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Exercise 2.4
Ques 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) 2x^{3} + x^{2}  5x + 2;
(ii) x^{3}  4x^{2} + 5x  2; 2, 1, 1
Sol: (i) ∵ p (x) = 2x^{3} + x^{2}  5x + 2
∴
⇒ 1/2 is a zero of p(x).
Again,
p (1) = 2 (1)^{3} + (1)^{2}  5 (1) + 2
= 2 + 1  5 + 2
= (2 + 2 + 1)  5
= 5  5 = 0
⇒ 1 is a zero of p (x).
Also p ( 2) = 2 ( 2)^{3} + ( 2)^{2}  5 ( 2) + 2
= 2( 8) + (4) + 10 + 2
=  16 + 4 + 10 + 2
=  16 + 16 = 0
⇒ 2 is a zero of p (x).
Relationship
∵ p(x) = 2x^{3} + x^{2}  5x + 2
∴ Comparing it with ax^{3} + bx^{2} + cx + d, we have :
a = 2, b = 1, c =  5 and d = 2
Also 1/2 , 1 and  2 are the zeroes of p(x)
Let α = 1/2 , β = 1 and γ =  2
∴ α + β +γ =
Sum of the zeroes =
⇒ α + β + γ = b/a
Sum of product of zeroes taken in pair:
Product of zeroes
Also
⇒
Thus, the relationship between the coefficients and the zeroes of p (x) is verified.
(ii) Here, p(x) = x^{3}  4x^{2} + 5x  2
∴ p (2) = (2)^{3}  4 (2)^{2} + 5 (2)  2
= 8  16 + 10  2
= 18  18 = 0
⇒ 2 is a zero of p(x)
Again p (1) = (1)^{3}  4 (1)^{2} + 5 (1)  2
= 1  4 + 5  2
= 6  6 = 0
⇒ 1 is a zero of p(x).
∴ 2, 1, 1 are zeroes of p(x).
Now,
Comparing p (x) = x^{3}  4 (x^{2}) + 5x  2 with ax^{3} + bx^{2} + cx + d = 0, we have
a = 1, b =  4, c = 5 and d =  2
∵ 2, 1, and 1 are the zeroes of p (x)
∴ Let α = 2
β = 1
γ = 1
Relationship, α + β + γ = 2 + 1 + 1 = 4
Sum of the zeroes =
⇒ α + β + γ = (b/a)
Sum of product of zeroes taken in pair:
αβ + βγ + γα = 2 (1) + 1 (1) + 1 (2)
= 2 + 1 + 2 = 5
and = c/a = 5/1 = 5
⇒ αβ + βγ + γα = c/a
Product of zeroes = αβγ = (2) (1) (1) = 2
⇒
Thus, the relationship between the zeroes and the coefficients of p(x) is verified.
Ques 2: Find the cubic polynomial with the sum, sum of the products of its zeroes taken two at a time and the product of its zeroes as 2,  7,  14 respectively.
Sol: Let the required cubic polynomial be ax^{3} + bx^{2} + cx + d and its zeroes be a, b and g.
∴
Since,
If a = 1, then b/a = 2 ⇒ b =  2
c/a =  7 ⇒ c =  7
d/a =  14 ⇒ d = 14
∴ The required cubic polynomial
= 1x^{3} + (2) x^{2} + (7) x + 14
= x^{3}  2x^{2}  7x + 14
Ques 3: If the zeroes of the polynomial x^{3}  3x^{2} + x + 1 are a  b, a and a + b then find ‘a’ and ‘b’.
Sol: We have
p (x) = x^{3}  3x^{2 }+ x + 1
comparing it with Ax^{3} + Bx^{2} + Cx + D.
We have
A = 1, B =  3, C = 1 and D = 1
∵ It is given that (a  b), a and (a + b) are the zeroes of the polynomial.
∴ Let, α = (a  b)
β = a
and γ = (a + b)
∴ α + β + γ =
⇒ (a  b) + a + (a + b) = 3
⇒ 3a = 3
⇒ a = 3/3 = 1
Again, αβγ = D/A = 1
⇒ (a  b) × a × (a + b) =  1
⇒ (1  b) × 1 × (1 + b) =  1 [∵ a = 1, proved above]
⇒ 1  b^{2} =  1
⇒ b^{2} = 1 + 1 = 2
⇒ b = √2
Thus, a = 1 and b = .√2
Ques 4: If two zeroes of the polynomial x^{4}  6x^{3}  26x^{2} + 138x  35 are 2 √3, find other zeroes.
Sol: Here, p(x) = x^{4}  6x^{3}  26x^{2} + 138x  35.
∵ Two of the zeroes of p (x) are :
or (x  2)^{2}  (√3)^{2}
or (x^{2} + 4  4x)  3
or x^{2}  4x + 1 is a factor of p(x).
Now, dividing p (x) by x^{2}  4x + 1, we have :
∴ (x^{2 } 4x + 1) (x^{2}  2x  35) = p (x)
⇒ (x^{2}  4x + 1) (x  7) (x + 5) = p (x)
i.e., (x  7) and (x + 5) are other factors of p(x).
∴ 7 and  5 are other zeroes of the given polynomial.
Ques 5: If the polynomial x^{4}  6x^{3} + 16x^{2}  25x + 10 is divided by another polynomial x^{2}  2x + k, the remainder comes out to be (x + α), find k and α.
Sol: Applying the division algorithm to the polynomials x^{4}  6x^{3} + 16x^{2}  25x + 10 and
x^{2}  2x + k, we have:
∴ Remainder = (2k  9) x  k (8  k) + 10
But the remainder = x + α
Therefore, comparing them, we have :
2k  9 = 1
⇒ 2k = 1 + 9 = 10
⇒ k = 10/2 = 5
and
α =  k (8  k) + 10
=  5 (8  5) + 10
=  5 (3) + 10
=  15 + 10
=  5
Thus, k = 5 and α =  5
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