Polynomials

Class 10 > Math > NCERT > Polynomials

LAQ for Chapter 2 Polynomials Class 10 Math NCERT

Important Questions

1

Polynomial x4 + 7x3 + 7x2 + px + q is exactly divisible by x2 + 7x + 12, then find the value of p and q.

Answer

We have f(x) = x4 + 7x3 + 7x2 + px + q
Now x2 + 7x + 12 = 0
x2 + 4x + 3x + 12 = 0
x(x + 4) + 3(x + 4) = 0
(x + 4)(x + 3) = 0
x = −4, −3
Since f(x) = x4 + 7x3 + 7x2 + px + q is exactly divisible by x2 + 7x + 12, then x = −4 and x = −3 must be its zeroes and these must satisfy f(x) = 0
So putting x = −4 and x = −3 in f(x) and equating to zero we get
f(−4) : (−4)4 + 7(−4)3 + 7(−4)2 + p(−4) + q = 0
256 – 448 + 112 – 4p + q = 0
−4p + q – 80 = 0
4p – q = −80 …(1)
f(−3) : (−3)4 + 7(−3)3 + 7(−3)2 + p(−3) + q = 0
81 – 189 + 63 – 3p + q = 0
−3p + q – 45 = 0
3p – q = −45 …(2)
Subtracting equation (2) from (1) we have
p = −35
Substituting the value of p in equation (1) we have
4(−35) – q = − 80
−140 – q = − 80
− q = 140 – 80
or – q = 60
q = −60
Hence, p = −35 and q = −60.
LAQ

2

If α and β are the zeroes of the polynomial p(x) = 2x2 + 5x + k satisfying the relation, α2 + β2 + αβ = 21/4, then find the value of k.

Answer

We have p(x) = 2x2 + 5x + k
Sum of zeroes,

Product of zeroes

According to the question,

Substituting values we have

Hence , k = 2
LAQ

3

If α and β are the zeroes of polynomial p(x) = 3x2 + 2x + 1, find the polynomial whose zeroes are .

Answer

We have p(x) = 3x2 + 2x + 1
Since α and β are the zeroes of polynomial 3x2 + 2x + 1,
We have
α + β = −2/3
and αβ =1/3
Let α1 and β1 be zeros of new polynomial q(x).
Then for q(x), sum of zeroes,

For q(x), product of the zeroes,

Hence, Required polynomial
q(x) = x2 – (α1 + β1)2x + α1β1
= x2 – 2x + 3
LAQ

4

If α and β are the zeroes of the polynomial x2 + 4x + 3, find the polynomial whose zeroes are 1 + β/α and 1 + α/β .

Answer

We have p(x) = x2 + 4x + 3
Since α and β are the zeroes of the quadratic polynomial x2 + 4x + 3,
So, α + β = −4
and αβ = 3
Let α1 and β1 be zeros of new polynomial q(x).
Then for q(x), sum of zeroes,

For q(x), product of the zeroes,

Hence, required polynomial
q(x) = x2 – (α1 + β1)2x + α1β1
LAQ

5

If α and β are zeroes of the polynomial p(x) = 6x2 – 5x + k such that α – β = 1/6 , Find the value of k.

Answer

We have p(x) = 6x2 – 5x + k
Since α and β are zeroes of
p(x) = 6x2 – 5x + k
Sum of zeroes,
α + β = −(−5/6) = 5/6 …(1)
Product of zeroes
αβ = k/6 …(2)
Given
α – β = 1/6 …(3)
Solving (1) and (3) we get α = ½ and β = 1/3 and substituting the values of (2) we have

Hence, k = 1.
LAQ

6

If β and 1/β are zeroes of the polynomial (a2 + a)x2 + 61x + 6a. Find the value of β and α.

Answer

We have p(x) = (a2 + a)x2 + 61x + 6
Since β and 1/β are the zeroes of polynomial, p(x)
Sum of zeroes,

Product of zeroes

a + 1 = 6
a = 5
Substituting this value of a in (1) we get

30β2 + 30 = −61β
30β2 + 61β + 30 = 0
LAQ

7

If α and β are the zeroes the polynomial 2x2 – 4x + 5, find the values of
(i) α2 + β2
(ii) 1/α + 1/β
(iii) (α – β)2
(iv) 1/α2 + 1/β2
(v) α3 + β3

Answer

We have p(x) = 2x2 – 4x + 5
If α and β are the zeroes of p(x) = 2x2 – 4x + 5, then

(i) α2 + β2 = (α + β)2 – 2αβ
= 22 – 2 × 5/2
= 4 – 5 = −1

(iii) (α – β)2 = (α – β)2 – 4αβ

4 – 10 = −6

(v) (α3 + β3) = (α – β)3 – 3αβ(α + β)
= 23 – 3 × 5/2 × 2 = 8 – 15 = −7
LAQ

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