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? The 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? 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. 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.

Polynomials in one variable The degree of a polynomial in one variable is the largest exponent of that variable. A constant has no variable. It is a 0 degree polynomial. This is a 1st degree polynomial. 1st degree polynomials are linear. This is a 2nd degree polynomial. 2nd degree polynomials are quadratic. This is a 3rd degree polynomial. 3rd degree polynomials are cubic.

Examples 5 Constant Monomial 2x - 4 1 Linear Binomial 3x2 + x 2 Polynomials Degree Classify by degree Classify by no. of terms. 5 Constant Monomial 2x - 4 1 Linear Binomial 3x2 + x 2 Quadratic x3 - 4x2 + 1 3 Cubic Trinomial Text Txt Text Text Text

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 Solutions: (i) p(x) =5x-4x²+3 p(0) =5(0)-4(0)2+3=0-0+3=3 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

Examples Divide the polynomial 3x4 – 4x3 – 3x –1 by x – 1. Solution: By long division, we have:

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

EXAMPLE Ques: 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+8z²-22xy+42yz-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

QUESTIONS 1) 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

QUESTIONS Evaluate 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)

QUESTIONS Verify: 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 )

QUESTIONS Factorize: 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 VIII x3 + 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

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

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

squares of dimension 1 and 1 to complete the rectangle .Now X 3 Add 21 small squares of dimension 1 and 1 to complete the rectangle .Now the area becomes X²+7X+3X+21 The area of rectangle =(x+3)(x+7) =x2+7x+3x+21 =x2+10x+21 X 7 x+7

Activity 2 to 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 c b a c b a

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 b a a c c b b

TYPE III :- two rectangles from yellow glazed paper of length a units and breadth c units as shown above. c c a a

iv) Paste all six rectangles and three squares on the cardboard as shown below. ab ac a ab b2 bc b ac bc c2 c