POLYNOMIALS

TOPICS COVERED

Polynomials in One Variable

Zeroes of a polynomial

Remainder Theorem

Factor Theorem

Algebraic Identities

Introduction An algebraic expression in which variables involved have only non-

negative integral powers is called a polynomial.

E.g.- (a) 2x3–4x2+6x–3 is a polynomial in one variable x.

(b) 8p7+4p2+11p3-9p is a polynomial in one variable p.

(c) 4+7x4/5+9x5 is an expression but not a polynomial

since it contains a term x4/5, where 4/5 is not

a non-negative integer.

In the polynomial x2 + 2x, the expressions x2 and 2x are called the terms of the polynomial.

Similarly, the polynomial 3y2 + 5y + 7 has three terms, namely, 3y2, 5y & 7.

Example : –x3 + 4x2 + 7x – 2. This polynomial has 4 terms, namely, –x3, 4x2, 7x and –2. Each term of a polynomial has a coefficient. So, in –x3 + 4x2 + 7x – 2, the coefficient of x3 is –1, the coefficient of x2 is 4, the coefficient of x is 7 and –2 is the coefficient of x0. (Remember, x0 = 1)

The coefficient of x in x2 – x + 7 is –1.

Examples of Polynomials in one variable Algebraic expressions like 2x, x2 + 2x, x3 –

x2 + 4x + 7 have only whole numbers as the exponents of the variable. Expressions of this form are called polynomials in one variable. In the examples above, the variable is x.

3y2 + 5y is a polynomial in the variable y. t2 + 4 is a polynomial in the variable t.

Degree of a Polynomial in one variable.

What is degree of the following binomial?

35 2 xThe answer is 2. 5x2 + 3 is a polynomial in x of degree 2.

In case of a polynomial in one variable, the highest power of the variable is called the degree of polynomial.

Degree of a Polynomial in two variables.

What is degree of the following polynomial?

49375 332 yxyxyx

In case of polynomials on more than one variable, the sum of powers of the variables in each term is taken up and the highest sum so obtained is called the degree of polynomial.

• The answer is five because if we add 2 and 3 , the answer is five which is the highest power in the whole polynomial.

E.g.- is a polynomial in x and y of degree 7.

92853 243 yxyxyx

Polynomials in one variable

The degree of a polynomial in one variable is the largest exponent of that variable.

14 x

A constant has no variable. It is a 0 degree polynomial.2This is a 1st degree polynomial. 1st degree polynomials are linear.

1425 2 xx This is a 2nd degree polynomial. 2nd degree polynomials are quadratic.

183 3 x This is a 3rd degree polynomial. 3rd degree polynomials are cubic.

Examples

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Polynomials Degree Classify by degree

Classify by no. of terms.

5 0 Constant Monomial

2x - 4 1 Linear Binomial

3x2 + x 2 Quadratic Binomial

x3 - 4x2 + 1 3 Cubic Trinomial

QUESTIONS

Which of the following expressions are polynomials in one variable and which are not ?

i) 4x ²-3x+7 ii) y+2/y Classify the following as linear ,quadratic and

cubic polynomials : i)x ²+x ii)y+y ²+4 iii)1+x iv)3t vi)t²

Zeroes of a Polynomial

A zero of a polynomial p(x) is a number c such that p(c)=0.

Example:

Q-Check whether -2 and +2 are the zeroes of the polynomial x+2?

ANS-Let p(x)=x+2

Then p(2)=2+2=4, p(-2)=(-)2 +2=0

Therefore -2 is a zero of the polynomial but +2 is not.

QUESTIONS

Find the value of the polynomial 5x-4x²+3 at i)x=0 ii) x= (-1) (iii)x=2

Solutions: (i) p(x) =5x-4x²+3 p(0) =5(0)-4(0)2+3=0-0+3=3 ii) p(-1)=5(-1)-4(-1)2+3 = -5-4+3 = -6 iii) p(2) =5(2)-4(2)2+3 =10-16+3 = -3

QUESTIONS Verify whether the following are zeroes of the

polynomials . i)p(x) =3x+1 ,x =(-1/3) Sol :p(x) =3x+1 p(-1/3) =3(-1/3) +1 = -1+1 =0 ∴ x= (-1/3) is the zero of 3x+1. ii) p(x) = (x+1) (x-2) ,x =-1 ,2 Sol : p(-1) =(-1+1)(-1-2) =0(-3)=0 Since p(-1) =0 ,so x= -1 is a zero of p(x) . p(2) = (2+1) (2-2) =3(0) =0 So ,x=2 is a zero of p(x) .

QUESTIONS Find the zero of the polynomial in each of

the following cases : i)p(x) =x+5 Sol : p(x) =x+5 P(x)=0 ,x+5 =0 ,x= -5 Thus ,a zero of (x+5 ) is (-5) ii) p(x) =3x P(x) =0 ,3x =0 ,x=0/3 =0 Thus ,a zero of (3x) is 0 .

QUESTIONS FOR PRACTICE 1) Find p(0) ,p(1) ,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) 2) Verify whether the following are zeroes of the

polynomial . i) p(x) =x²-1 ,x =1 , -1 ii ) p(x) = 2x+1 ,x =1/2 3)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 .

Remainder Theorem

Statement: Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number. If p(x) is divided by the linear polynomial x – a, then the remainder is p(a).

.

Proof : Let p(x) be any polynomial with degree greater than or equal to 1. Suppose that when p(x) is divided by x – a, the quotient is q(x) and the remainder is r(x), i.e., p(x) = (x – a) q(x) + r(x)

Since the degree of x – a is 1 and the degree of r(x) is less than the degree of x – a, the degree of r(x) = 0. This means that r(x) is a constant, say r.

So, for every value of x, r(x) = r.

Therefore, p(x) = (x – a) q(x) + r

In particular, if x = a, this equation gives us

p(a) = (a – a) q(a) + r = r,

which proves the theorem

Examples1. Divide the polynomial 3x4 – 4x3 – 3x –1 by x – 1.

Solution: By long division, we have:

Here, the remainder is – 5. Now, the zero of x – 1 is 1. So, putting x = 1 in p(x), we see that p(1) = 3(1)4 – 4(1)³ – 3(1) – 1 = 3 – 4 – 3 – 1 = – 5, which is the remainder.

2. Find the remainder when x4 + x3 – 2x2 + x + 1 is divided by x – 1.

Solution : Here, p(x) = x4 + x3 – 2x2 + x + 1, and the zero of x – 1 is 1.

So, p(1) = (1)4 + (1)3 – 2(1)2 + 1 + 1= 2

So, by the Remainder Theorem, 2 is the remainder when x4 + x3 – 2x2 + x + 1 is divided by x – 1.

Practice questions

1.Find the remainder when x³+3x²+3x+1 is divided by

i)x+1 ii)x-1/2 iii)x iv)x+л v)5+2x

2. Find the remainder when x³-ax²+6x-a is divided by x-a.

3.Check whether 7+3x is a factor of 3x³+7x.

Factor Theorem

Statement: If p(x) is a polynomial of degree n > 1 and a is any real number, then

(i) x – a is a factor of p(x), if p(a) = 0, and

(ii) p(a) = 0, if x – a is a factor of p(x).

This actually follows immediately from the Remainder Theorem

Examples

Examine whether x + 2 is a factor of x3 + 3x2 + 5x + 6 and of 2x + 4.

Solution : The zero of x + 2 is –2. Let p(x) = x3 + 3x2 + 5x + 6 and s(x) = 2x + 4.

Then, p(–2) = (–2)3 + 3(–2)2 + 5(–2) + 6

= –8 + 12 – 10 + 6

= 0

So, by the Factor Theorem, x + 2 is a factor of x3 + 3x2 + 5x + 6.

Again, s(–2) = 2(–2) + 4 = 0

So, x + 2 is a factor of 2x + 4.

In fact, it can also be checked without applying the Factor Theorem, since 2x + 4 = 2(x + 2).

Practice Questions

1.Determine whether x+1 is a factor of the following polynomials,

x³-x²-(2+2)x+22. Find the value of k ,if x-1 is a factor of p(x)

in each of the following cases:

i)2x²+kx+2 ii)kx²-2x+1

Factorizing a Polynomial

Example: Q-Factorise p(y)=y2 -5y +6 by using factor

theorem. Ans- factors of p(y), we find the factors of 6

factors of 6 are 1,2,3

Now, p(2)=22 -(5X2) +6=0

So, y-2 is a factor of p(y).

Also, p(3)=32 –(5X3) +6=0

So y-3 is also a factor.

Therefore, y2 -5y +6=(y-2) (y-3)

Practice Questions

1.factorize:

i)12x² -7x+1 ii)2x²+7x+3

ii)6x²+5x-6 iv)3x²-x-4

Factorize: x³-23x²+142x-120

Soln: let p(x)=x³-23x²+142x-120

Factors of -120 are:

± 1 ,±2,±3,±4,±5,±6,±8,±10,±12,±24,±30,±40,±60,±120.

By hit and trial we get p(1)=0. so x-1 is a factor of p(x).

Therefore p(x)=x³-x²-22x²+22x+120x-120

=x²(x-1)-22x(x-1)+120(x-1)

=(x-1)(x²-22x+120)

{taken (x-1) common}

further x²-22x+120 can be factorized to

= x²-12x-10x+120

=x(x-12)-10x(x-12)

=(x-12) (x-10)

Hence, p(x)=(x-1) (x-12) (x-10)

Practice Questions

Factorize:

i) x³-2x²-x+2 ii) x³-3x²-9x-5

ii) x³+13x²+32x+20 iv) 2y³ +y²-2y-1

Algebraic Identities

An algebraic identity is an algebraic equation that is true for all values of the variables occurring in it.

Identity I : (x + y)2 = x2 + 2xy + y2

Identity II : (x – y)2 = x2 – 2xy + y2

Identity III : x2 – y2 = (x + y) (x – y)

Identity IV : (x + a) (x + b) = x2 + (a + b)x + ab

Practice Questions

1.Use suitable identities to find the product of

i)(3x+4) (3x-5) ii)(y²+3/2)(y²-3/2)

2.Evaluate the following without multiplying directly.

i)103x107 ii)95x96 iii)104x96

3.Factorise using appropriate identities:

i)9x²+6xy+y² ii)4y²-4y+1

iii)x²-(y²/100)

Identity V: (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Remark :We call the right hand side expression the expanded form of the left hand side expression. The expansion of (x + y + z)2 consists of three square terms and three product terms.

Identity V: (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

We shall compute (x + y + z)2 by using Identity I.

Let x + y = t. Then,

(x + y + z)2 = (t + z)2

= t2 + 2tz + t2 (Using Identity I)

= (x + y)2 + 2(x + y)z + z2 (Substituting value of t)

= x2 + 2xy + y2 + 2xz + 2yz + z2

= x2 + y2 + z2 + 2xy + 2yz + 2zx(Rearranging)

So, we get the following identity:

(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Example:Ques: Write (3a+4b+5c)2 in expanded form.

Soln: comparing the given expression with (x+y+z) 2, we

find that,

x=3a, y=4b, z=5c.

Therefore, using Identity V, we have

(3a+4b+5c) 2

=(3a) 2+(4b) 2+(5c) 2+2(3a)(4b)+ 2(4b)(5c)+2(5c)(3a)

=9a2+16b2+25c2+24ab+40bc+30ac.

EXAMPLEQues: Factorize 4x2 +y2+z2-4xy-2yz+4zx

Ans: 4x2+y2+z2-4xy-2yz+4zx

= (2x) 2 +(-y) 2+(z) 2+2(2x)(-y)+ 2(-y)(z)+2(2x)(z)

= [2x+(-y)+z] 2

= (2x-y+z) 2

Practice Questions

1.Expand each of the following:

i)(x+2y+4z)² ii)(2x-y+z)²

iii)(3a-7b-c)² iv)[(1/4)a-(1/2)b+1]²

2.Factorize:

i)2x²+y²+8z²-22xy+42yz-8xz

ii) 4x²+9y²+16z²+12xy-24yz-16xz

Identity VI: (x + y)3 = x3 + y3 + 3xy (x+ y)

(x + y)3 = (x + y) (x + y)2

= (x + y)(x2 + 2xy + y2)

=x( x2 + 2xy + y2) + y(x2 + 2xy + y2)

= x3 + 2x2y + xy2 + x2y + 2xy2 + y3

= x3 + 3x2y + 3xy2 + y3

= x3 + y3 + 3xy(x + y)

So, we get the following identity:

(x + y)3 = x3 + y3 + 3xy (x+ y)

Identity VII: (x - y)3 = x3 - y3 - 3xy (x- y)

Identity VI : (x + y)3 = x3 + y3 + 3xy (x+ y)

By replacing y by –y in the Identity VI, we get

Identity VII : (x - y)3 = x3 - y3 - 3xy (x- y)

= x3 – 3x2y + 3xy2 – y3

QUESTIONS1) Write the following cubes in expanded form :

i)(2x+1)³ =(2x)³+1³+3(2x)(1)[(2x)+1]

=8x³+1+6x[2x+1]

= 8x³+1+12x²+6x

=8x³+12x²+6x+1

ii)(2a-3b)³ =(2a)³-(3b)³-3(2a)(3b)[(2a)-(3b)]

= 8a³-27b³-18ab(2a-3b)

=8a³-27b³- [36a²b-54ab²]

=8a³-27b³-36a²b+54ab²

QUESTIONSEvaluate the following using suitable identities: i)(99)³= (100-1)³ = (100)³-1³-3(100)(1)[100-1]

=1000000-1-300[100-1]

=1000000-1-30000+300

=1000300-30001=970299

ii)(102)³ =(100+2)³

=(100)³+(2)³+3(100)(2)[100+2]

=1000000+8+600[100+2]

=1000000+8+60000+1200

=1061208

QUESTIONS

Factorize:

i)8a³+b³+12a²b+6ab²

=(2a)³+(b)³+3(2a)²b+3(2a)(b)²

=(2a+b)³=(2a+b)(2a+b)(2a+b)

ii)8a³-b³-12a²b+6ab²

=(2a)³-(b)³-3(2a)²(b)+3(2a)(b)²

=(2a-b)³=(2a-b)(2a-b)(2a-b)

QUESTIONSVerify:

i)x³+y³=(x+y)(x²-xy+y²)

ii)x³-y³=(x-y)(x²+xy+y²)

SOLN: i)RHS=(x+y)(x²-xy+y²)

=x(x²-xy+y²)+y(x²-xy+y²)

=x³-x²y+xy²+x²y-xy²+y³ =x³+y³ =LHS

ii)RHS=(x-y)(x²+xy+y²)

=x(x²+xy+y²)-y(x²+xy+y²)

=x³+x²y+xy²-x²y-xy²-y³

=x³-y ³=LHS

)

QUESTIONSFactorize:

i)27y³+125z³=(3y)³+(5z)³

=(3y+5z)[(3y)²-(3y)(5z)+(5z)²]

=(3y+5z)(9y²-15yz+25z²)

ii) 64m³-343n³=(4m)³-(7n)³

=(4m-7n)[(4m)²+(4m)(7n)+(7n)²]

=(4m-7n)(16m²+28mn+49n²)

Identity VIIIx3 + y3 + z3- 3xyz = (x + y + z)(x2 + y2 + z2– xy – yz –

zx)Now consider (x + y + z)(x2+ y2 + z2– xy – yz – zx)

On expanding, we get the product as

x(x2 + y2 + z2– xy – yz – zx) + y(x2 + y2 + z2– xy – yz – zx) + z(x2 + y2 + z2– xy – yz – zx)

= x3 + xy2 + xz2 – x2y – xyz – zx2 + x2y + y3 + yz2 – xy2 – y2z – xyz + x2z + y2z + z3 – xyz – yz2 – xz2

= x3 + y3 + z3- 3xyz (On simplification)

QUESTIONS

Factorize:

27x³+y³+z³-9xyz

=(3x)³+(y)³+(z)³-3(3x)(y)(z)

=(3x+y+z)[(3x)²+(y)²+(z)²-(3x)(y)-(y)(z)-(z)(3x)

=(3x+y+z)(9x²+y²+z²-3xy-yz-3zx)

QUESTIONS

If x+y+z=0,show that x³+y³+z³=3xyz

SOLN: Since x+y+z=0 (given)

LHS = x³+y³+z³=(x³+y³+z³-3xyz)+3xyz

=(x+y+z)(x²+y²+z²-xy-yz-zx)+3xyz

=(0)(x²+y²+z²-xy-yz-zx)+3xyz

=0+3xyz =3xyz =RHS

QUESTIONS

Without actually calculating the cubes ,find the values of the following:

(-12)³+(7)³+(5)³

SOLN: Let x=(-12) ,y=7 ,z=5

then,x+y+z=(- 12)+7+5= -12+12=0

we know that if x+y+z=0,then x³+y³+z³=3xyz

(-12)³+(7)³+(5)³=3(-12)(7)(5)

= -1260

ACTIVITY: interpret geometrically the factors of a quadratic expression of the type ax²+bx+c (where a=1),using square grids.Take a=1 ,b=10 and c=21 to get the polynomial x 2 +10x +21 Take a square grid of dimension (10X10) which represents x2 [here x=10] . Find two numbers whose sum is 10 and product is 21 i.e.,7 and 3 .

X

X

X

7

X+7

Add 7 strips of dimensions x and 1 .Now the area of the rectangle formed is x²+7x

X 3

XX

7

Then add 3 strips of dimemsions x and 1 . Now the total area = (x2+7x)+3x

X 3

x+7

X

7

Add 21 smallsquares of dimension 1 and 1 to complete the rectangle .Now the area becomesX²+7X+3X+21The area of rectangle =(x+3)(x+7)=x2+7x+3x+21 =x2+10x+21

Activity 2to verify the algebraic identity: (a+b+c)+

Activity 2 : to verify the algebraic identity:(a+b+c)2= a2+ b2+ c2+ 2ab+ 2bc+ 2ca

i)Take a cardboard of convenient size and paste it on a white sheet.

ii)Cut three squares of sides a unit, b units and c units from red glazed paper

a

a

b

b

c

c

iii)Cut six rectangles from glazed paper as shown below

TYPE I :- Two rectangles from green glazed paper of length a units and breadth b units .

TYPE II :- two rectangles from blue glazed paper of length b units and breadth c units

b

a a

b

b

c c

b

TYPE III :- two rectangles from yellow glazed paper of length a units

and breadth c units as shown above.

c

a

c

a

a2 ab ac

b2

c2

bcab

ac bc

a b c

a

b

c

iv) Paste all six rectangles and three squares on the cardboard as shown below.