Maths Questions Ch 2: Solutions

Ans1. The graph y = p(x) intersects the x axis at four points.
So, the polynomial y = p(x) has 4 zeroes. (1 Mark)
Ans2. Let p (x) = x3 + 4x2 + 2x − 1
Then p (1) = (1)3 + 4 (1)2 + 2(1) – 1
= 1 + 4 + 2 -1
= 6
and p (-1) = (-1)3 + 4 (-1)2 + 2(-1) – 1
= -1 + 4 – 2 – 1
= 0
So, -1 is the zero of the polynomial 3 2 x + 4x + 2x − 1 and 1 is not. (1 Mark)
Ans3. Let the quadratic polynomial be ax2+bx + c, and its zeroes be a and b.
We have,
a + b = 3 =
b
a
−
ab = -1 =
c
a
(1 Mark)

c = -a and b = -3a
If a = 1, then c = -1 and b = -3
So, one quadratic polynomial which satisfies the given condition is x2- 3x -1.
So, any quadratic equation of the form k(x2- 3x -1), where k is a real
number, will satisfy the above condition. (1 Mark)
Ans 4. We have,
p(x) = 2x2 + 3x + 1
q(x) = 2x-1
r(x) = 3
From division algorithm we know that,
If p(x) and g(x) are the two polynomials with g(x) ¹ 0, we can find
polynomial q(x) and r(x) such that
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2
p(x) = g(x) x q(x) + r(x) (1 Mark)
On substitution, we get
2x2 + 3x + 1 = g (x) x (2x -1) + 3

2x2 + 3x – 2 = g(x) x (2x-1)

2x2 + 4x – x – 2 = g(x) x (2x-1)

2x (x + 2) -1 (x+ 2) = g(x) x (2x-1)

(2x-1) ( x+ 2) = g(x) x (2x-1)

g(x) = x + 2 (1 Mark)
Ans 5. We have,
x2 − 5 = (x − 5) (x + 5)
[Using identity, 2 2 a − b = (a − b)(a + b) ]
x2-5 =0
x = 5 ,- 5
Therefore, the zeroes of the polynomial 2 x − 5 are 5 and - 5 (1 Mark)
Now,
Sum of zeroes = 5 - 5 = 0 = 2
(coefficient of x)
coefficient of x
−
(1 Mark)
Product of zeroes = 5 x 5 = 5 = 2
cons tant term
coefficient of x
(1 Mark)
Hence the relationship of the coefficients and zeroes of polynomial is verified.
Ans6. We have,
q(x) = x-1
p(x) = 3 x − 2x + 1
1
Mark
2
 
 
 
Dividing p(x) by q (x), we get
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3
2
3
3 2
2
2
x x 1
x 1 x 2x 1
x x
( ) ( )
x 2x 1
x x
( ) ( )
x 1
x 1
( ) ( )
0
+ −
− − +
−
− +
− +
−
− +
− +
− +
+ −
1
1 Marks
2
 
 
 
Here we get, Remainder = 0 (1 Mark)
Therefore, q (x) = x-1 is a factor of the polynomial p(x) = x3 − 2x + 1
Ans7.
We have polynomial p(x) = 2 3x − 5x + 2 and zeroes are a – b and a + b.
We know that,
Sum of zeroes = 2
(coefficient of x)
coefficient of x
−
So, a – b + a + b =
5
3
2a =
5
3

a =
5
6
1
1 Marks
2
 
 
 
Product of zeroes = 2
cons tant term
coefficient of x
So, (a – b) (a + b) =
2
3
a2 – b2 =
2
3
Substituting the value of a in eqn (2)
b2 =
25
36
-
2
3
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4
=
25 24
36
−
=
1
36
1
1 Marks
2
 
 
 
b =
1
6
Therefore, a =
5
6
and b =
1
6
Ans8.
Since the two zeroes are
5
3
and -
5
3
, (x +
5
3
)(x -
5
3
)= 2 5
x
3
− is a factor
of the given polynomial. (1 Mark)
Now, we will find the other two polynomials by long division.
2
2 4 3 2
4 2
3 2
3
2
2
3x 6x 3
5
x 3x 6x 2x 10x 5
3
3x 5x
( ) ( )
6x 3x 10x 5
6x 10x
( ) ( )
3x 5
3x 5
( ) ( )
0
+ +
− + − − −
−
− +
+ − −
−
− +
−
−
− +
1
1 Marks
2
 
 
 
So, 3x4 + 6x3 − 2x2 − 10x − 5 = ( 2 5
x
3
− ) ( 2 3x + 6x + 3)
1
Mark
2
 
 
 
Factorising 2 3x + 6x + 3 gives,
2 3x + 6x + 3= 2 3x + 3x + 3x + 3
= 3x (x+1) + 3(x+1)
1
1 Marks
2
 
 
 
= (3x+3) (x+1)
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5
So, its zeroes are -1 and -3.
1
Mark
2
 
 
 
Therefore, zeroes of the polynomial 3x4 + 6x3 − 2x2 − 10x − 5 are
5
3
,-
5
3
, -
1 and -3. (1 Mark)