POLYNOMIALS (Exercise 2.1 and examples)

POLYNOMIALS

NOTE:

1. If p(x) is a polynomial in x, the highest power of x in p(x) is called  the degree of the polynomial p(x).

2. A polynomial of degree 1 is called a linear polynomial. Eg. 2x - 3

3. A polynomial of degree 2 is called a quadratic polynomial. 
    Eg  2x²-3x+4

4. A polynomial of degree 3 is called a cubic polynomial. 
    Eg. 5x³-4x²+x -Ö2

5. If p(x) is a polynomial in x, and if k is any real number, then the value obtained by replacing x by k in p(x), is called the value of p(x) at x = k and is denoted by p(k)

6. A real number k is said to be a zero of a polynomial p(x), if p(k) = 0.

How to find the zeroes of a linear polynomial:

If k is a zero of p(x) = 2x + 3, then p(k) = 0 gives us,

2k + 3  = 0

k = -3

       2

In general, if k is a zero of p(x) = ax + b,

then p(k) = ak + b = 0, i.e., k = -b

                                                a

So, the zero of the linear polynomial  ax + b is  -b  =  - (Constant term)

                                                                      a        Coefficient of x

Geometrical meaning of zeroes of a polynomial :

Consider a linear polynomial ax + b, a ≠ 0. This graph is a straight line.

Eg. y = 2x + 3 is a straight line passing through the points (-2,-1) and (2,7)

i.e.,

x	-2	2
y = 2x + 3	-1	7

Fig 2.1

From Fig. 2.1, you can see that the graph of y = 2x + 3 intersects the  
x-axis mid-way between x = –1 and x = – 2, 
that is, at the point (-3, 0) .

                               2

You also know that the zero of  2x + 3 is - 3

                                                                2

 Thus, the zero of  the polynomial 2x + 3 is the x-coordinate of the point where the graph of y = 2x + 3 intersects the x-axis.

In general, for a linear polynomial ax + b, a ≠ 0, the graph of 
 y = ax + b is a straight line which intersects the  x-axis at exactly one 
point, namely (- b , 0)

                        a

 Therefore, the linear polynomial ax + b, a ≠ 0, has exactly one zero, namely, the  x-coordinate of the point where the graph of  y = ax + b intersects the x-axis.

Note : In general, given a polynomial p(x) of degree n, the graph of 
y = p(x) intersects the x- axis at most n points. Therefore, a polynomial
p(x) of degree n has at most n zeroes.

EXERCISE 2.1

1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Fig. 2.10

(i)

number of zeroes  = No Zeroes  

(ii)

number of zeroes = 1

(iii)

 number of zeroes = 3

(iv)


 number of zeroes = 2
(v)


 number of zeroes = 4
(vi)


 number of zeroes = 3

Example 2 : Find the zeroes of the quadratic polynomial x² + 7x + 10, and verify the relationship between the zeroes and the coefficients.

Solution : We have,

x² + 7x + 10 = (x + 2)(x + 5)

The value of x² + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0,

 i.e., when x = – 2 or x = –5.

\the zeroes of x² + 7x + 10 are – 2 and – 5.

           sum of zeroes = - 2  + (-5) = -7 = -7  =  - (Coefficient of x)

                                                               1          Coefficient of x²

          product of zeroes =  -2  x -5  = 10  = 10  =   Constant term

                                                                   1       Coefficient of x²

Example 3 : Find the zeroes of the polynomial x² – 3 and verify the relationship

between the zeroes and the coefficients.

Solution :  x² – 3  can be written as,

x² – 3 =  (x −Ö 3 )(x + Ö3 )  [a² – b² = (a – b)(a + b)]

i.e., the value of x² – 3 is zero when  x = Ö3 or x = – Ö3

\the zeroes of x² – 3 are Ö3 and − Ö3

sum of zeroes =  Ö3 - Ö3 = 0 =  - (Coefficient of x)

                                                  Coefficient of x²

product of zeroes = (Ö3) (- Ö3) = -3  = Constant term

                                                   1     Coefficient of x²

Example 4 : Find a quadratic polynomial, the sum and product of whose zeroes are    – 3 and 2, respectively.

Solution : Let the quadratic polynomial be ax² + bx + c, and its zeroes be a and b.

a + b = – 3 = - b

                       a

and a b = 2 =  c

                      a

From above,   a = 1, b = 3 and c = 2.

So, one quadratic polynomial which fits the given conditions is 
x² + 3x + 2. You can check that any other quadratic polynomial that fits these conditions will be of the form k (x² + 3x + 2), where k is real.

Example 5 : Verify that 3, –1,- 1   are the zeroes of the cubic

                                                3

polynomial  p(x) = 3x³ – 5x² – 11x – 3, and then verify the relationship between the zeroes and the  coefficients.

Solution :  Given polynomial is  p(x) = 3x³ – 5x² – 11x – 3

Comparing it with ax³ + bx² + cx + d we get,

a = 3, b = – 5, c = –11, d = – 3

Also,  p(3) = (3 × 3³) – (5 × 3²) – (11 × 3) – 3 = 81 – 45 – 33 – 3 = 0,

p(–1) = 3 × (–1)³ – 5 × (–1)² – 11 × (–1) – 3 = –3 – 5 + 11 – 3 = 0,

and p( - 1 )  =  3 ( -1 )3 - 5 x ( -1 )2 - 11 ( -1 )  - 3

             3              3                 3               3

                   =  -1  -  5 + 11  - 3  =  - 2  +  2  =  0

                         9     9     3               3      3

\ 3, –1 and - 1  are the zeroes of                

                     3

the polynomial  3x³ – 5x² – 11x – 3

So, we take a = 3, b = –1 and γ = - 1

                                                      3

a + b + γ = 3 - 1 - 1  =  5   =  - (-5) =  -b

                              3     3            3       a

        

ab + bγ + γa = 3 ( -1)  + (-1) (-1 )  + (-1) ( 3)
                                                3          3      

                      =  -3  + 1  - 1   =  -11  =  - c

                                  3               3         a

abγ  =  3  x  -1  x  -1  =  1  =  - ( -3 )  = - d

                              3                    3         a