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) 4x2 - 3x + 7      (ii) y2 + √2        (iii) 3√t + t√2           (iv) y + 2/y              (v) x10 + y3 + t50

Answer:

(i) 4x2 - 3x + 7

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

(ii) y2 + √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) x10 + y3 + t50

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 x2 in each of the following:

(i) 2 + x2 + x               (ii) 2 - x2 + x3              (iii) π/2 * x2 + x               (iv) √2x - 1

Answer:

(i) 2 + x2 + x = 2 + 1 * x2 + x               

So, the coefficient of x2 is 1

(ii) 2 - x2 + x3 = 2 + (-1)x2 + x3            

So, the coefficient of x2 is -1

(iii) π/2 * x2 + x               

So, the coefficient of x2 is π/2

(iv) √2x - 1 = √2x + 0 * x2 - 1

 So, the coefficient of x2 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 = x35 + 7

A binomial of degree 100 = 2x100

Question 4:

Write the degree of each of the following polynomials:

(i) 5x3 + 4x2 + 7x                 (ii) 4 - y2                (iii) 5t - √7               (iv) 3

Answer:

(i) The degree of 5x3 + 4x2 + 7x is 3                 

(ii) The degree of 4 - y2 is 2              

(iii) The degree of 5t - √7 = 5t1 - √7 is 1               

(iv) The degree of 3 = 3x0 is 0

 

 Question 5:

Classify the following as linear, quadratic and cubic polynomials:

(i) x2 + x     (ii) x - x3        (iii) y + y2 + 4            (iv) 1 + x       (v) 3t         (vi) r2           (vii) 7x3

Answer:

(i) x2 + x is a quadratic polynomial.    

(ii) x - x3 is a cubic polynomial.      

(iii) y + y2 + 4 is a quadratic polynomial.            

(iv) 1 + x is a linear polynomial.     

(v) 3t is a linear polynomial.       

(vi) r2 is a quadratic polynomial.         

(vii) 7x3 is a cubic polynomial.

 

                                                    Exercise 2.2

Question 1:

Find the value of the polynomial 5x - 4x2 + 3 at

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

Answer:

Let p(x) = 5x - 4x2 + 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 * 22 + 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) = y2 – y + 1                        (ii) p(t) = 2 + t + 2t2 – t3

 (iii) p(x) = x3                                  (iv) p(x) = (x – 1) (x + 1)

Answer.

(i)   p(y) = y2 – 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 + 2t2 – t3

        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) = x3

         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) = x2 – 1, x = 1, –1                                     

(iv) p(x) = (x + 1) (x – 2), x = – 1, 2      (v) p(x) = x2, x = 0                         (vi) p(x) = lx + m, x = -m/l

(vii) p(x) = 3x2 - 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) = 12 – 1 = 1 – 1 = 0

Hence, x = 1 is a zero of the polynomial p(x) = x2 – 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) = x2 – 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) = x2, x = 0

Put x = 0, we get

p(0) = 02 = 0

Hence, x = 0 is a zero of the polynomial p(x) = x2                          

(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 3x2 – 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 3x2 – 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 x3 + 3x2 + 3x + 1 is divided by

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

Answer:

Let p(x) = x3 + 3x2 + 3x + 1

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

x = -1

Using remainder theorem, when p(x) = x3 + 3x2 + 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) = x3 + 3x2 + 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) = x3 + 3x2 + 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) = x3 + 3x2 + 3x + 1 is divided by x + π, remainder is given

by p(-π).

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

                    = -π3 + 3 π3 - 3 π + 1

Hence, the remainder is -π3 + 3 π3 - 3 π + 1.                  

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

     2x = -5

=> x = -5/2

Using remainder theorem, when p(x) = x3 + 3x2 + 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 x3 – ax2 + 6x – a is divided by x – a.

Answer:

Let p(x) = x3 – ax2 + 6x – a

Put x – a = 0, we get

x = a

Using remainder theorem, when p(x) = x3 – ax2 + 6x – a is divided by x - a, remainder is given

by p(a).

Now, p(a) = a3 – a * a2 + 6 * a – a

                   = a3 – a3 + 6a – a

                   = 5a

Hence, the remainder is 5a.  

Question 3:

Check whether 7 + 3x is a factor of 3x2 + 7x

Answer:

Given, 3x2 + 7x = 3 * x * x + 7 * x

                            = x(3 * x + 7)

                            = x(3x + 7)

                            = x(7 + 3x)

Hence, 7 + 3x is a factor of 3x2 + 7x

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