Polynomials

 Exercise 2.1
ANS: 1
(i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.
(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.
(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.
(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.
(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.
(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

Exercise 2.2
ANS: 1

The value of is zero when x − 4 = 0 or x + 2 = 0, i.e., when x= 4 or x = −2
Therefore, the zeroes of are 4 and −2.
Sum of zeroes =
Product of zeroes

The value of 4s² − 4s + 1 is zero when 2s − 1 = 0, i.e.,
Therefore, the zeroes of 4s² − 4s + 1 areand.
Sum of zeroes =
Product of zeroes

The value of 6x² − 3 − 7x is zero when 3x + 1 = 0 or 2x − 3 = 0, i.e., or
Therefore, the zeroes of 6x² − 3 − 7xare.
Sum of zeroes =
Product of zeroes =

The value of 4u² + 8u is zero when 4u= 0 or u + 2 = 0, i.e., u = 0 or u = −2
Therefore, the zeroes of 4u² + 8u are 0 and−2.
Sum of zeroes =
Product of zeroes =

The value of t² − 15 is zero when or , i.e., when
Therefore, the zeroes of t² − 15 are and.
Sum of zeroes =
Product of zeroes =

The value of 3x² − x − 4 is zero when 3x − 4 = 0 or x + 1 = 0, i.e., when or x = −1
Therefore, the zeroes of 3x² − x −4 are and−1.
Sum of zeroes =
Product of zeroes
ANS: 2

Let the polynomial be , and its zeroes beand .

Therefore, the quadratic polynomial is 4x² −x − 4.

Let the polynomial be , and its zeroes beand .

Therefore, the quadratic polynomial is 3x² −x+ 1.

Let the polynomial be , and its zeroes beand .

Therefore, the quadratic polynomial is .

Let the polynomial be , and its zeroes beand .

Therefore, the quadratic polynomial is .

Let the polynomial be , and its zeroes beand .

Therefore, the quadratic polynomial is .

Let the polynomial be .

Therefore, the quadratic polynomial is.

Exercise 2.3
ANS: 1


Quotient = x − 3
Remainder = 7x − 9


Quotient = x² + x − 3
Remainder = 8


Quotient = −x² − 2
Remainder = −5x +10

ANS: 2

=

Since the remainder is 0,
Hence, is a factor of .


Since the remainder is 0,
Hence, is a factor of .


Since the remainder ,
Hence, is not a factor of .

ANS: 3

Since the two zeroes are ,
is a factor of.
Therefore, we divide the given polynomial by .

We factorize

Therefore, its zero is given by x + 1 = 0
x = −1
As it has the term , therefore, there will be 2 zeroes at x = −1.
Hence, the zeroes of the given polynomial are,−1 and −1.

ANS: 4

g(x) = ? (Divisor)
Quotient = (x −2)
Remainder = (− 2x + 4)
Dividend = Divisor × Quotient + Remainder

g(x) is the quotient when we divideby



ANS: 5
According to the division algorithm, if p(x) and g(x) are two polynomials with
g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that
p(x) = g(x) × q(x) + r(x),
wherer(x) = 0 or degree of r(x) < degree of g(x)
Degree of a polynomial is the highest power of the variable in the polynomial.
(i) degp(x) = deg q(x)
Degree of quotient will be equal to degree of dividend when divisor is constant ( i.e., when any polynomial is divided by a constant).
Let us assume the division of by 2.
Here, p(x) =
g(x) = 2
q(x) = and r(x) = 0
Degree of p(x) and q(x) is the same i.e., 2.
Checking for division algorithm,
p(x) = g(x) × q(x) +r(x)
= 2()
=
Thus, the division algorithm is satisfied.
(ii) degq(x) = deg r(x)
Let us assume the division of x³+ x by x²,
Here, p(x) = x³ + x
g(x) = x²
q(x) = x and r(x) = x
Clearly, the degree of q(x) and r(x) is the same i.e., 1.
Checking for division algorithm,
p(x) = g(x) × q(x) +r(x)
x³ + x = (x² ) × x+ x
x³ + x = x³ + x
Thus, the division algorithm is satisfied.
(iii)degr(x) = 0
Degree of remainder will be 0 when remainder comes to a constant.
Let us assume the division of x³+ 1by x².
Here, p(x) = x³ + 1
g(x) = x²
q(x) = x and r(x) = 1
Clearly, the degree of r(x) is 0.
Checking for division algorithm,
p(x) = g(x) × q(x) +r(x)
x³ + 1 = (x² ) × x+ 1
x³ + 1 = x³ + 1
Thus, the division algorithm is satisfied.
Exercise 2.4
ANS: 1
(i)


Therefore, , 1, and −2 are the zeroes of the given polynomial.
Comparing the given polynomial with , we obtain a = 2, b = 1, c = −5, d = 2

Therefore, the relationship between the zeroes and the coefficients is verified.
(ii)


Therefore, 2, 1, 1 are the zeroes of the given polynomial.
Comparing the given polynomial with , we obtain a = 1, b = −4, c = 5, d =−2.
Verification of the relationship between zeroes and coefficient of the given polynomial

Multiplication of zeroes taking two at a time = (2)(1) + (1)(1) + (2)(1) =2 + 1 + 2 = 5
Multiplication of zeroes = 2 × 1 × 1 = 2
Hence, the relationship between the zeroes and the coefficients is verified.

ANS: 2
Let the polynomial be and the zeroes be .
It is given that

If a= 1, then b = −2, c = −7, d = 14
Hence, the polynomial is .
ANS: 3

Zeroes area − b, a + a + b
Comparing the given polynomial with , we obtain
p = 1, q = −3, r = 1, t = 1

The zeroes are .

Hence, a= 1 and b = or .

ANS: 4
Given that 2 +and 2­­are zeroes of the given polynomial.
Therefore,=x² + 4 ­­− 4x − 3
= x² ­− 4x + 1 is a factor of the given polynomial
For finding the remaining zeroes of the given polynomial, we will find the quotient by dividing by x² ­− 4x + 1.

Clearly,=
It can be observed that is also a factor of the given polynomial.
And =
Therefore, the value of the polynomial is also zero when or
Or x= 7 or −5
Hence, 7 and −5 are also zeroes of this polynomial.

ANS: 5
By division algorithm,
Dividend = Divisor × Quotient + Remainder
Dividend −Remainder = Divisor × Quotient
will be perfectly divisible by .
Let us divide by

It can be observed thatwill be 0.
Therefore, = 0 and = 0
For = 0,
2 k =10
And thus, k = 5
For = 0
10 − a − 8 × 5 + 25 = 0
10 − a − 40 + 25 = 0
− 5 − a = 0
Therefore, a = −5
Hence, k = 5 and a = −5