Joy Of Mathematics Class X (CBSE): 2014

Monday, February 24, 2014

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

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

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

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) .

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

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)

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

and a b = 2 = c

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

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

the polynomial 3x³ – 5x² – 11x – 3

So, we take a = 3, b = –1 and γ = - 1

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

Wednesday, February 19, 2014

REAL NUMBERS (CLASS X - CBSE) Exercise 1.4

Previous Index

Exercise 1.4

1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating decimal expansion.

(i) 13

3125

Solution : q = 3125 = 5 x 5 x 5 x 5 x 5 = 5⁵

Therefore, denominator is of the form 2ⁿ5^m, where m = 5 and n = 0.

It means rational number 13 has a terminating decimal expansion.

3125

(ii) 17

Solution : q = 8 = 2 x 2 x 2 = 2³

Therefore, denominator is of the form 2ⁿ5^m, where m = 0 and n = 3.

It means rational number 17 has a terminating decimal expansion.

(iii) 64

455

Solution : q = 455 = 5 x 91

Therefore, denominator is not of the form 2ⁿ5^m. It means rational number 64 has 455

a non-terminating repeating decimal expansion.

(iv) 15 = 3

1600 320

Solution : q = 320 = 2 x 2 x 2 x 2 x 2 x 2 x 5

Therefore, denominator is of the form 2ⁿ5^m, where m = 1 and n = 6.

It means rational number 15 has a terminating decimal expansion.

1600

(v) 29

343

Solution : q = 343 = 7 x 7 x 7

Therefore, denominator is not of the form 2ⁿ5^m. It means rational number 29

343

has non-terminating repeating decimal expansion.

(vi) 23

2³5²

Solution : q = 2³ x 5²

Therefore, denominator is of the form 2ⁿ5^m, where m = 2 and n = 3.

It means rational number 23 has terminating decimal expansion.

2³5²

(vii) 129

2²5⁷7⁵

Solution : q = 2² x 5⁷ x 7⁵

Therefore, denominator is not of the form 2ⁿ5^m.It means rational number 129 2²5⁷7⁵

has non-terminating repeating decimal expansion.

(viii) 6 = 2

15 5

Solution : q = 5 = 5¹

Therefore, denominator is of the form 2ⁿ5^m, where m = 1 and n = 0.

It means rational number 6 has terminating decimal expansion.

(ix) 35 = 7

50 10

Solution : q = 10 = 2 x 5 = 2¹5¹

Therefore, denominator is of the form 2ⁿ5^m, where m = 1 and n = 1..

It means rational number 35 has terminating decimal expansion.

(x) 77 = 11

210 30

Solution : q = 30 = 5 x 3 x 2

Therefore, denominator is not of the form 2ⁿ5^m.It means rational number 77

210

has non-terminating repeating decimal expansion.

2. Write down the decimal expansions of those rational numbers in question 1 above which have terminating decimal expansions.

(i) 13 = 0.00416

3125

(ii) 17 = 2.125

(iv) 15 = 3 = 0.009375

600 320

(vi) 23 = 0.115

2³5²

(viii) 6 = 2 = 0.4

15 5

(ix) 35 = 7 = 0.7

50 10

3. The following real numbers have decimal expansions as given below.
In each case, decide whether they are rational or not.
If they are rational and are of the form p ,
q
what can you say about the prime factors of q?

(i) 43.123456789

It is a rational number because decimal expansion is terminating
and it can be expressed in p form, where the factors of
q
q are of the form 2ⁿ5^m, where n and m are non-negative integers.

(ii) 0.120120012000120000...

It is irrational as the decimal expansion is neither terminating nor non-terminating repeating.

(iii) 43.̅1̅2̅3̅4̅5̅6̅7̅8̅9̅

It is a rational number because decimal expansion is
non terminating and repeating and it can be expressed in p form,

q
where the factors of q are not of the form 2ⁿ5^m,
where n and m are non-negative integers.

Previous Index

Joy Of Mathematics Class X (CBSE)

Monday, February 24, 2014

CLASS X - CBSE MATHEMATICS INDEX

CHAPTER 2 - POLYNOMIALS (CLASS X - CBSE)

POLYNOMIALS (Exercise 2.1 and examples)

Wednesday, February 19, 2014

REAL NUMBERS (CLASS X - CBSE) Exercise 1.4