## Polynomials

## Exercise 2.1

### Q.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.

#### (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.

Maths NCERT solutions for Class 10 Chapter 2 Polynomials

Tweet

## Exercise 2.2

### Q.1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

#### (i) x2 – 2x – 8

(i) X2 – 2x – 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

(ii) 4s2 – 4s + 1 = (2s-1)2 The value of 4s2 – 4s + 1 is zero when 2s – 1 = 0, i.e., s = ½ Therefore, the zeroes of 4s2 – 4s + 1 are ½ and ½.

Sum of zeroes = ½ + ½ = 1 = -(-4)/4 = -(Coefficient of s)/Coefficient of s2

Product of zeroes = ½ × ½ = ¼ = Constant term/Coefficient of s2.

#### (iii) 6×2 – 3 – 7x

(iii) 6×2 – 3 – 7x

= 6×2 – 7x – 3

= (3x + 1) (2x – 3)

The value of 6×2 – 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 6×2 – 3 – 7x are -1/3 and 3/2.Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of x)/Coefficient of x2

Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of x2.

#### (iv) 4u2 + 8u

(iv) 4u2 + 8u

= 4u2 + 8u + 0

= 4u(u + 2)

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 = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of u)/Coefficient of u2

Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of u2.

#### (v) T2 – 15

(v) T2 – 15

= t2 – 0.t – 15

= (t – √15) (t + √15)

The value of t2 – 15 is zero when t – √15 = 0 or t + √15 = 0, i.e., when t = √15 or t = -√15

Therefore, the zeroes of t2 – 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 u2.

#### (vi) 3×2 – x – 4

(vi) 3×2 – x – 4

= (3x – 4) (x + 1)

The value of 3×2 – 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 3×2 – x – 4 are 4/3 and -1.

Sum of zeroes = 4/3 + (-1) = 1/3 = -(-1)/3 = -(Coefficient of x)/Coefficient of x2

Product of zeroes = 4/3 × (-1) = -4/3 = Constant term/Coefficient of x2.

### Q.2. 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

(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 4×2 – 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

αß = 1/3 = c/a

If a = 3, then b = -3√2, c = 1

Therefore, the quadratic polynomial is 3×2 -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 x2 + √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 x2 – 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 4×2 + 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 x2 – 4x +1.

Maths NCERT solutions for Class 10 Chapter 2 Polynomials

Tweet

## Exercise 2.3

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

#### (i) p(x) = x3 – 3×2 + 5x – 3, g(x) = x2 – 2 Quotient = x-3 and remainder 7x – 9

(i) p(x) = x^{3} – 3x^{2} + 5x – 3, g(x) = x^{2} – 2

Quotient = x-3 and remainder 7x – 9

#### (ii) p(x) = x4 – 3×2 + 4x + 5, g(x) = x2 + 1 – x Quotient = x2 + x – 3 and remainder 8

(ii) p(x) = x^{4} – 3x^{2} + 4x + 5, g(x) = x^{2} + 1 – x

Quotient = x^{2} + x – 3 and remainder 8

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

(iii) p(x) = x^{4} – 5x + 6, g(x) = 2 – x^{2}

Quotient = -x^{2} -2 and remainder -5x +10

### Q. 2 Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

#### (i) t^{2} – 3, 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12

t^{2} – 3 exactly divides 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12 leaving no remainder. Hence, it is a factor of 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12.

#### (ii) x^{2} + 3x + 1, 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

x^{2} + 3x + 1 exactly divides 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2 leaving no remainder. Hence, it is factor of 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2.

#### (iii) x^{3} – 3x + 1, x^{5} – 4x^{3} + x^{2} + 3x + 1

x^{3} – 3x + 1 didn’t divides exactly x^{5} – 4x^{3} + x^{2} + 3x + 1 and leaves 2 as remainder. Hence, it not a factor of x^{5} – 4x^{3} + x^{2} + 3x + 1.

### Q.3. Obtain all other zeroes of 3x⁴ + 6x³– 2x² – 10x – 5, if two of its zeroes are √(5/3) And – √(5/3).

p(x) = 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5

Since the two zeroes are √(5/3) and – √(5/3).

We factorize x^{2} + 2x + 1

= (x + 1)^{2}

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

x = -1

As it has the term (x + 1)^{2} , therefore, there will be 2 zeroes at x = – 1.

Hence, the zeroes of the given polynomial are √(5/3) and – √(5/3), – 1 and – 1.

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

In the given question, Dividend = x3 – 3×2 + x + 2 Quotient = x – 2 Remainder = -2x + 4 Divisor = g(x) We know that, Dividend = Quotient × Divisor + Remainder ⇒ x3 – 3×2 + x + 2 = (x – 2) × g(x) + (-2x + 4)⇒ x3 – 3×2 + x + 2 – (-2x + 4) = (x – 2) × g(x) ⇒ x3 – 3×2 + 3x – 2 = (x – 2) × g(x) ⇒ g(x) = (x3 – 3×2 + 3x – 2)/(x – 2) ∴ g(x) = (x2 – x + 1)

### Q.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)

(i) Let us assume the division of 6x^{2} + 2x + 2 by 2

Here, p(x) = 6x^{2} + 2x + 2

g(x) = 2

q(x) = 3x^{2} + 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, 6x^{2} + 2x + 2 = 2x (3x^{2} + x + 1)

Hence, division algorithm is satisfied.

#### (ii) deg q(x) = deg r(x)

(ii) Let us assume the division of x^{3}+ x by x^{2},

Here, p(x) = x^{3} + x

g(x) = x^{2}

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)

x^{3} + x = (x^{2} ) × x + x

x^{3} + x = x^{3} + x

Thus, the division algorithm is satisfied.

#### (iii) deg r(x) = 0

(iii) Let us assume the division of x^{3}+ 1 by x^{2}.

Here, p(x) = x^{3} + 1

g(x) = x^{2}

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)

x^{3} + 1 = (x^{2} ) × x + 1

x^{3} + 1 = x^{3} + 1

Thus, the division algorithm is satisfied.

Maths NCERT solutions for Class 10 Chapter 2 Polynomials

Tweet