f(x) = x3 + 3px2 + 3qx + r Let, a – d, a, a + d be the zeroes of the polynomial. Then, The sum of zeroes = – b/a a + a – d + a + d = -3p 3a = -3p a = -p Since, a is the zero of the polynomial f(x), Therefore, f(a) = 0 f(a) = a3 + 3pa2 + 3qa + r = 0 Therefore, f(a) = 0f(a) = 0 ⇒ a3 + 3pa2 + 3qa + r = 0 = ⇒ (−p)3 + 3p(−p)2 + 3q(−p) + r = 0 = − p3 + 3p3 – pq + r = 0 = 2p3 – pq + r = 0 Question: 5 If zeroes of the polynomial f(x) = ax3 + 3bx2 + 3cx + d are tin A.P., prove that 2b3 – 3abc + a2d = 0. Solution: f(x) = x3 + 3px2 + 3qx + r Let, a – d, a, a + d be the zeroes of the polynomial. Then, The sum of zeroes = - b/a a + a – d + a + d = – 3b/a Since, f(a) = 0 ⇒ a(a2) + 3b(a)2 + 3c(a) + d = 0 ⇒ a(a2) + 3b(a)2 + 3c(a) + d = 0 Question: 6 If the zeroes of the polynomial f(x) = x3 – 12x2 + 39x + k are in A.P., find the value of k. Solution: f(x) = x3 – 12x2 + 39x + k Let, a-d, a, a + d be the zeroes of the polynomial f(x). The sum of the zeroes = 12 3a = 12 a = 4 Now, f(a) = 0 f(a) = a3 – 12a2 + 39a + k f(4) = 43 – 12(4)2 + 39(4) + k = 0 f(4) = 43 –12(4)2 + 39(4) + k = 0 64 – 192 + 156 + k = 0 k = – 28
Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and coefficients in each of the following cases:
(i) f(x) = 2x3+ x2– 5x + 2; 1/2, 1, – 2
(ii) g(x) = x3– 4x2 + 5x – 2; 2, 1, 1
(i) f(x) = 2x3 + x2 – 5x + 2; 1/2, 1, – 2
(a) By putting x = 1/2 in the above equation, we will get
(b) By putting x = 1 in the above equation, we will get
f(1) = 2(1)3 + (1)2 – 5(1) + 2
= 2 + 1 – 5 + 2 = 0
(c) By putting x = -2 in the above equation, we will get
f(−2) = 2(−2)3 + (−2)2 – 5(−2) + 2
= -16 + 4 + 10 + 2 = – 16 + 16 = 0
Now,
Sum of zeroes = α + β + γ = - b/a
Product of the zeroes = αβ + βγ + αγ = c/a
Hence, verified.
(ii) g(x) = x3 – 4x2 + 5x – 2; 2, 1, 1
(a) By putting x = 2 in the given equation, we will get
g(2) = (2)3 – 4(2)2 + 5(2) – 2
= 8 – 16 + 10 – 2 = 18 – 18 = 0
(b) By putting x = 1 in the given equation, we will get
g(1) = (1)3 – 4(1)2 + 5(1) – 2
= 1 – 4 + 5 – 2
= 0
Sum of zeroes= α + β + γ =-b/a
⇒ 2 + 1 + 1 = −(−4)
4 = 4
2 × 1 + 1 × 1 + 1 × 2 = 5
2 + 1 + 2 = 5
5 = 5
αβγ = –(−2)
2 × 1 × 1 = 2
2 = 2
Find a cubic polynomial with the sum, sum of the product of its zeroes is taken two at a time, and product of its zeroes as 3, – 1 and – 3 respectively.
Any cubic polynomial is of the form ax3 + bx2 + cx + d:
= x3 – (sum of the zeroes) x2 + (sum of the products of its zeroes) x – (product of the zeroes)
= x3 – 3x2 + (−1)x + (−3)
= k[x3 – 3x2 – x – 3]
k is any non-zero real numbers.
If the zeroes of the polynomial f(x) = 2x3 – 15x2 + 37x – 30, find them.
Let, α = a – d, β = a and γ = a + d be the zeroes of the polynomial.
f(x) = 2x3 – 15x2 + 37x – 30
And, a (a2 + d2) = 15
Find the condition that the zeroes of the polynomial f(x) = x3 + 3px2 + 3qx + r may be in A.P.
f(x) = x3 + 3px2 + 3qx + r
Let, a – d, a, a + d be the zeroes of the polynomial.
Then,
The sum of zeroes = – b/a
a + a – d + a + d = -3p 3a = -3p a = -p Since, a is the zero of the polynomial f(x),
Therefore, f(a) = 0
f(a) = a3 + 3pa2 + 3qa + r = 0
Therefore, f(a) = 0f(a) = 0
⇒ a3 + 3pa2 + 3qa + r = 0
= ⇒ (−p)3 + 3p(−p)2 + 3q(−p) + r = 0
= − p3 + 3p3 – pq + r = 0
= 2p3 – pq + r = 0
If zeroes of the polynomial f(x) = ax3 + 3bx2 + 3cx + d are tin A.P., prove that 2b3 – 3abc + a2d = 0.
The sum of zeroes = - b/a
a + a – d + a + d = – 3b/a
Since, f(a) = 0
⇒ a(a2) + 3b(a)2 + 3c(a) + d = 0
If the zeroes of the polynomial f(x) = x3 – 12x2 + 39x + k are in A.P., find the value of k.
f(x) = x3 – 12x2 + 39x + k
Let, a-d, a, a + d be the zeroes of the polynomial f(x).
The sum of the zeroes = 12
3a = 12
a = 4
f(a) = 0
f(a) = a3 – 12a2 + 39a + k f(4) = 43 – 12(4)2 + 39(4) + k = 0
f(4) = 43 –12(4)2 + 39(4) + k = 0
64 – 192 + 156 + k = 0
k = – 28
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Chapter 2: Polynomials Exercise – 2.1...
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