Class 10 Mathematics Chapter 2 Polynomials

Algebra is that branch of mathematics which treats the relation of numbers.

In algebra, two types of symbols are used: constants and variable (literals).

What is Constant?

Constant is a symbol whose value always remains the same, whatever the situation be.

For example: 5, – 9, $latex \frac{3}{8}$, π, , etc.

What is Variable?

Variable is a symbol whose value changes according to the situation.

For example : x, y, z, ax, a + x, 5y, – 7x, etc.

Algebraic Expression 

  1. An algebraic expression is a collection of terms separated by plus (+) or minus (–) sign. For example : 3x + 5y, 7y – 2x, 2x – ay + az, etc.
  2. The various parts of an algebraic expression that are separated by ‘+’ or ‘–’ sign are called terms.

For example :

Algebraic expression No. of Terms Terms

–32x

1 –32x
2x + 3y 2 2x and 3y
ax – 5y + cz 3 ax, –5y and cz
$latex \frac{3}{x}+\frac{y}{7}-\frac{{xy}}{8}+9$ 4

$latex \frac{3}{x},\,\frac{y}{7},\,-\frac{{xy}}{8}$ and 9 & so on.

How Many Types of Algebraic Expressions

Monomial

An algebraic expression having only one term is called a monomial. For ex. 8y, –7xy, 4×2, abx, etc. ‘mono’ means ‘one’.

Binomial

An algebraic expression having two terms is called a Binomial.

For ex.

8x + 3y, 8x + 3, 8 + 3y, a + bz, 9 – 4y,

2×2 – 4z, 6y2 – 5y, etc. ‘bi’ means ‘two’.

Trinomial

An algebraic expression having three terms is called a Trinomial.

For ex.

ax – 5y + 8z, 3×2 + 4x + 7, 9y2 – 3y + 2x, etc.

‘tri means ‘three’.

Multinomial

An algebraic expression having two or more terms is called a Multinomial.        

 Factor and Coefficient

What is Factor?

Each combination of the constants and variables, which form a term, is called a Factor.

For examples

(i)   7, x and 7x are factors of 7x, in which
7 is constant (numerical) factor and x is variable (literal) factor.

(ii)  In –5x2y, the numerical factor is –5 and literal factors are : x, y, xy, x2 and x2y.

²  Coefficient :

Any factor of a term is called the coefficient of the remaining term.

For example :

(i)   In 7x ; 7 is coefficient of x

(ii)  In –5x2y; 5 is coefficient of ­–x2y; –5 is coefficient of x2y.

Ex. 1     Write the coefficient of :

(i)   x2 in 3×3 – 5×2 + 7

(ii)  xy in 8xyz

(iii) –y in 2y2 – 6y + 2

(iv) x0 in 3x + 7

Sol.       (i) –5

            (ii) 8z

(iii) 6

(iv) Since x0 = 1, Therefore

3x + 7 = 3x + 7×0

coefficient of x0 is 7.Ø

        

      The greatest power (exponent) of the terms of a polynomial is called degree of the polynomial.

For example :  

(a)  In polynomial 5×2 – 8×7 + 3x :

(i)   The power of term 5×2 = 2

(ii)  The power of term –8×7 = 7

(iii) The power of 3x = 1

Since, the greatest power is 7, therefore degree of the polynomial 5×2 – 8×7 + 3x is 7

 

(b)  The degree of polynomial :

(i)   4y3 – 3y + 8 is 3

(ii) 7p + 2 is 1(p = p1)

(iii) 2m – 7m8 + m13 is 13 and so on.

v EXAMPLES v

Ex.2     Find which of the following algebraic expression is a polynomial.

(i) 3×2 – 5x              (ii) x +

(iii) – 8              (iv) z5 –  + 8

Sol.       (i)   3×2 – 5x = 3×2 – 5×1

It is a polynomial.

(ii)  x +  = x1 + x–1

It is not a polynomial.

(iii) – 8 = y1/2 – 8

Since, the power of the first term () is , which is not a whole number.

(iv) z5 –  + 8 = z5 – z1/3 + 8

Since, the exponent of the second term is
1/3, which in not a whole number. Therefore, the given expression is not a polynomial.

Ex.3     Find the degree of the polynomial :

(i)   5x – 6×3 + 8×7 + 6×2

(ii)  2y12 + 3y10 – y15 + y + 3

(iii) x

(iv) 8

Sol.       (i)   Since the term with highest exponent (power) is 8×7 and its power is 7.

\   The degree of given polynomial is 7.

(ii)  The highest power of the variable is 15

Þ   degree = 15.

(iii) x = x1   Þ    degree is 1.

(iv) 8 = 8×0 Þ    degree = 0

(A) Based on degree :

      If degree of polynomial is

  Examples
1. One   Linear x + 3, y – x + 2, x –3
2. Two Quadratic 2x2 –7,x2+y2 –2xy, x2 +1+ 3y
 

3.

 

Three

 

Cubic

 

x3 + 3x2 –7x+8, 2x2+5x3+7,

 

4.

 

Four

 

bi-quadratic

 

x4 + y4 + 2x2y2, x4 + 3,…

(B) Based on Terms :

If number of terms in polynomial is

  Examples
1. One Monomial 7x, 5x9, x16, xy, ……
 

2.

 

Two

 

Binomial

 

2 + 7y6, y3 + x14, 7 + 5x9,…

 

3.

 

Three

 

Trinomial

 

x3 –2x + y, x31+y32+ z33,…..

Note :   (1) Degree of constant polynomials

(Ex.5, 7, –3, 8/5, …) is zero.

(2)  Degree of zero polynomial (zero = 0
= zero polynomial) is not defined.

      If a polynomial has only one variable then it is called polynomial in one variable.

Ex. P(x) = 2x3 + 5x – 3               Cubic trinomial

Q(x) = 7x7 – 5x5 – 3x3 + x + 3 polynomial of

degree 7

R(y) = y                              Linear, monomial

S(t) = t2 + 3                          Quadratic Binomial

      Note : General form of a polynomial in one variable x of degree ‘n’ is anxn + an–1xn–1 + an–2xn–2 + ….+ a2x2 + a1x + a0, an ¹ 0, where an, an–1,… a2, a1, a0 all are constants.

            \ for linear       ax + b,                 a ¹ 0  

            for quadratic      ax2 + bx + c,         a ¹ 0

for cubic            ax3 + bx2 + cx + d, a ¹ 0

Ø    

(i)   Remainder obtained on dividing polynomial p(x) by x – a is equal to p(a) .

(ii)  If a polynomial p(x) is divided by (x + a) the remainder is the value of p(x) at x = –a.

(iii) (x – a) is a factor of polynomial p(x) if p(a) = 0   

(iv)   (x + a) is a factor of polynomial p(x) if p(–a) = 0

(v) (x – a)(x – b) is a factor of polynomial p(x),

if p(a) = 0 and p(b) = 0.

v EXAMPLES v

Ex.4     Find the remainder when 4×3 – 3×2 + 2x – 4 is divided by

            (a) x – 1      (b) x + 2            (c) x +

Sol.       Let p(x) =  4×3 – 3×2 + 2x – 4

(a) When p(x)  is divided by (x – 1), then by remainder theorem, the required remainder will be p(1)

p(1) = 4 (1)3 – 3(1)2 + 2(1) – 4

= 4 × 1 – 3 × 1 + 2 × 1 – 4

= 4 – 3 + 2 – 4 = – 1

(b)  When p(x) is divided by (x + 2), then by remainder theorem, the required remainder will be p (–2).

p(–2) = 4 (–2)3 – 3 (–2)2 + 2(–2) – 4

= 4 × (–8) – 3 × 4 – 4 – 4

= – 32 – 12 – 8 = – 52

(c) When p(x) is divided by,  then by remainder theorem, the required remainder will be

p = 4 – 3  + 2 – 4

= 4 ×  – 3 ×  – 2 ×– 4

= –– – 1– 4 = – – 5

=  =

Ø                      For a polynomial f(x) = 3×2 – 4x + 2.

To find its value at x = 3;

replace x by 3 everywhere.

So, the value of f(x) = 3×2 – 4x + 2 at x = 3 is

f(3) = 3 × 32 – 4 × 3 + 2

= 27 – 12 + 2 = 17.

Similarly, the value of polynomial

f(x) = 3×2 – 4x + 2,

(i)   at x = –2 is f(–2) = 3(–2)2 –4(–2) + 2

= 12 + 8 + 2 = 22

(ii)  at x = 0 is f(0) = 3(0)2 – 4(0) + 2

= 0 – 0 + 2 = 2

(iii) at x =  is f = 3– 4 + 2

=  – 2 + 2 =

Ex.5      Find the value of the polynomial 5x – 4×2 + 3 at:

(i) x = 0                   (ii) x = –1

Sol.       Let p(x) = 5x – 4×2 + 3.

(i)   At x = 0, p(0) = 5 × 0 – 4 × (0)2 + 3

= 0 – 0 + 3 = 3

(ii)  At x = –1, p(–1) = 5(–1) – 4(–1)2 + 3

= –5 – 4 + 3 = – 6

Ø If for x = a, the value of the polynomial p(x) is 0 i.e., p(a) = 0; then x = a is a zero of the polynomial p(x).

For example :

(i)   For polynomial p(x) = x – 2; p(2) = 2 – 2 = 0

\ x = 2 or simply 2 is a zero of the polynomial

p(x) = x – 2.

(ii)  For the polynomial g(u) = u2 – 5u + 6;

g(3) = (3)2 – 5 × 3 + 6 = 9 – 15 + 6 = 0

\ 3 is a zero of the polynomial g(u)

= u2 – 5u + 6.

Also, g(2) = (2)2 – 5 × 2 + 6 = 4 – 10 + 6 = 0

\ 2 is also a zero of the polynomial

g(u) = u2 – 5u + 6

(a)  Every linear polynomial has one and only one zero.

(b)  A given polynomial may have more than one zeroes.

(c)  If the degree of a polynomial is n; the largest number of zeroes it can have is also n.

For example :

If the degree of a polynomial is 5, the polynomial can have at the most 5 zeroes; if the degree of a polynomial is 8; largest number of zeroes it can have is 8.

(d)  A zero of a polynomial need not be 0.

For example : If f(x) = x2 – 4,

then f(2) = (2)2 – 4 = 4 – 4 = 0

Here, zero of the polynomial f(x) = x2 – 4 is 2 which itself is not 0.

(e)  0 may be a zero of a polynomial.

For example : If f(x) = x2 – x,

then f(0) = 02 – 0 = 0

Here 0 is the zero of polynomial

f(x) = x2 – x.

v EXAMPLES v

Ex.6     Verify whether the indicated numbers are zeroes of the polynomial corresponding to them in the following cases :

(i)   p(x) = 3x + 1, x = –

(ii) p(x) = (x + 1) (x – 2), x = – 1, 2

(iii) p(x) = x2, x = 0

(iv) p(x) = lx + m, x = –

(v)  p(x) = 2x + 1, x =

Sol.       (i)   p(x) = 3x + 1

Þ   p = 3 × – + 1 = –1 + 1 = 0

\   x = –  is a zero of p(x) = 3x + 1.

(ii)  p(x) = (x + 1) (x – 2)

Þ   p(–1) = (–1 + 1) (–1 – 2) = 0 × –3 = 0

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

\   x = –1 and x = 2 are zeroes of the given polynomial.

(iii) p(x) = x2      Þ   p(0) = 02 = 0

\   x = 0 is a zero of the given polynomial

(iv) p(x) = lx + m Þ p = l + m

= – m + m = 0

\   x = –  is a zero of the given polynomial.

(v)  p(x) = 2x + 1 Þ p = 2 ×  + 1

= 1 + 1 = 2 ¹ 0

\   x =  is not a zero of the given polynomial.

Ex.7     Find the zero of the polynomial in each of the following cases :

(i) p(x) = x + 5               (ii) p(x) = 2x + 5

(iii) p(x) = 3x – 2

Sol.       To find the zero of a polynomial p(x) means to solve the polynomial equation p(x) = 0.

(i)   For the zero of polynomial p(x) = x + 5

p(x) = 0       Þ   x + 5 = 0     Þ   x = –5

\ x = –5 is a zero of the polynomial
p(x) = x + 5.

(ii)  p(x) = 0       Þ   2x + 5 = 0

Þ   2x = –5 and x =

\   x =  is a zero of p(x) = 2x + 5.

(iii) p(x) = 0 Þ   3x – 2 = 0

Þ   3x = 2 and x = .

\ x =  is zero of p(x) = 3x – 2

                        Let us consider linear polynomial ax + b. The graph of y = ax + b is a straight line.

      For example : The graph of y = 3x + 4 is a straight line passing through (0, 4) and (2, 10).

 

 

(i)   Let us consider the graph of y = 2x – 4 intersects the x-axis at x = 2. The zero 2x – 4 is 2. Thus, the zero of the polynomial 2x – 4 is the x-coordinate of the point where the graph y = 2x – 4 intersects the x-axis.

 

(ii)  A general equation of a linear polynomial is
ax + b. The graph of y = ax + b is a straight line which intersects the x-axis at .

Zero of the polynomial ax + b is the x-coordinate of the point of intersection of the graph with x-axis.

(iii) Let us consider the quadratic polynomial
x2 – 4x + 3. The graph of x2 – 4x + 3 intersects the x-axis at the point (1, 0) and (3, 0). Zeroes of the polynomial x2 – 4x + 3 are the
x-coordinates of the points of intersection of the graph with x-axis.

The shape of the graph of the quadratic polynomials is È and the curve is known as parabola.

(iv) Now let us consider one more polynomial
–x2 + 2x + 8. Graph of this polynomial intersects the x-axis at the points
(4, 0), (–2, 0).    Zeroes of the polynomial –x2 + 2x + 8 are the x-coordinates of the points at which the graph intersects the x-axis. The shape of the graph of the given quadratic polynomial is Ç and the curve is known as parabola.

 

The zeroes of a quadratic polynomial
ax2 + bx + c he x-coordinates of the points where the graph of y = ax2 + bx + c intersects the x-axis.

Cubic polynomial : Let us find out geometrically how many zeroes a cubic has.

Let consider cubic polynomial

x3 – 6×2 + 11x – 6.

            Case 1 :

The graph of the cubic equation intersects the
x-axis at three points (1, 0), (2, 0) and (3, 0). Zeroes of the given polynomial are the
x-coordinates of the points of intersection with the x-axis.

            Case 2 :

The cubic equation x3 – x2 intersects the x-axis at the point (0, 0) and (1, 0). Zero of a polynomial x3 – x2 are the x-coordinates of the point where the graph cuts the x-axis.

Zeroes of the cubic polynomial are 0 and 1.

            Case 3 :

y = x3

Cubic polynomial has only one zero.

In brief : A cubic equation can have 1 or 2 or 3 zeroes or any polynomial of degree three can have at most three zeroes.

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

v EXAMPLES v

Ex.8     Which of the following correspond to the graph to a linear or a quadratic polynomial and find the number of zeroes of polynomial.

(i)            (ii)

(iii)         (iv)

(v)     (vi)

(vii)      (viii)

(ix)    (x)

Sol. (i)   The graph is a straight line so the graph is of a linear polynomial. The number of zeroes is one as the graph intersects the x-axis at one point only.

(ii)  The graph is a parabola. So, this is the graph of quadratic polynomial. The number of zeroes is zero as the graph does not intersect the x-axis.

(iii) Here the polynomial is quadratic as the graph is a parabola. The number of zeroes is one as the graph intersects the x-axis at one point only (two coincident points).

(iv) Here, the polynomial is quadratic as the graph is a parabola. The number of zeroes is two as the graph intersects the x-axis at two points.

(v)  The polynomial is linear as the graph is straight line. The number of zeroes is zero as the graph does not intersect the x-axis.

(vi) The polynomial is quadratic as the graph is a parabola. The number of zeroes is 1 as the graph intersects the x-axis at one point (two coincident points) only.

(vii)The polynomial is quadratic as the graph is a parabola. The number of zeroes is zero, as the graph does not intersect the x-axis.

(viii) Polynomial is neither linear nor quadratic as the graph is neither a straight line nor a parabola is one as the graph intersects the x-axis at one point only.

(ix) Here, the polynomial is quadratic as the graph is a parabola. The number of zeroes is one as the graph intersects the x-axis at one point only (two coincident points).

(x)  The polynomial is linear as the graph is a straight line. The number of zeroes is one as the graph intersects the x-axis at only one point.

      Consider quadratic polynomial

P(x) = 2×2 – 16x + 30.

Now, 2×2 – 16x + 30 = (2x – 6) (x – 3)

= 2 (x – 3) (x – 5)

The zeroes of P(x) are 3 and 5.

Sum of the zeroes

= 3 + 5 = 8 =  = –

Product of the zeroes

= 3 × 5 = 15 =  =

So if  ax2 + bx + c,  a ¹ 0 is a quadratic polynomial and a, b are two zeroes of polynomial then

v EXAMPLES v

Ex.9     Find the zeroes of the quadratic polynomial
6×2 – 13x + 6 and verify the relation between the zeroes and its coefficients.

Sol.      We have, 6×2 – 13x + 6 = 6×2 – 4x – 9x + 6

= 2x (3x – 2) –3 (3x – 2)

= (3x – 2) (2x – 3)

So, the value of 6×2 – 13x + 6 is 0, when
(3x – 2) = 0 or (2x – 3) = 0 i.e.,

When    x =    or

Therefore, the zeroes of 6×2 – 13x + 6 are

and .

Sum of the zeroes

= +  =  =  =

Product of the zeroes

=  ×  =  =

Ex.10    Find the zeroes of the quadratic polynomial
4×2 – 9 and verify the relation between the zeroes and its coefficients.

Sol.      We have,

4×2 – 9 = (2x)2 – 32 = (2x – 3) (2x + 3)

So, the value of 4×2 – 9 is 0, when

2x – 3 = 0    or   2x + 3 = 0

i.e., when   x =   or   x = –.

Therefore, the zeroes of 4×2 – 9 are  & –.

Sum of the zeroes

=  –  = 0 =  =

Product of the zeroes

=   =  =

Ex.11    Find the zeroes of the quadratic polynomial
9×2 – 5 and verify the relation between the zeroes and its coefficients.

Sol.      We have,

9×2 – 5 = (3x)2 – ()2 = (3x – ) (3x +)

So, the value of 9×2 – 5 is 0,

when 3x – = 0 or 3x + = 0

i.e., when x =     or   x = .

Sum of the zeroes

=  –  = 0 =  =

Product of the zeroes

=  =  =

Ex.12    If a and b are the zeroes of ax2 + bx + c, a ¹ 0 then verify the relation between the zeroes and its coefficients.

Sol.      Since a and b are the zeroes of polynomial

ax2 + bx + c.

            Therefore,    (x – a), (x – b) are the factors of the polynomial ax2 + bx + c.

Þ   ax2 + bx + c = k (x – a) (x – b)

Þ   ax2 + bx + c = k {x2 – (a + b) x + ab}

Þ ax2 + bx + c = kx2 – k (a + b) x + kab …(1)

Comparing the coefficients of x2, x and constant terms of (1) on both sides, we get

a = k, b = – k (a + b) and c = kab

Þ   a + b = – and  ab =

a + b =        and  ab =       [Q k = a]

Sum of the zeroes =  =

Product of the zeroes =  =

Ex. 13   Prove relation between the zeroes and the coefficient of the quadratic polynomial
ax2 + bx + c.

Sol.      Let a and b be the zeroes of the polynomial
ax2 + bx + c

\   a =              ….(1)

b =              ….(2)

By adding (1) and (2), we get

a + b =  +

=  = –  =

Hence, sum of the zeroes of the polynomial
ax2 + bx + c is –

By multiplying (1) and (2), we get

ab =

=  =

=  =

=

Hence, product of zeroes =

 

Ex.14    find the zeroes of the quadratic polynomial
x2 – 2x – 8 and verify a relationship between zeroes and its coefficients.

Sol.      x2 – 2x – 8 = x2 – 4x + 2x – 8

= x (x – 4) + 2 (x – 4) = (x – 4) (x + 2)

So, the value of x2 – 2x – 8 is zero when
x – 4 = 0 or x + 2 = 0 i.e., when x = 4 or x = – 2.

So, the zeroes of x2 – 2x – 8 are 4, – 2.

Sum of the zeroes

= 4 – 2 = 2 =  =

Product of the zeroes

= 4 (–2) = –8 =  =

Ex.15    Verify that the numbers given along side of the cubic polynomials are their zeroes. Also verify the relationship between the zeroes and the coefficients. 2×3 + x2 – 5x + 2 ; , 1, – 2

Sol.      Here, the polynomial p(x) is

2×3 + x2 – 5x + 2

Value of the polynomial 2×3 + x2 – 5x + 2

when  x = 1/2

= 2+– 5+2 =+–+ 2 = 0

So, 1/2 is a zero of p(x).

On putting x = 1 in the cubic polynomial

2×3 + x2 – 5x + 2

= 2(1)3 + (1)2 –­ 5(1) + 2 = 2 + 1 – 5 + 2 = 0

On putting x = – 2 in the cubic polynomial

2×3 + x2 – 5x + 2

= 2(–2)3 + (–2)2 – 5 (–2) + 2

= – 16 + 4 + 10 + 2 = 0

Hence, , 1, – 2 are the zeroes of the given polynomial.

Sum of the zeroes of p(x)

=  + 1 – 2 = –  =

Sum of the products of two zeroes taken at a time

=  × 1 +  × (–2) + 1 × (–2)

= – 1 – 2 = – =

Product of all the three zeroes

= × (1) × (–2) = –1

==

² Symmetric function :

An algebraic expression in a and b, which remains unchanged, when a and b are interchanged is known as symmetric function in a and b.

For example, a2 + b2 and a3 + b3 etc. are symmetric functions. Symmetric function is to be expressed in terms of (a + b) and ab. So, this can be evaluated for a given quadratic equation.

²  Some useful relations involving a and b :

  1. a2 + b2 = (a + b)2 – 2ab
  2. (a – b)2 = (a + b)2 – 4ab
  3. a2 – b2 = (a + b) (a – b) = (a + b)

 

  1. a3 + b3 = (a + b)3 – 3ab (a + b)
  2. a3 – b3 = (a – b)3 + 3ab (a – b)
  3. a4 + b4 =[(a + b)2 – 2ab]2 –2(ab)2
  4. a4 – b4 = (a2 + b2) (a2 – b2) then use (1) and (3)

v EXAMPLES v

Ex.16    If a and b are the zeroes of the polynomial
ax2 + bx + c. Find the value of

(i) a – b            (ii) a2 + b2.

Sol.      Since a and b are the zeroes of the polynomial ax2 + bx + c.

\   a + b = – ;     ab  =

(i)   (a – b)2 = (a + b)2 – 4ab

=  –   =  –  =

a – b =

(ii)  a2 + b2 = a2 + b2 + 2ab – 2ab

= (a + b)2 – 2ab

=  – 2 =

Ex.17    If a and b are the zeroes of the quadratic polynomial ax2 + bx + c. Find the value of

(i) a2 – b2          (ii) a3 + b3.

Sol.      Since a and b are the zeroes of ax2 + bx + c

\   a + b = ,      ab =

(i)   a2 – b2 = (a + b) (a – b)

= –

= –  = –

= –

(ii) a3 + b3 = (a + b) (a2 + b2 – ab)

= (a + b) [(a2 + b2 + 2ab) – 3ab]

= (a + b) [(a + b)2 – 3ab]

=

==

=

 

v EXAMPLES v

Ex.18    Form the quadratic polynomial whose zeroes are 4 and 6.

Sol.      Sum of the zeroes = 4 + 6 = 10

Product of the zeroes = 4 × 6 = 24

Hence the polynomial formed

= x2 – (sum of zeroes) x + Product of zeroes

= x2 – 10x + 24

Ex.19    Form the quadratic polynomial whose zeroes are –3, 5.

Sol.      Here, zeroes are – 3 and 5.

Sum of the zeroes = – 3 + 5 = 2

Product of the zeroes = (–3) × 5 = – 15

Hence the polynomial formed

= x2 – (sum of zeroes) x + Product of zeroes

= x2 – 2x – 15

Ex.20    Find a quadratic polynomial whose sum of zeroes and product of zeroes are respectively-

(i) , – 1    (ii)  ,     (iii) 0,

Sol.      Let the polynomial be ax2 + bx + c and its zeroes be a and b.

(i)   Here, a + b =  and a . b = – 1

Thus the polynomial formed

= x2 – (Sum of zeroes) x + Product of zeroes

= x2 –  x – 1 = x2 –  – 1

The other polynomial are k

If k = 4, then the polynomial is 4×2 – x – 4.

(ii)  Here, a + b =, ab =

Thus the polynomial formed

= x2 – (Sum of zeroes) x + Product of zeroes

= x2 – () x +   or   x2 –x +

Other polynomial are k

If k = 3, then the polynomial is

3×2 – 3x + 1

(iii) Here,    a + b = 0     and     a.b =

Thus the polynomial formed

= x2 – (Sum of zeroes) x + Product of zeroes

= x2 – (0) x + = x2 +

Ex.21    Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time, and product of its zeroes as 2,
– 7 and –14, respectively.

Sol.      Let the cubic polynomial be

ax3 + bx2 + cx + d

Þ   x3 +  x2 + x +                  ….(1)

and its zeroes are a, b and g, then

a + b + g = 2 = –

ab + bg + ag = – 7 =

abg = – 14 = –

Putting the values of ,  and  in (1),

we get

x3 + (–2) x2 + (–7)x + 14

Þ   x3 – 2×2 – 7x + 14

Ex.22    Find the cubic polynomial with the sum, sum of the product of its zeroes taken two at a time and product of its zeroes as 0, –7 and –6 respectively.

Sol.      Let the cubic polynomial be

ax3 + bx2 + cx + d

Þ   x3 + x2 +  x +                   ….(1)

and its zeroes are a, b, g. Then

a + b + g = 0 = –

ab + bg + ag = – 7 =

abg = – 6 =

Putting the values of ,  and  in (1),

we get

x3 – (0) x2 + (–7) x + (–6)

or   x3 – 7x + 6

Ex.23    If a and b are the zeroes of the polynomials
ax2 + bx + c then form the polynomial whose zeroes are  and .

Sol.      Since a and b are the zeroes of ax2 + bx + c

So a + b = , ab =

Sum of the zeroes = +  =

=  =

Product of the zeroes

=  ==

But required polynomial is

x2 – (sum of zeroes) x + Product of zeroes

Þ   x2 –  x +

or   x2 +  x +

or c

Þ   cx2 + bx + a

Ex.24    If a and b are the zeroes of the polynomial
x2 + 4x + 3, form the polynomial whose zeroes are 1 +  and 1 + .

Sol.      Since a and b are the zeroes of the polynomial x2  + 4x + 3.

            Then,    a + b = – 4,  ab = 3

Sum of the zeroes

= 1 +  + 1 +  =

=  =  =  =

Product of the zeroes

=  = 1 +  +  +

= 2 +  =

=  =  =

But required polynomial is

x2 – (sum of zeroes) x + product of zeroes

or   x2 –  x +     or k

or   3    (if  k = 3)

Þ   3×2 – 16x + 16

 

Ø   

Step 1:

First arrange the term of dividend and the divisor in the decreasing order of their degrees.

Step 2 :

To obtain the first term of quotient divide the highest degree term of the dividend by the highest degree term of the divisor.

Step 3 :

To obtain the second term of the quotient, divide the highest degree term of the new dividend obtained as remainder by the highest degree term of the divisor.

Step 4 :

Continue this process till the degree of remainder is less than the degree of divisor.

²  Division Algorithm for Polynomial

If p(x) and g(x) are any two polynomials with

g(x) ¹ 0, then we can find polynomials q(x) and r(x) such that

p(x) = q(x) × g(x) + r(x)

where r(x) = 0 or degree of r(x) < degree of g(x).

The result is called Division Algorithm for polynomials.

 

v EXAMPLES v

Ex.25    Divide 3×3 + 16×2 + 21x + 20  by  x + 4.

Sol.

 

Quotient = 3×2 + 4x + 5

Remainder = 0

Ex.26    Apply the division algorithm to find the quotient and remainder on dividing p(x) by g(x) as given below :

p(x) = x3 – 3×2 + 5x – 3, g(x) = x2 – 2

Sol.      We have,

p(x) = x3 – 3×2 + 5x – 3 and g(x) = x2 – 2

 

We stop here since

degree of (7x – 9) < degree of (x2 – 2)

So, quotient = x – 3, remainder = 7x – 9

Therefore,

Quotient × Divisor + Remainder

=    (x – 3) (x2 – 2) + 7x – 9

=    x3 – 2x – 3×2 + 6 + 7x – 9

=    x3 – 3×2 + 5x – 3 = Dividend

Therefore, the division algorithm is verified.

Ex.27    Apply the division algorithm to find the quotient and remainder on dividing p(x) by g(x) as given below

p(x) = x4 – 3×2 + 4x + 5, g (x) = x2 + 1 – x

Sol.      We have,

p(x) = x4 – 3×2 + 4x + 5, g(x) = x2 + 1 – x

 

 

 

We stop here since

degree of (8) < degree of (x2 – x + 1).

So, quotient = x2 + x – 3, remainder = 8

Therefore,

Quotient × Divisor + Remainder

= (x2 + x – 3) (x2 – x + 1) + 8

= x4 – x3 + x2 + x3 – x2 + x – 3×2 + 3x – 3 + 8

= x4 – 3×2 + 4x + 5    = Dividend

Therefore the Division Algorithm is verified.

Ex.28    Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm.t2 – 3; 2t4 + 3t3 – 2t2 – 9t – 12.

Sol.      We divide 2t4 + 3t3 – 2t2 – 9t – 12 by t2 – 3

Here, remainder is 0, so t2 – 3 is a factor of
2t4 + 3t3 – 2t2 – 9t – 12.

2t4 + 3t3 – 2t2 – 9t – 12 = (2t2 + 3t + 4) (t2 – 3)

Ex.29    Obtain all the zeroes of

            3×4 + 6×3 – 2×2 – 10x – 5, if two of its zeroes are  and – .

Sol.      Since two zeroes are  and –,

x = , x = –

Þ    = x2 –  or 3×2 – 5 is a factor of the given polynomial.

Now, we apply the division algorithm to the given polynomial and 3×2 – 5.

 

So, 3×4 + 6×3 – 2×2 – 10x – 5

= (3×2 – 5) (x2 + 2x + 1) + 0

Quotient = x2 + 2x + 1 = (x + 1)2

Zeroes of (x + 1)2 are –1, –1.

Hence, all its zeroes are , – , –1, –1.

Ex.30    On dividing x3 – 3×2 + x + 2 by a polynomial
g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Find g(x).

Sol.      p(x) = x3 – 3×2 + x + 2

q(x) = x – 2 and r (x) = –2x + 4

By Division Algorithm, we know that

p(x) = q(x) × g(x) + r(x)

Therefore,

x3 – 3×2 + x + 2 = (x – 2) × g(x) + (–2x + 4)

Þ   x3 – 3×2 + x + 2 + 2x – 4 = (x – 2) × g(x)

Þ   g(x) =

On dividing x3 – 3×2 + 3x – 2 by x – 2,
we get g(x)

Hence, g(x) = x2 – x + 1.

 

Ex.31    Give examples of polynomials p(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 q(x) = 0

Sol. (i)   Let q(x) = 3×2 + 2x + 6, degree of q(x) = 2

p(x) = 12×2 + 8x + 24, degree of p(x) = 2

Here, deg p(x) = deg q(x)

(ii)  p(x) = x5 + 2×4 + 3×3 + 5×2 + 2

q(x) = x2 + x + 1,            degree of q(x) = 2

g(x) = x3 + x2 + x + 1

r(x) = 2×2 – 2x + 1,   degree of r(x) = 2

Here, deg q(x) = deg r(x)

(iii) Let p(x) = 2×4 + 8×3 + 6×2 + 4x + 12

q(x) = 2, degree of q(x) = 0

g(x) = x4 + 4×3 + 3×2 + 2x + 6

r(x) = 0

Here, deg q(x) = 0

Ex.32    If the zeroes of polynomial x3 – 3x2 + x + 1 are a – b, a , a + b. Find a and b.  

Sol.      Q a – b, a, a + b are zeros

            \ product (a – b) a(a + b) = –1

Þ (a2 – b2) a = –1     …(1)

and sum of zeroes is (a – b) + a + (a + b) = 3

Þ 3a = 3 Þ a = 1           …(2)

by (1) and (2)

(1 – b2)1 = –1

Þ 2 = b2 Þ b = ±

\ a = –1 & b = ±                  Ans.

Ex.33    If two zeroes of the polynomial

            x4 – 6x3 –26x2 + 138x – 35 are 2 ± ,
find other zeroes.

Sol.       Q   2 ±are zeroes.

\   x = 2 ±

Þ   x – 2 = ±(squaring both sides)

Þ   (x – 2)2 = 3  Þ  x2 + 4 – 4x – 3 = 0

Þ   x2 – 4x + 1 = 0 , is a factor of given polynomial

\ other factors

=

 

\ other factors = x2 – 2x – 35

= x2 – 7x + 5x – 35 = x(x – 7) + 5(x – 7)

= (x – 7) (x + 5)

\ other zeroes are (x – 7) = 0 Þ x = 7

x + 5 = 0 Þ x = – 5              Ans.

Ex.34    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 & a.

Sol.

 

According to questions, remainder is x + a

\ coefficient of x = 1

Þ 2k  – 9 = 1

Þ k = (10/2) = 5

Also constant term = a

Þ k2 – 8k + 10 = a Þ (5)2 – 8(5) + 10 = a

Þ a = 25 – 40 + 10

Þ a = – 5

\ k = 5, a = –5                          Ans.

 

 

 

exercise # 1

 

Factorize each of the following expression

Q.1       x2 – x – 42

 

Q.2       6 – 5y – y2

 

Q.3       a2 + 46a + 205

 

Q.4       ab + ac – b2 – bc

 

Q.5       p4 – 81q4

 

Use remainder theorem to find remainder,
when
p(x) is divided by q(x) in following questions.

Q.6       p(x) = 2x2 – 5x + 7, q(x) = x – 1

 

Q.7       p(x) = x9 – 5x4 + 1 , q(x) = x + 1

 

Q.8       p(x) = 2x3 – 3x2 + 4x – 1 , q(x) = x + 2

 

Q.9       Find positive square root of 36x2 + 60x + 25

 

Q.10     Simplify :

 

Q.11     (x2 + 4y)2 + 21 (x2 + 4y) + 98

 

Q.12     Find the value of k if (x – 2) is a factor of
2x3– 6x2+ 5x + k.

 

Q.13     Find the value of k if (x + 3) is a factor of
3x2 + kx + 6.

 

Q.14      p(x) = 3x6 – 7x5 + 7x4 – 3x3 + 2x2 – 2, q(x) = x – 1

 

Q.15     For what value of k is y3 + ky + 2k – 2 exactly divisible by (y + 1) ?

 

 

Q.16     If x + 1 and x – 1 are factors of

mx3 + x2 – 2x + n, find the value of m and n.

 

Q.17     Find the zeros of the polynomial

f(x) = 2x2 + 5x – 12 and verify the relation between its zeroes and coefficients.

Q.18     Find the zeroes of the polynomial f(x) = x2 – 2 and verify the relation between its zeroes and coefficients.

 

Q.19     Obtain the zeroes of the quadratic polynomial x2 – 8x + 4and verify the relation between its zeroes and coefficients.

 

Q.20     Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and the product of its zeroes as
2, –7 and –14 respectively.

 

Q.21     Find a cubic polynomial whose zeroes are 3, 5 and – 2.

 

Q.22     Divide 5x3 – 13x2 + 21x – 14 by (3 – 2x + x2) and verify the division algorithm.

 

Q.23     What real number should be subtracted from the polynomial (3x3 + 10x2 – 14x + 9) so that (3x – 2) divides it exactly?

 

Q.24     Find all the zeroes of (2x4 – 3x3 – 5x2 + 9x –3), it being given that two of its zeroes are
and – .

 

 

 

answer key

  1. Verty Short Answer Type :
  2. (x + 6) (x – 7) 2. (6 + y) (1 – y)             3. (a + 41) (a + 5)                       4. (a – b) (b + c)
  3. (p + 3q) (p – 3q) (p2 + 9q2) 6. 4   7. – 5                                8. –37
  4. Short Answer Type :
  5. 6x + 5 10. (a +b) 11. (x2 + 4y + 7) (x2 + 4y + 14)     12. –2
  6. 11 15. 3
  7. Long Answer Type :
  8. m = 2, n = – 1 17. –4,                       18.                              19.
  9. x3 – 2x2 – 7x + 14 21. x3 – 6x2 – x + 30 22. quotient = 5x – 3, Remainder = –5
  10. 5 24.
    exercise # 2

 

Q.1       If  = 3, then find value of   .

Q.2       If  =, then  find  .

Q.3       If  = 4, then find  .

Q.4       If (x – 2) is a factor of (x2 + 3qx – 2q), then find the value of q.

Q.5       If x3 + 6x2 + 4x + k is exactly divisible by
(x + 2), then find the value of k.

Q.6       Let f(x) = x3 – 6x2 + 11x – 6. Then, which one of the following is not factor of f(x) ?

            (A) x – 1                   (B) x – 2

(C) x + 3                   (D) x – 3

Q.7       If x100 + 2x99 + k is divisible by (x + 1), then find the value of k.

Q.8       On dividing (x3 – 6x + 7) by (x + 1), find the remainder.

Q.9       Find the value of expression (16x2 + 24x + 9) for x = – .

Q.10     If 2x3 + 5x2 – 4x – 6 is divided by 2x + 1, then find remainder.

 

Q.11     If p(x) = x2 – 2x – 3, then find

            (i) p(3); (ii) p(–1)

 

Q.12     Find the zeros of the quadratic polynomial
(6x2 – 7x – 3) and verify the relation between its zeros and coefficients.

 

Q.13     Find the zeros of the quadratic polynomial
(5u2 + 10u) and verify the relation between the zeros and the coefficients.

 

Q.14     Find the quadratic polynomial whose zeros are  and . Verify the relation between the coefficients and the zeros of the polynomial.

 

Q.15     Find the quadratic polynomial, sum of whose zeros is 8 and their product is 12. Hence, find the zeros of the polynomial.

 

Q.16     Find the quadratic polynomial , the sum of whose zeros is –5 and their product is 6. Hence, find the zeros of the polynomial.

 

Q.17     Find  the quadratic polynomial, the sum of whose zeros is 0 and their product is –1. Hence, find the zeros of the polynomial.

 

Q.18     Find a quadratic polynomial whose one zero is 5 + .

 

Q.19     On dividing (x3 – 3x2 + x + 2) by a polynomial g(x), the quotient and remainder are (x – 2) and (–2x + 4) respectively. Find g(x).

 

Q.20     If the polynomial (x4 + 2x3 + 8x2 + 12x + 18) is divided by another polynomial (x2 + 5), the remainder comes out to be (px + q). Find the value of p and q.

 

Q.21     Obtain all zeros of the polynomial
(2x3 – 4x – x2 + 2), if two of its zeros are  and –.

 

Q.22     If 1 and –2 are two zeros of the polynomial
(x3 – 4x2 – 7x + 10), find its third zero.

 

Q.23     Find all the zeros of the polynomial
(2x4 – 11x3 + 7x2 + 13x – 7), it being given that two if its zeros are (3 + ) and (3 – ).

 

Q.24     If a, b are the zeros of the polynomial
f(x) = x2 – 5x + k such that a – b = 1, find the value of k.

 

Q.25     Show that the polynomial f(x) = x4 + 4x2 + 6 has no zero.

 

Q.26     Use remainder theorem to find the value of k, it being given that when x3 + 2x2 + kx + 3 is divided by (x – 3), then the remainder is 21.

 

 

 

answer key

  1. 7 2. 9 4. – 1    5. – 8                6. (C)   
  2. 1 8. 12 9. 0                               10. – 3
  3. (i) 0 , (ii) 0 12.                       13. – 2, 0                       14. 12x2 – 5x – 2
  4. (x2 – 8x + 12), {6, 2} 16. (x2 + 5x + 6), {– 3, – 2}
  5. (x2 – 1), {1, –1} 18. x2 – 10x + 18 19. x2 – x + 1                 20. p = 2, q = 3
  6. 22. 5 23. , , – 1
  7. k = 6 26. k = – 9

 

 

 

 

 

           

 

 

 

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