# Xpert Solutions For Class X Math Chapter 2: Polynomials

Excercise 2.1

1. The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Sol: (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

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x²– 2x– 8
(ii) 4s² – 4s + 1
(iii) 6x² – 3 – 7x
(iv) 4u² + 8u
(v) t² – 15
(vi) 3x² – x – 4

Sol:

(i) x² – 2x – 8

To make factors given below the method:
The Product of  Factors should be: 1 × -8  = -8   (Coefficient of  X²  × constant value)
And   The sum of factors using above value should be   -2 (Coefficient of X) 2, 2, 2       (Therefore    -4 + 2 = -2,   -4 × 2 = -8)
= (x – 4) (x + 2)
The value of x² – 2x – 8 is zero when x – 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = -2
Therefore, the zeroes of x² – 2x – 8 are 4 and -2.
Sum of zeroes = 4 + (-2) = 2 = -(-2)/1 = -(Coefficient of x)/Coefficient of x²
Product of zeroes = 4 × (-2) = -8 = -8/1 = Constant term/Coefficient of x²

(ii) 4s² – 4s + 1
= (2s-1)²                  (Using Formula Here:  (a-b)² =  a² – 2ab + b² )

The value of 4s² – 4s + 1 is zero when 2s – 1 = 0, i.e., s = 1/2

Therefore, the zeroes of 4s² – 4s + 1 are 1/2 and 1/2.
Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -(Coefficient ofs)/Coefficient of s²
Product of zeroes = 1/2 × 1/2 = 1/4 = Constant term/Coefficient of s² .

(iii) 6 – 3 – 7x
6 – 7– 3
= 6 +2x -9x – 3
[ i.e 6 × -3 = -18, L.C.M = 2 × 3 × 3, Product of -9 and 2 is -18 & Sum is -7]
= 2x(3x+1 ) – 3x(3x+1)
= (3x + 1) (2x – 3)
The value of 6 – 3 – 7x is zero when 3x + 1 = 0 or 2x – 3 = 0, i.e.,x = -1/3 or x = 3/2

Therefore, the zeroes of 6 – 3 – 7x are -1/3 and 3/2.
Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of x)/Coefficient of
Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of

(iv) 4 + 8u
4 + 8
= 4u(u + 2)
The value of 4 + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = – 2
Therefore, the zeroes of 4 + 8u are 0 and – 2.
Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of u)/Coefficient of
Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of .

(v)  – 15
– 0.t – 15
= (– √15) (t + √15)
The value of  – 15 is zero when t – √15 = 0 or t + √15 = 0, i.e., when t = √15 or = -√15
Therefore, the zeroes of  – 15 are √15 and -√15.Sum of zeroes = √15 + -√15 = 0 = -0/1 = -(Coefficient of t)/Coefficient of t2
Product of zeroes = (√15) (-√15) = -15 = -15/1 = Constant term/Coefficient of .

(vi) 3 – x – 4
= (3x – 4) (x + 1)
The value of 3 – x – 4 is zero when 3x – 4 = 0 and x + 1 = 0,i.e., when x = 4/3 or x = -1
Therefore, the zeroes of 3 – x – 4 are 4/3 and -1.

Sum of zeroes = 4/3 + (-1) = 1/3 = -(-1)/3 = -(Coefficient of x)/Coefficient of

Product of zeroes = 4/3 × (-1) = -4/3 = Constant term/Coefficient of .

1. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) 1/4 , -1
(ii) √2 , 1/3
(iii) 0, √5
(iv) 1,1
(v) -1/4 ,1/4
(vi) 4,1
Sol:
(i) 1/4 , -1
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 1/4 = –b/a
αß = -1 = -4/4 = c/a
If a = 4, then b = -1, c = -4
Therefore, the quadratic polynomial is 4 – x -4.

(ii) √2 , 1/3
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = √2 = -3√2/3 = –b/a  [multiply value  by 3/3 to equal value of a with αß  ]
αß = 1/3 = c/a
If a = 3, then b = -3√2, c = 1
Therefore, the quadratic polynomial is 3 -3√2x +1.
or

k { x² – (sum of zeros)x + Product of zeroes }
k { x² – (√2)x + 1/3 }
k{ 3x² – 3√2x + 1 }

(iii) 0, √5
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 0 = 0/1 = –b/a
αß = √5 = √5/1 = c/a
If a = 1, then b = 0, c = √5
Therefore, the quadratic polynomial is  + √5.

(iv) 1, 1
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 1 = 1/1 = –b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -1, c = 1
Therefore, the quadratic polynomial is  – x +1.

(v) -1/4 ,1/4
Let the polynomial be ax² bx + c, and its zeroes be α and ß
α + ß = -1/4 = –b/a
αß = 1/4 = c/a
If a = 4, then b = 1, c = 1
Therefore, the quadratic polynomial is 4 + x +1.

(vi) 4,1
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 4 = 4/1 = –b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -4, c = 1
Therefore, the quadratic polynomial is  – 4x +1.

Exercise 2.3

1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

Sol
(i) p(x) =  – 3x² + 5x – 3, g(x) = x² – 2 Quotient = x-3 and remainder 7x – 9

(ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x Quotient = x² + – 3 and remainder 8

(iii) p(x) = x⁴ – 5x + 6, g(x) = 2 – x² Quotient = –x² -2 and remainder -5x +10

1. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

Sol

(i) t² – 3,  2t⁴ + 3 – 2t² – 9t – 12

t² – 3 exactly divides 2t⁴ + 3 – 2t² – 9t – 12 leaving no remainder. Hence, it is a factor of 2t⁴ + 3 – 2t² – 9t – 12.

(ii) x² + 3x + 1, 3x⁴ + 5 – 7x² + 2x + 2 x² + 3x + 1 exactly divides 3x⁴ + 5 – 7x² + 2x + 2 leaving no remainder. Hence, it is factor of 3x⁴ + 5 – 7x² + 2x + 2.

(iii)  – 3x + 1, x⁵ – 4 + x² + 3x + 1 – 3x + 1 didn’t divides exactly  x⁵ – 4 + x² + 3x + 1 and leaves 2 as remainder. Hence, it not a factor of  x⁵ – 4 + x² + 3x + 1

1. Obtain all other zeroes of 3x+ 6x³– 2x² – 10x – 5, if two of its zeroes are √(5/3) and – √(5/3).

Sol:

p(x) = 3x+ 6x³– 2x² – 10x – 5
Since the two zeroes are √(5/3) and – √(5/3). We factorize x² + 2+ 1 = (+ 1)²     [ From the other equ ( x²-5/3) factor are  √(5/3), – √(5/3) already given]

Therefore, its zero is given by x + 1 = 0   or   x = -1

As it has the term (+ 1)² , therefore, there will be 2 zeroes at x = – 1.
Hence,the zeroes of the given polynomial are √(5/3), – √(5/3), – 1 and – 1.

1.  On dividing x³– 3x²x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4, respectively. Find g(x).

Sol:

Here in the given question,

Dividend = x³– 3x²x + 2

Quotient = x – 2

Remainder = -2x + 4

Divisor = g(x)

We know that,

Dividend = Quotient × Divisor + Remainder

⇒ x³– 3x²x + 2 = (x – 2) × g(x) + (-2x + 4)
⇒ x³– 3x²x + 2 – (-2x + 4) = (x – 2) × g(x)
⇒ x3 – 3x2 + 3x – 2 = (x – 2) × g(x)
⇒ g(x) =  (x3 – 3x2 + 3x – 2)/(x – 2) ∴ g(x) = (x2 – x + 1)

5.Give examples of polynomial p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0

Sol:

(i) Let us assume the division of 6x² + 2x + 2 by 2
Here, p(x) = 6x² + 2x + 2
g(x) = 2
q(x) = 3x² + x + 1
r(x) = 0
Degree of p(x) and q(x) is same i.e. 2.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
Or, 6x² + 2x + 2 = 2x (3x² + + 1)
Hence, division algorithm is satisfied.
(ii) Let us assume the division of x³x 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) Let us assume the division of x³+ 1 by 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 ) × + 1
x³ + 1 = x³ + 1
Thus, the division algorithm is satisfied. 