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 k1, k2 and k3
Then, k1 + k2 +k3 = -(-2)/1 = 2 = -b/a
k1k2+k2k3+k1k3= -7 = -7/1 = c/a
k1k2k3 = -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
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
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)
So (x + 5) and (x - 7) are other factors of p(x).
- 5 and 7 are other zeroes of the given polynomial.
If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a.
On dividing x4 – 6x3 + 16x2 – 25x + 10 by x2 – 2x + k
Remainder = (2k - 9)x - (8 - k)k + 10
But the remainder is given as x+ a.
On comparing their coefficients,
2k - 9 = 1
⇒ k = 10
⇒ k = 5 and,
-(8-k)k + 10 = a
⇒ a = -(8 - 5)5 + 10 =- 15 + 10 = -5
Hence, k = 5 and a = -5
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