Polynomial and Ans

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UNIT-2
POLYNOMIALS
It is not once nor twice but times without number that the same ideas make their appearance in the world.
1. Find the value for K for which x4 + 10x3 + 25x2 + 15x + K exactly divisible by x + 7. (Ans : K= - 91)
4

4

2

Ans: Let P(x) = x + 10x + 25x + 15x + K and g(x) = x + 7
Since P(x) exactly divisible by g(x)

r (x) = 0
x 3 + 3 x 2 + 4 x − 13
now x + 7 x 4 + 10 x 3 + 25 x 2 + 15 x + K
x 4 + 7 x3
------------3x3 + 25 x2
3x3 + 21x2
------------------4x2 + 15 x
4x2 + 28x
------------------13x + K
- 13x - 91
---------------K + 91
-----------∴ K + 91 = 0
K= -91
2. If two zeros of the polynomial f(x) = x4 - 6x3 - 26x2 + 138x – 35 are 2 ± √3.Find the other zeros.
(Ans:7, -5)

Ans: Let the two zeros are 2 + 3 and 2 - 3
Sum of Zeros
=2+ 3 +2- 3
=4
Product of Zeros = ( 2+ 3 )(2 - 3 )
=4–3
=1
Quadratic polynomial is x2 – (sum) x + Product

11

x2 – 2x – 35
x2 – 4x + 1 x 4 − 6 x3 − 26 x 2 + 138 x − 35

x 4 − 4 x3 + x 2
-----------------2x3 – 27x2 + 138x
- 2x3 + 8x2 – 2x
-----------------------35x2 + 140x – 35
-35x2 + 140x – 35
-----------------------0
-----------------------∴ x2 – 2x – 35 = 0
(x – 7)(x + 5) = 0
x = 7, -5

other two Zeros are 7 and -5

3. Find the Quadratic polynomial whose sum and product of zeros are √2 + 1,

1
2 +1

.

Ans: sum = 2 2
Product = 1
Q.P =
X2 – (sum) x + Product
∴ x2 – (2 2 ) x + 1
4. If α,β are the zeros of the polynomial 2x2 – 4x + 5 find the value of a) α2 + β 2 b) (α - β)2.
(Ans: a) -1 , b) –6)

Ans: p (x) = 2 x2 – 4 x + 5
−b 4
α+β=
= =2
a2
c5
αβ= =
a2
2
2
α + β = (α + β)2 – 2 α β
Substitute then we get, α 2+ β2 = -1
(α - β)2 = (α + β)2 - 4 α β
Substitute, we get = (α - β)2 = - 6

12

5. If α,β are the zeros of the polynomial x2 + 8x + 6 frame a Quadratic polynomial

1

whose zeros are a)

α

and

1

β

b) 1 +

β
α

Sum =

1

+

α

Product =

1

α

1

β
x

=

1

β

α
β

.
4
1
32
32
(Ans: x2+ x + , x2- x + )
3
6
3
3

Ans: p (x) = x2 + 8 x + 6
α + β = -8 and α β = 6

a) Let two zeros are

,1+

1

α

and

1

β

α + β −8 −4
=
=
α .β
6
3
=

1
1
=
α .β 6

Required Q.P is
x2 +

b) Let two Zeros are 1+

4
1
x+
3
6

β
α
and 1 +
α
β

β
α
+1+
α
β
α
β
=2+
+
β
α
2
2
α +β
= 2+
αβ
(α + β ) 2 − 2αβ
after solving this problem,
= 2+
αβ

sum = 1+

We get

=

32
3

Product = ( 1 +

β
α
)(1+ )
α
β

α
β
+
+1
β
α
α2 +β2
=2+
αβ
= 1+

Substitute this sum,

13

We get =

32
3

Required Q.P. is x2 -

32
32
x+
3
3

6. On dividing the polynomial 4x4 - 5x3 - 39x2 - 46x – 2 by the polynomial g(x) the quotient is x2 - 3x – 5 and the remainder is -5x + 8.Find the polynomial g(x). (Ans:4 x2+7x+2)

Ans: p(x) = g (x) q (x) + r (x)
p( x) − r ( x)
g(x) =
q( x)
let p(x) = 4x4 – 5x3 – 39x2 – 46x – 2
q(x) = x2 – 3x – 5 and r (x) = -5x + 8
now p(x) – r(x) = 4x4 – 5x3 – 39x2 – 41x - 10
p( x) − r ( x)
when
= 4x2 + 7x +2
q( x)
2
∴ g(x) = 4x + 7x + 2
7.

If the squared difference of the zeros of the quadratic polynomial x2 + px + 45 is equal to 144 , find the value of p.
(Ans: ± 18).

Ans: Let two zeros are α and β where α > β
According given condition
(α - β)2 = 144
Let p(x) = x2 + px + 45
−b − p
α+β=
=
=-p
a
1
c 45
αβ = =
= 45
a
1
now (α - β)2 = 144
(α + β)2 – 4 αβ = 144
(-p)2 – 4 (45) = 144
Solving this we get p = ± 18
8. If α,β are the zeros of a Quadratic polynomial such that α + β = 24, α - β = 8. Find a Quadratic polynomial having α and β as its zeros.
(Ans: k(x2– 24x + 128))

Ans: α+β = 24
α-β=8
----------2 α = 32

14

32
= 16, ∴ α = 16
2
Work the same way to α+β = 24
α=

So, β = 8
Q.P is x2 – (sum) x + product
= x2 – (16+8) x + 16 x 8
Solve this,
it is k (x2 – 24x + 128)
If α & ß are the zeroes of the polynomial 2x2 ─ 4x + 5, then find the value of a. α2 + ß2 b. 1/ α + 1/ ß c. (α ─ ß)2 d. 1/α2 + 1/ß2...
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