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...