Class 9 - Maths - Polynomials

**Question 1:**

Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.

(i) 4x^{2} - 3x + 7 (ii) y^{2} + √2 (iii) 3√t + t√2 (iv) y + 2/y (v) x^{10} + y^{3} + t^{50}

Answer:

(i) 4x^{2} - 3x + 7

There is only one variable x with whole number power. So, this polynomial is in one variable.

(ii) y^{2} + √2

There is only one variable y with whole number power. So, this polynomial is in one variable.

(iii) 3√t + t√2

There is only one variable t but in 3√t, power of t is 1/2 which is not a whole number. So this is

not a polynomial.

(iv) y + 2/y

There is only one variable y but in 2/y, power of y is -1/2 which is not a whole number. So this

is not a polynomial.

(v) x^{10} + y^{3} + t^{50}

There are three variables x, y and z and these powers are whole number. So this is not a

polynomial.

**Question 2:**

Write the coefficients of x^{2} in each of the following:

(i) 2 + x^{2} + x (ii) 2 - x^{2} + x^{3} (iii) π/2 * x^{2} + x (iv) √2x - 1

Answer:

(i) 2 + x^{2} + x = 2 + 1 * x^{2} + x

So, the coefficient of x^{2} is 1

(ii) 2 - x^{2} + x^{3} = 2 + (-1)x^{2} + x^{3}

So, the coefficient of x^{2} is -1

(iii) π/2 * x^{2} + x

So, the coefficient of x^{2} is π/2

(iv) √2x - 1 = √2x + 0 * x^{2} - 1

So, the coefficient of x^{2} is 0

**Question 3:**

Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Answer:

A binomial of degree 35 = x^{35} + 7

A binomial of degree 100 = 2x^{100}

**Question 4:**

Write the degree of each of the following polynomials:

(i) 5x^{3} + 4x^{2} + 7x (ii) 4 - y^{2} (iii) 5t - √7 (iv) 3

Answer:

(i) The degree of 5x^{3} + 4x^{2} + 7x is 3

(ii) The degree of 4 - y^{2} is 2

(iii) The degree of 5t - √7 = 5t^{1} - √7 is 1

(iv) The degree of 3 = 3x^{0} is 0

** Question 5:**

Classify the following as linear, quadratic and cubic polynomials:

(i) x^{2} + x (ii) x - x^{3} (iii) y + y^{2} + 4 (iv) 1 + x (v) 3t (vi) r^{2} (vii) 7x^{3}

Answer:

(i) x^{2} + x is a quadratic polynomial.

(ii) x - x^{3} is a cubic polynomial.

(iii) y + y^{2} + 4 is a quadratic polynomial.

(iv) 1 + x is a linear polynomial.

(v) 3t is a linear polynomial.

(vi) r^{2} is a quadratic polynomial.

(vii) 7x^{3 }is a cubic polynomial.

**Exercise 2.2**

**Question 1:**

Find the value of the polynomial 5x - 4x^{2} + 3 at

(i) x = 0 (ii) x = -1 (iii) x = 2

Answer:

Let p(x) = 5x - 4x^{2} + 3

(i) Put x = 0, we get

p(0) = 5 * 0 – 4 * 0 + 3 = 0 – 0 + 3 = 3

(ii) x = -1

p(-1) = 5 * (-1) - 4 * (-1)^{2} + 3 = -5 – 4 + 3 = -9 + 3 = -6

(iii) x = 2

p(2) = 5 * 2 - 4 * 2^{2} + 3 = 10 - 4 * 4 + 3 = 10 - 16 + 3 = 13 - 16 = -3

**Question 2:**

Find p(0), p(1) and p(2) for each of the following polynomials:

(i) p(y) = y^{2} – y + 1 (ii) p(t) = 2 + t + 2t^{2} – t^{3}

(iii) p(x) = x^{3} (iv) p(x) = (x – 1) (x + 1)

Answer.

(i) p(y) = y^{2} – y + 1

p(0) = (0)^{2} – (0) + 1 = 0 – 0 + 1 = 1

p(1) = (1)^{2} – (1) + 1 = 1 – 1 + 1 = 1

p(2) = (2)^{2} – 2 + 1 = 4 – 2 + 1 =

(ii) p(t) = 2 + t + 2t^{2} – t^{3}

p(0) = 2 + (0) + 2(0)^{2} – (0)^{3}

= 2 + 0 + 0 – 0 = 2

p(1) = 2 + (1) + 2(1)^{2} – (1)^{3}

= 2 + 1 + 2 – 1 = 4

p(2) = 2 + 2 + 2(2)^{2} – (2)^{3}

= 2 + 2 + 8 – 8 = 4

(iii) p(x) = x^{3}

p(0) = (0)^{3} = 0

p(1) = (1)^{3} = 1

p(2) = (2)^{3} = 8

(iv) p(x)= (x – 1)(x + 1)

p(0) = (0 – 1)(0 + 1) = –1 * 1 = –1

p(1) = (1 – 1)(1 + 1) = 0 * 2 = 0

p(2) = (2 – 1)(2 + 1) = 1 * 3 = 3

**Question 3:**

Verify whether the following are zeroes of the polynomial, indicated against them.

(i) p(x) = 3x + 1, x = -1/3 (ii) p(x) = 5x – π, x = 4/5 (iii) p(x) = x^{2} – 1, x = 1, –1

(iv) p(x) = (x + 1) (x – 2), x = – 1, 2 (v) p(x) = x^{2}, x = 0 (vi) p(x) = lx + m, x = -m/l

(vii) p(x) = 3x^{2} - 1, x = -1/√3, 2/√3 (viii) p(x) = 2x + 1, x = 1/2

Answer:

(i) Put x = -1/3, we get

p(-1/3) = 3 * (-1/3) + 1 = -1 + 1 = 0

Hence, x = -1/3 is a zero of the polynomial p(x) = 3x + 1

(ii) put x = 4/5, we get

p(x) = 5 * 4/5 – π = 4 – π

Hence, x = 4/5 is a zero of the polynomial p(x) = 5x – π

(iii) Put x = 1, we get

p(1) = 1^{2} – 1 = 1 – 1 = 0

Hence, x = 1 is a zero of the polynomial p(x) = x^{2} – 1

Put x = -1, we get

p(1) = (-1)^{2} – 1 = 1 – 1 = 0

Hence, x = -1 is a zero of the polynomial p(x) = x^{2} – 1

(iv) Put x = -1, we get

p(-1) = (-1 + 1) (-1 – 2) = 0 * (-3) = 0

Hence, x = -1 is a zero of the polynomial p(x) = (x + 1)(x - 2)

Put x = 2, we get

p(2) = (2 + 1) (2 – 2) = 3 * 0 = 0

Hence, x = 2 is a zero of the polynomial p(x) = (x + 1)(x - 2)

(v) p(x) = x^{2}, x = 0

Put x = 0, we get

p(0) = 0^{2} = 0

Hence, x = 0 is a zero of the polynomial p(x) = x^{2}

(vi) Put x = -m/I, we get

p(-m/l) = l * (-m/l) + m = -m + m = 0

Hence, x = -m/l is a zero of the polynomial p(x) = lx + m

(vii) Put x = -1/√3, we get

p(-1/√3) = 3 * (-1/√3)^{2} – 1 = 3 * 1/3 – 1 = 1 – 1 = 0

Hence, x = -1/√3 is a zero of the polynomial 3x^{2} – 1

Put x = 2/√3, we get

p(2/√3) = 3 * (2/√3)^{2} – 1 = 3 * 4/3 – 1 = 4 – 1 = 3

It in not equal to zero. Hence, x = 2/√3 is not a zero of the polynomial 3x^{2} – 1

(viii) Put x = 1/2, we get

p(1/2) = 2 * (1/2) + 1 = 1 + 1 = 2

It in not equal to zero. Hence, x = 1/2 is not a zero of the polynomial 2x + 1.

**Question 4:**

Find the zero of the polynomial in each of the following cases:

(i) p(x) = x + 5 (ii) p(x) = x – 5 (iii) p(x) = 2x + 5 (iv) p(x) = 3x – 2

(v) p(x) = 3x (vi) p(x) = ax, a ≠ 0 (vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.

Answer:

To find the zero of a polynomial p(x), we have to equate it to zero i.e. p(x) = 0

(i) p(x) = 0

=> x + 5 = 0

=> x = -5

Hence, x = -5 is zero of the given polynomial.

(ii) p(x) = 0

=> x – 5 = 0

=> x = 5

Hence, x = 5 is zero of the given polynomial.

(iii) p(x) = 0

=> 2x + 5 = 0

=> 2x = -5

=> x = -5/2

Hence, x = -5/2 is zero of the given polynomial.

(iv) p(x) = 0

=> 3x – 2 = 0

=> 3x = 2

=> x = 2/3

Hence, x = 2/3 is zero of the given polynomial.

(v) p(x) = 0

=> 3x = 0

=> x = 0/3

=> x = 0

Hence, x = 0 is zero of the given polynomial.

(vi) p(x) = 0

=> ax = 0

=> x = 0/a

=> x = a

Hence, x = -5 is zero of the given polynomial where a ≠ 0.

(vii) p(x) = 0

=> cx + d = 0

=> cx = -d

=> c = -d/c

Hence, x = -d/c is zero of the given polynomial where c ≠ 0.

**Exercise 2.3**

**Question 1:**

Find the remainder when x^{3} + 3x^{2} + 3x + 1 is divided by

(i) x + 1 (ii) x – 1/2 (iii) x (iv) x + π (v) 5 + 2x

Answer:

Let p(x) = x^{3} + 3x^{2} + 3x + 1

(i) put x + 1 = 0, we get

x = -1

Using remainder theorem, when p(x) = x^{3} + 3x^{2} + 3x + 1 is divided by x + 1, remainder is given

by p(-1).

Now, p(-1) = (-1)^{3} + 3 * (-1)^{2} + 3 * (-1) + 1

= -1 + 3 – 3 + 1

= -4 + 4

= 0

Hence, the remainder is 0.

(ii) put x – 1/2 = 0, we get

x = 1/2

Using remainder theorem, when p(x) = x^{3} + 3x^{2} + 3x + 1 is divided by x – 1/2, remainder is given

by p(1/2).

Now, p(1/2) = (1/2)^{3} + 3 * (1/2)^{2} + 3 * (1/2) + 1

= 1/8 + 3 * 1/4 + 3/2 + 1

= 1/8 + 3/4 + 3/2 + 1

= (1 * 1 + 3 * 2 + 3 * 4 + 1 * 8)/8 [LCM(8, 4, 2, 1) = 8]

= (1 + 6 + 12 + 8)/8

= 27/8

Hence, the remainder is 27/8

(iii) put x = 0

Using remainder theorem, when p(x) = x^{3} + 3x^{2} + 3x + 1 is divided by x, remainder is given

by p(0).

Now, p(-1) = (0)^{3} + 3 * (0)^{2} + 3 * (0) + 1

= 0 + 0 + 0 + 1

= 1

Hence, the remainder is 1.

(iv) put x + π = 0

Using remainder theorem, when p(x) = x^{3} + 3x^{2} + 3x + 1 is divided by x + π, remainder is given

by p(-π).

Now, p(-π) = (-π)^{3} + 3 * (-π)^{2} + 3 * (-π) + 1

= -π^{3} + 3 π^{3} - 3 π + 1

Hence, the remainder is -π^{3} + 3 π^{3} - 3 π + 1.

(v) put 5 + 2x = 0, we get

2x = -5

=> x = -5/2

Using remainder theorem, when p(x) = x^{3} + 3x^{2} + 3x + 1 is divided by 5 + 2x, remainder is given

by p(-5/2).

Now, p(-5/2) = (-5/2)^{3} + 3 * (-5/2)^{2} + 3 * (-5/2) + 1

= -125/8 + 3 * 25/4 - 15/2 + 1

= -125/8 + 75/4 - 15/2 + 1

= (-1 * 125 + 75 * 2 - 15 * 4 + 1 * 8)/8 [LCM(8, 4, 2, 1) = 8]

= (-125 + 150 - 60 + 8)/8

= -27/8

Hence, the remainder is 27/8

**Question 2:**

Find the remainder when x^{3} – ax^{2} + 6x – a is divided by x – a.

Answer:

Let p(x) = x^{3} – ax^{2} + 6x – a

Put x – a = 0, we get

x = a

Using remainder theorem, when p(x) = x^{3} – ax^{2} + 6x – a is divided by x - a, remainder is given

by p(a).

Now, p(a) = a^{3} – a * a^{2} + 6 * a – a

= a^{3} – a^{3} + 6a – a

= 5a

Hence, the remainder is 5a.

**Question 3:**

Check whether 7 + 3x is a factor of 3x^{2} + 7x

Answer:

Given, 3x^{2} + 7x = 3 * x * x + 7 * x

= x(3 * x + 7)

= x(3x + 7)

= x(7 + 3x)

Hence, 7 + 3x is a factor of 3x^{2} + 7x

.