class-X Maths Book | Chapter-2 | Slides By MANAV |

20 MATHEMATICS
2
2.1 Introduction
In Class IX, you have studied polynomials in one variable and their degrees. Recall
that 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). For example, 4x + 2 is a polynomial in the variable x of
degree 1, 2y2
– 3y + 4 is a polynomial in the variable y of degree 2, 5x3
– 4x2
+ x – 2
is a polynomial in the variablex of degree 3 and 7u6
–
4 23
4 8
2
u u u+ + − is a polynomial
in the variable u of degree 6. Expressions like
1
1x −
, 2x + , 2
1
2 3x x+ +
etc., are
not polynomials.
A polynomial of degree 1 is called a linear polynomial. For example, 2x – 3,
3 5,x + 2y+ ,
2
11
x − , 3z + 4,
2
1
3
u + , etc., are all linear polynomials. Polynomials
such as 2x + 5 – x2
, x3
+ 1, etc., are not linear polynomials.
A polynomial of degree 2 is called a quadratic polynomial. The name ‘quadratic’
has been derived from the word ‘quadrate’, which means ‘square’. 2 2,2 3
5
x x+ −
y2
– 2, 2
2 3 ,x x− + 2 2 22 1
2 5, 5 , 4
3 3 7
u
u v v z− + − + are some examples of
quadratic polynomials (whose coefficients are real numbers). More generally, any
quadratic polynomial in x is of the form ax2
+ bx + c, where a, b, c are real numbers
and a ≠ 0. A polynomial of degree 3 is called a cubic polynomial. Some examples of
POLYNOMIALS
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POLYNOMIALS 21
a cubic polynomial are 2 – x3
, x3
, 3
2 ,x 3 – x2
+ x3
, 3x3
– 2x2
+ x – 1. In fact, the most
general form of a cubic polynomial is
ax3
+ bx2
+ cx + d,
where, a, b, c, d are real numbers and a ≠ 0.
Now consider the polynomial p(x) = x2
– 3x – 4. Then, putting x = 2 in the
polynomial, we get p(2) = 22
– 3 × 2 – 4 = – 6. The value ‘– 6’, obtained by replacing
x by 2 in x2
– 3x – 4, is the value of x2
– 3x – 4 at x = 2. Similarly, p(0) is the value of
p(x) at x = 0, which is – 4.
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).
What is the value of p(x) = x2
–3x – 4 at x = –1? We have :
p(–1) = (–1)2
–{3 × (–1)} – 4 = 0
Also, note that p(4) = 42
– (3 × 4) – 4 = 0.
As p(–1) = 0 and p(4) = 0, –1 and 4 are called the zeroes of the quadratic
polynomial x2
– 3x – 4. More generally, a real number k is said to be a zero of a
polynomial p(x), if p(k) = 0.
You have already studied in Class IX, how to find the zeroes of a linear
polynomial. For example, if k is a zero of p(x) = 2x + 3, then p(k) = 0 gives us
2k + 3 = 0, i.e., k =
3
2
− ⋅
In general, if k is a zero of p(x) = ax + b, then p(k) = ak + b = 0, i.e.,
b
k
a
−
= ⋅
So, the zero of the linear polynomial ax + b is
(Constant term)
Coefficient of
b
a x
− −
= .
Thus, the zero of a linear polynomial is related to its coefficients. Does this
happen in the case of other polynomials too? For example, are the zeroes of a quadratic
polynomial also related to its coefficients?
In this chapter, we will try to answer these questions. We will also study the
division algorithm for polynomials.
2.2 Geometrical Meaning of the Zeroes of a Polynomial
You know that a real number k is a zero of the polynomial p(x) if p(k) = 0. But why
are the zeroes of a polynomial so important? To answer this, first we will see the
geometrical representations of linear and quadratic polynomials and the geometrical
meaning of their zeroes.
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22 MATHEMATICS
Consider first a linear polynomial ax + b, a ≠ 0. You have studied in Class IX that the
graph of y = ax + b is a straight line. For example, the graph of y = 2x + 3 is a straight
line passing through the points (– 2, –1) and (2, 7).
x –2 2
y = 2x + 3 –1 7
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, , 0
b
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.
Now, let us look for the geometrical meaning of a zero of a quadratic polynomial.
Consider the quadratic polynomial x2
– 3x – 4. Let us see what the graph* of
y = x2
– 3x – 4 looks like. Let us list a few values of y = x2
– 3x – 4 corresponding to
a few values for x as given in Table 2.1.
* Plotting of graphs of quadratic or cubic polynomials is not meant to be done by the students,
nor is to be evaluated.
Fig. 2.1
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POLYNOMIALS 23
Table 2.1
x – 2 –1 0 1 2 3 4 5
y = x2
– 3x – 4 6 0 – 4 – 6 – 6 – 4 0 6
If we locate the points listed
above on a graph paper and draw
the graph, it will actually look like
the one given in Fig. 2.2.
In fact, for any quadratic
polynomial ax2
+ bx + c, a ≠ 0, the
graph of the corresponding
equation y = ax2
+ bx + c has one
of the two shapes either open
upwards like or open
downwards like depending on
whether a > 0 or a < 0. (These
curves are called parabolas.)
You can see from Table 2.1
that –1 and 4 are zeroes of the
quadratic polynomial. Also
note from Fig. 2.2 that –1 and 4
are the x-coordinates of the points
where the graph of y = x2
– 3x – 4
intersects the x-axis. Thus, the
zeroes of the quadratic polynomial
x2
– 3x – 4 are x-coordinates of
the points where the graph of
y = x2
– 3x – 4 intersects the
x-axis.
This fact is true for any quadratic polynomial, i.e., the zeroes of a quadratic
polynomial ax2
+ bx + c, a ≠ 0, are precisely the x-coordinates of the points where the
parabola representing y = ax2
+ bx + c intersects the x-axis.
From our observation earlier about the shape of the graph of y = ax2
+ bx + c, the
following three cases can happen:
Fig. 2.2
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24 MATHEMATICS
Case (i) : Here, the graph cuts x-axis at two distinct points A and A′.
The x-coordinates of A and A′ are the two zeroes of the quadratic polynomial
ax2
+ bx + c in this case (see Fig. 2.3).
Fig. 2.3
Case (ii) : Here, the graph cuts the x-axis at exactly one point, i.e., at two coincident
points. So, the two points A and A′ of Case (i) coincide here to become one point A
(see Fig. 2.4).
Fig. 2.4
The x-coordinate of A is the only zero for the quadratic polynomial ax2
+ bx + c
in this case.
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POLYNOMIALS 25
Case (iii) : Here, the graph is either completely above the x-axis or completely below
the x-axis. So, it does not cut the x-axis at any point (see Fig. 2.5).
Fig. 2.5
So, the quadratic polynomial ax2
+ bx + c has no zero in this case.
So, you can see geometrically that a quadratic polynomial can have either two
distinct zeroes or two equal zeroes (i.e., one zero), or no zero. This also means that a
polynomial of degree 2 has atmost two zeroes.
Now, what do you expect the geometrical meaning of the zeroes of a cubic
polynomial to be? Let us find out. Consider the cubic polynomial x3
– 4x. To see what
the graph of y = x3
– 4x looks like, let us list a few values of y corresponding to a few
values for x as shown in Table 2.2.
Table 2.2
x –2 –1 0 1 2
y = x3
– 4x 0 3 0 –3 0
Locating the points of the table on a graph paper and drawing the graph, we see
that the graph of y = x3
– 4x actually looks like the one given in Fig. 2.6.
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26 MATHEMATICS
We see from the table above
that – 2, 0 and 2 are zeroes of the
cubic polynomial x3
– 4x. Observe
that – 2, 0 and 2 are, in fact, the
x-coordinates of the only points
– 4x
intersects the x-axis. Since the curve
meets the x-axis in only these 3
points, their x-coordinates are the
only zeroes of the polynomial.
Let us take a few more
examples. Consider the cubic
polynomials x3
and x3
– x2
. We draw
the graphs of y = x3
and y = x3
– x2
in Fig. 2.7 and Fig. 2.8 respectively.
Fig. 2.7 Fig. 2.8
Fig. 2.6
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POLYNOMIALS 27
Note that 0 is the only zero of the polynomial x3
. Also, from Fig. 2.7, you can see
that 0 is the x-coordinate of the only point where the graph of y = x3
intersects the
x-axis. Similarly, since x3
– x2
= x2
(x – 1), 0 and 1 are the only zeroes of the polynomial
x3
– x2
. Also, from Fig. 2.8, these values are the x-coordinates of the only points
– x2
intersects the x-axis.
From the examples above, we see that there are at most 3 zeroes for any cubic
polynomial. In other words, any polynomial of degree 3 can have at most three zeroes.
Remark : In general, given a polynomial p(x) of degree n, the graph of y = p(x)
intersects the x-axis at atmost n points. Therefore, a polynomial p(x) of degree n has
at most n zeroes.
Example 1 : Look at the graphs in Fig. 2.9 given below. Each is the graph of y = p(x),
where p(x) is a polynomial. For each of the graphs, find the number of zeroes of p(x).
Fig. 2.9
Solution :
(i) The number of zeroes is 1 as the graph intersects the x-axis at one point only.
(ii) The number of zeroes is 2 as the graph intersects the x-axis at two points.
(iii) The number of zeroes is 3. (Why?)
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28 MATHEMATICS
(iv) The number of zeroes is 1. (Why?)
(v) The number of zeroes is 1. (Why?)
(vi) The number of zeroes is 4. (Why?)
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
2.3 Relationship between Zeroes and Coefficients of a Polynomial
You have already seen that zero of a linear polynomial ax + b is b
a
− . We will now try
to answer the question raised in Section 2.1 regarding the relationship between zeroes
and coefficients of a quadratic polynomial. For this, let us take a quadratic polynomial,
say p(x) = 2x2
– 8x + 6. In Class IX, you have learnt how to factorise quadratic
polynomials by splitting the middle term. So, here we need to split the middle term
‘– 8x’ as a sum of two terms, whose product is 6 × 2x2
= 12x2
. So, we write
2x2
– 8x + 6 = 2x2
– 6x – 2x + 6 = 2x(x – 3) – 2(x – 3)
= (2x – 2)(x – 3) = 2(x – 1)(x – 3)
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POLYNOMIALS 29
So, the value of p(x) = 2x2
– 8x + 6 is zero when x – 1 = 0 or x – 3 = 0, i.e., when
x = 1 or x = 3. So, the zeroes of 2x2
– 8x + 6 are 1 and 3. Observe that :
Sum of its zeroes = 2
( 8) (Coefficient of )
1 3 4
2 Coefficient of
x
x
− − −
+ = = =
Product of its zeroes = 2
6 Constant term
1 3 3
2 Coefficient of x
× = = =
Let us take one more quadratic polynomial, say, p(x) = 3x2
+ 5x – 2. By the
method of splitting the middle term,
3x2
+ 5x – 2 = 3x2
+ 6x – x – 2 = 3x(x + 2) –1(x + 2)
= (3x – 1)(x + 2)
Hence, the value of 3x2
+ 5x – 2 is zero when either 3x – 1 = 0 or x + 2 = 0, i.e.,
when x =
1
3
or x = –2. So, the zeroes of 3x2
+ 5x – 2 are
1
3
and – 2. Observe that :
Sum of its zeroes = 2
1 5 (Coefficient of )
( 2)
3 3 Coefficient of
x
x
− −
+ − = =
Product of its zeroes = 2
1 2 Constant term
( 2)
3 3 Coefficient of x
−
× − = =
In general, if α* and β* are the zeroes of the quadratic polynomial p(x)=ax2
+bx + c,
a ≠ 0, then you know that x – α and x – β are the factors of p(x). Therefore,
ax2
+ bx + c = k(x – α) (x – β), where k is a constant
= k[x2
– (α + β)x + α β]
= kx2
– k(α + β)x + k α β
Comparing the coefficients of x2
, x and constant terms on both the sides, we get
a = k, b = – k(α + β) and c = kαβ.
This gives ααααα + βββββ =
–b
a
,
αβαβαβαβαβ =
c
a
* α,β are Greek letters pronounced as ‘alpha’ and ‘beta’ respectively. We will use later one
more letter ‘γ’ pronounced as ‘gamma’.
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30 MATHEMATICS
i.e., sum of zeroes = α + β = 2
(Coefficient of )
Coefficient of
b x
a x
−
− = ,
product of zeroes = αβ = 2
Constant term
Coefficient of
c
a x
= .
Let us consider some examples.
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)
So, the value of x2
+ 7x + 10 is zero when x + 2 = 0 or x + 5 = 0, i.e., when x = – 2 or
x = –5. Therefore, the zeroes of x2
+ 7x + 10 are – 2 and – 5. Now,
sum of zeroes = 2
(7) –(Coefficient of ) ,–2 (–5) – (7)
1 Coefficient of
x
x
−
+ = = =
product of zeroes = 2
10 Constant term
( 2) ( 5) 10
1 Coefficient of x
− × − = = = ⋅
Example 3 : Find the zeroes of the polynomial x2
– 3 and verify the relationship
between the zeroes and the coefficients.
Solution : Recall the identity a2
– b2
= (a – b)(a + b). Using it, we can write:
x2
– 3 = ( )( )3 3x x− +
So, the value of x2
– 3 is zero when x = 3 or x = – 3⋅
Therefore, the zeroes of x2
– 3 are 3 and 3− ⋅
Now,
sum of zeroes = 2
(Coefficient of ) ,3 3 0
Coefficient of
x
x
−
− = =
product of zeroes = ( )( ) 2
3 Constant term
3 3 – 3
1 Coefficient of x
−
− = = = ⋅
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POLYNOMIALS 31
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 α and β.
We have
α + β = – 3 =
b
a
− ,
and αβ = 2 =
c
a
.
If a = 1, then 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.
Let us now look at cubic polynomials. Do you think a similar relation holds
between the zeroes of a cubic polynomial and its coefficients?
Let us consider p(x) = 2x3
– 5x2
– 14x + 8.
You can check that p(x) = 0 for x = 4, – 2,
1
2
⋅ Since p(x) can have atmost three
zeroes, these are the zeores of 2x3
– 5x2
– 14x + 8. Now,
sum of the zeroes =
2
3
1 5 ( 5) (Coefficient of )
4 ( 2)
2 2 2 Coefficient of
x
x
− − −
+ − + = = = ,
product of the zeroes = 3
1 8 – Constant term
4 ( 2) 4
2 2 Coefficient of x
−
× − × = − = = .
However, there is one more relationship here. Consider the sum of the products
of the zeroes taken two at a time. We have
{ }
1 1
4 ( 2) ( 2) 4
2 2
   
× − + − × + ×   
   
=
14
– 8 1 2 7
2
−
− + = − = = 3
Coefficient of
Coefficient of
x
x
.
In general, it can be proved that if α, β, γ are the zeroes of the cubic polynomial
ax3
+ bx2
+ cx + d, then
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32 MATHEMATICS
ααααα + βββββ + γγγγγ =
–b
a
,
αβαβαβαβαβ + βγβγβγβγβγ + γαγαγαγαγα =
c
a
,
α β γα β γα β γα β γα β γ =
–d
a
.
Let us consider an example.
Example 5* : Verify that 3, –1,
1
3
− 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 : Comparing the given polynomial with ax3
+ bx2
+ cx + d, we get
a = 3, b = – 5, c = –11, d = – 3. Further
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,
3 2
1 1 1 1
3 5 11 3
3 3 3 3
p
       
− = × − − × − − × − −       
       
,
=
1 5 11 2 2
– 3 – 0
9 9 3 3 3
− + − = + =
Therefore, 3, –1 and
1
3
− are the zeroes of 3x3
– 5x2
– 11x – 3.
So, we take α = 3, β = –1 and γ =
1
3
− ⋅
Now,
1 1 5 ( 5)
3 ( 1) 2
3 3 3 3
b
a
− − − 
α + β + γ = + − + − = − = = = 
 
,
1 1 1 11
3 ( 1) ( 1) 3 3 1
3 3 3 3
c
a
−   
αβ+ βγ +γα = × − + − × − + − × = − + − = =   
   
,
1 ( 3)
3 ( 1) 1
3 3
d
a
− − − 
αβγ = × − × − = = = 
 
.
* Not from the examination point of view.
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POLYNOMIALS 33
EXERCISE 2.2
1. Find the zeroes of the following quadratic polynomials and verify the relationship between
the zeroes and the coefficients.
(i) x2
– 2x – 8 (ii) 4s2
– 4s + 1 (iii) 6x2
– 3 – 7x
(iv) 4u2
+ 8u (v) t2
– 15 (vi) 3x2
– x – 4
2. Find a quadratic polynomial each with the given numbers as the sum and product of its
zeroes respectively.
(i)
1 , 1
4
− (ii)
1
2 ,
3
(iii) 0, 5
(iv) 1,1 (v)
1 1,
4 4
− (vi) 4,1
2.4 DivisionAlgorithm forPolynomials
You know that a cubic polynomial has at most three zeroes. However, if you are given
only one zero, can you find the other two? For this, let us consider the cubic polynomial
x3
– 3x2
– x + 3. If we tell you that one of its zeroes is 1, then you know that x – 1 is
a factor of x3
– 3x2
– x + 3. So, you can divide x3
– 3x2
– x + 3 by x – 1, as you have
learnt in Class IX, to get the quotient x2
– 2x – 3.
Next, you could get the factors of x2
– 2x – 3, by splitting the middle term, as
(x + 1)(x – 3). This would give you
x3
– 3x2
– x + 3 = (x – 1)(x2
– 2x – 3)
= (x – 1)(x + 1)(x – 3)
So, all the three zeroes of the cubic polynomial are now known to you as
1, – 1, 3.
Let us discuss the method of dividing one polynomial by another in some detail.
Before noting the steps formally, consider an example.
Example 6 : Divide 2x2
+ 3x + 1 by x + 2.
Solution : Note that we stop the division process when
either the remainder is zero or its degree is less than the
degree of the divisor. So, here the quotient is 2x – 1 and
the remainder is 3. Also,
(2x – 1)(x + 2) + 3 = 2x2
+ 3x – 2 + 3 = 2x2
+ 3x + 1
i.e., 2x2
+ 3x + 1 = (x + 2)(2x – 1) + 3
Therefore, Dividend = Divisor × Quotient + Remainder
Let us now extend this process to divide a polynomial by a quadratic polynomial.
x + 2 2 + 3 + 1x x
2
2 + 4x x2
2 1x –
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34 MATHEMATICS
+ x2
+ 2x + 5 by 1 + 2x + x2
.
Solution : We first arrange the terms of the
dividend and the divisor in the decreasing order
of their degrees. Recall that arranging the terms
in this order is called writing the polynomials in
standard form. In this example, the dividend is
already in standard form, and the divisor, in
standard form, is x2
+ 2x + 1.
Step 1 : To obtain the first term of the quotient, divide the highest degree term of the
dividend (i.e., 3x3
) by the highest degree term of the divisor (i.e., x2
). This is 3x. Then
carry out the division process. What remains is – 5x2
– x + 5.
Step 2 : Now, to obtain the second term of the quotient, divide the highest degree term
of the new dividend (i.e., –5x2
) by the highest degree term of the divisor (i.e., x2
). This
gives –5. Again carry out the division process with –5x2
– x + 5.
Step 3 : What remains is 9x + 10. Now, the degree of 9x + 10 is less than the degree
of the divisor x2
+ 2x + 1. So, we cannot continue the division any further.
So, the quotient is 3x – 5 and the remainder is 9x + 10. Also,
(x2
+ 2x + 1) × (3x – 5) + (9x + 10) = 3x3
+ 6x2
+ 3x – 5x2
– 10x – 5 + 9x + 10
= 3x3
+ x2
+ 2x + 5
Here again, we see that
Dividend = Divisor × Quotient + Remainder
What we are applying here is an algorithm which is similar to Euclid’s division
algorithm that you studied in Chapter 1.
This says that
If p(x) and g(x) are any two polynomials with g(x) ≠≠≠≠≠ 0, then we can find
polynomials q(x) and r(x) such that
p(x) = g(x) × q(x) + r(x),
where r(x) = 0 or degree of r(x) < degree of g(x).
This result is known as the Division Algorithm for polynomials.
Let us now take some examples to illustrate its use.
– x3
– 3x + 5 by x – 1 – x2
, and verify the division algorithm.
x2
+ 2 + 1x
3x – 5
3 + 6x x3
x2
+3
– – –
–5 –x2
x + 5
–5 – 10x2
x – 5
+ + +
9x + 10
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POLYNOMIALS 35
Solution : Note that the given polynomials
are not in standard form. To carry out
division, we first write both the dividend and
divisor in decreasing orders of their degrees.
So, dividend = –x3
+ 3x2
– 3x + 5 and
divisor = –x2
+ x – 1.
Division process is shown on the right side.
We stop here since degree (3) = 0 < 2 = degree (–x2
+ x – 1).
So, quotient = x – 2, remainder = 3.
Now,
Divisor × Quotient + Remainder
= (–x2
+ x – 1) (x – 2) + 3
= –x3
+ x2
– x + 2x2
– 2x + 2 + 3
= –x3
+ 3x2
– 3x + 5
= Dividend
In this way, the division algorithm is verified.
Example 9 : Find all the zeroes of 2x4
– 3x3
– 3x2
+ 6x – 2, if you know that two of
its zeroes are 2 and 2− .
Solution : Since two zeroes are 2 and 2− , ( )( )2 2x x− + = x2
– 2 is a
factor of the given polynomial. Now, we divide the given polynomial by x2
– 2.
First term of quotient is
4
2
2
2
2
x
x
x
=
Second term of quotient is
3
2
3
3
x
x
x
−
= −
Third term of quotient is
2
2
1
x
x
=
2019-20

36 MATHEMATICS
So, 2x4
– 3x3
– 3x2
+ 6x – 2 = (x2
– 2)(2x2
– 3x + 1).
Now, by splitting –3x, we factorise 2x2
– 3x + 1 as (2x – 1)(x – 1). So, its zeroes
are given by x =
1
2
and x = 1. Therefore, the zeroes of the given polynomial are
1 ,2, 2, and 1.
2
−
EXERCISE 2.3
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
– 3x2
+ 5x – 3, g(x) = x2
– 2
(ii) p(x) = x4
– 3x2
+ 4x + 5, g(x) = x2
+ 1 – x
(iii) p(x) = x4
– 5x + 6, g(x) = 2 – x2
2. Check whether the first polynomial is a factor of the second polynomial by dividing the
second polynomial by the first polynomial:
(i) t2
– 3, 2t4
+ 3t3
– 2t2
– 9t – 12
(ii) x2
+ 3x + 1, 3x4
+ 5x3
– 7x2
+ 2x + 2
(iii) x3
– 3x + 1, x5
– 4x3
+ x2
+ 3x + 1
3. Obtain all other zeroes of 3x4
+ 6x3
– 2x2
– 10x – 5, if two of its zeroes are
5 5
and –
3 3
⋅
4. On dividing x3
– 3x2
+ x + 2 by a polynomial g(x), the quotient and remainder were x – 2
and –2x + 4, respectively. Find g(x).
5. Give examples of polynomialsp(x), g(x),q(x) and r(x), which satisfy the division algorithm
and
(i) deg p(x) = deg q(x) (ii) deg q(x) = deg r(x) (iii) deg r(x) = 0
EXERCISE 2.4 (Optional)*
1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes.
Also verify the relationship between the zeroes and the coefficients in each case:
(i) 2x3
+ x2
– 5x + 2;
1 , 1, – 2
2
(ii) x3
– 4x2
+ 5x – 2; 2, 1, 1
2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a
time, and the product of its zeroes as 2, –7, –14 respectively.
*These exercises are not from the examination point of view.
2019-20

POLYNOMIALS 37
3. If the zeroes of the polynomial x3
– 3x2
+ x + 1 are a – b, a, a + b, find a and b.
4. If two zeroes of the polynomial x4
– 6x3
– 26x2
+ 138x – 35 are 2 3,± find other zeroes.
5. If the polynomial x4
– 6x3
+ 16x2
– 25x + 10 is divided by another polynomial x2
– 2x + k,
the remainder comes out to be x + a, find k and a.
2.5 Summary
In this chapter, you have studied the following points:
1. Polynomials of degrees 1, 2 and 3 are called linear, quadratic and cubic polynomials
respectively.
2. A quadratic polynomial in x with real coefficients is of the form ax2
+ bx + c, where a, b, c
are real numbers with a ≠ 0.
3. The zeroes of a polynomial p(x) are precisely the x-coordinates of the points, where the
graph of y = p(x) intersects the x-axis.
4. A quadratic polynomial can have at most 2 zeroes and a cubic polynomial can have
at most 3 zeroes.
5. If α and β are the zeroes of the quadratic polynomial ax2
+ bx + c, then
b
a
α +β = − ,
c
a
αβ = .
6. If α, β, γ are the zeroes of the cubic polynomial ax3
+ bx2
+ cx + d, then
b
a
−
α +β + γ = ,
c
a
αβ+ β γ + γα = ,
and
d
a
−
αβγ = .
7. The division algorithm states that given any polynomial p(x) and any non-zero
polynomial g(x), there are polynomials q(x) and r(x) such that
p(x) = g(x) q(x) + r(x),
where r(x) = 0 or degree r(x) < degree g(x).
2019-20

class-X Maths Book | Chapter-2 | Slides By MANAV |

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