Page 1 :
Polynomials and Factorisation, , 02, , 2.1 I NTRODUCTION, There are six rows and each row has six plants in a garden. How many plants are there in, total ? If there are ‘x’ plants, planted in ‘x’ rows then how many plants will be there, in the garden? Obviously it is x2., The cost of onions is `10 per kg. Inder purchased, p kg., Raju purchased q kg. and Hanif purchased r kg., How much each would have paid? The payments would, be `10p, `10q and `10r respectively. All such examples, show the use of algebraic expression., We also use algebraic expressions such as ‘s2’ to, find area of a square, ‘lb’ for area of a rectangle and ‘lbh’, for volume of a cuboid. What are the other algebraic expressions that we use?, Algebraic expressions such as 3xy, x2+2x, x3- x2 + 4x + 3, r2, ax + b etc. are called, polynomials. Note that, all algebraic expressions we have considered so far only have nonnegative integers as exponents of the variables., Can you find the polynomials among the given algebraic expressions:, x2,, , 1, , x 2 + 3,, , 3, 2x2 ! + 5;, x, , x2 + xy + y2, , 1, , 1, , From the above x 2 + 3 is not a polynomials because the first term x 2 is a term with an, 1, 3, ) and also 2x2 ! + 5 is not a polynomial, x, 2, + 5. Here the second term (3x!1) has a negative, , exponent that is not a non-negative integer (i.e., , because it can be written as 2x2 ! 3x!1, exponent. (i.e., !1). An algebraic expression in which the variables involved have only nonnegative integral powers is called a polynomial.

Page 2 :
28, , IX-CLASS MATHEMATICS, , THINK, DISCUSS AND WRITE, Which of the following expressions are polynomials ? Which are not ? Give reasons., (i), , 4x2 + 5x - 2, , (ii), , y2 - 8, , (iii), , (v), , 3x 2 " 5 y, , (vi), , 1, x "1, , (vii), , 5, , 2x2 ", , (iv), x, , 3, !5, x, , (viii) 3 xyz, , We shall start our study with polynomials in various forms. In this chapter we will also, learn factorisation of polynomials using Remainder Theorem and Factor Theorem and their use, in the factorisation of polynomials., , 2.2 POL, S, YNOMIAL, YNOMIALS, OLYNOMIAL, , IN ONE, , V ARIABLE, , Let us begin by recalling that a variable is denoted by a symbol that can take any real, value. We use the letters x, y, z etc. to denote variables. We have algebraic expressions, 3, x .... all in one variable x. These expressions, 4, are of the form (a constant) # (some power of variable). Now,, suppose we want to find the perimeter of a square we use the, formula P = 4s., , s, , 2x, 3x, !x,, , Here ‘4’ is a constant and ‘s’ is a variable, representing the, side of a square. The side could vary for different squares., , s, , s, , s, , Observe the following table:, Side of square, , Perimeter, , (s), , (4s), , 4 cm, , P = 4 # 4 = 16 cm, , 5 cm, , P = 4 # 5 = 20 cm, , 10 cm, , P = 4 # 10 = 40 cm, , Here the value of the constant i.e. ‘4’ remains the same throughout this situation. That is,, the value of the constant does not change in a given problem, but the value of the variable (s), keeps changing., Suppose we want to write an expression which is of the form ‘(a constant) # (a variable)’, and we do not know, what the constant is, then we write the constants as a, b, c ... etc. So, , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 3 :
POLYNOMIALS AND FACTORISATION, , 29, , these expressions in general will be ax, by, cz, .... etc. Here a, b, c ... are arbitrary constants., You are also familiar with other algebraic expressions like x2, x2 + 2x + 1, x3 + 3x2 ! 4x + 5., All these expressions are polynomials in one variable., , DO THESE, • Write two polynomials with variable ‘x’, • Write three polynomials with variable ‘y’, • Is the polynomial 2x2 + 3xy + 5y2 in one variable ?, • Write the formulae of area and volume of different solid shapes. Find out the variables, and constants in them., , 2.3 DEGREE OF THE POL, YNOMIAL, POLYNOMIAL, Each term of the polynomial consists of the products of a constant, called the coefficient, of the term and a finite number of variables raised to non-negative integral powers. Degree of a, term is the sum of the exponent of is variable factors. And the degree of a polynomial is the, largest degree of its variable term., Lets find the terms, their coefficients and the degree of polynomials:, (i) 3x2 + 7x + 5, , (ii) 3x2y2 + 4xy + 7, , In the polynomial 3x2 + 7x + 5, each of the expressions 3x2, 7x and 5 are terms. Each, term of the polynomial has a coefficient, so in 3x2 + 7x + 5, the coefficient of x2 is 3, the, coefficient of x is 7 and 5 is the coefficient of x0 (Remember x0=1), You know that the degree of a polynomial is the highest degree of its variable term., As the term 3x2 has the highest degree among all the other terms in that expression, Thus, the degree of 3x2 + 7x + 5 is ‘2’., Now can you find coefficient and degree of polynomial 3x2y3 + 4xy + 7., The coefficient of x2y3 is 3, xy is 4 and x0y0 is 7. The sum of the exponents of the, variables in term 3x2y3 is 2 + 3 = 5 which is greater than that of the other terms. So the degree, of polynomial 3x2y3 + 4xy + 7 is 5., Now think what is the degree of a constant? As the constant contains no variable, it can be, written as product of x0. For example, degree of 5 is zero as it can be written as 5x0. Now that, you have seen what a polynomial of degree 1, degree 2, or degree 3 looks like, can you write, , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 4 :
30, , IX-CLASS MATHEMATICS, , down a polynomial in one variable of degree n for any natural number n? A polynomial in one, variable x of degree n is an expression of the form, anxn + an–1xn–1 + . . . + a1x + a0, where a0, a1, a2, . . ., an are constants and an $ 0., In particular, if a0 = a1 = a2 = a3 = . . . = an = 0 (i.e. all the coefficients are zero), we get, the zero polynomial, which is the number ‘0’., Can you say the degree of zero? It is not defined as we can’t write it as a product of a, variable raised to any power., , DO THESE, 1. Write the degree of each of the following polynomials, (i), (iii), , 7x3 + 5x2 + 2x ! 6, , (ii) 7 ! x + 3x2, , 5p !, , (v) !5xy2, , 3, , (iv) 2, , 2. Write the coefficient of x2 in each of the following, (i), , 15 ! 3x + 2x2 (ii) 1 ! x2, , (iii), , x2 ! 3x + 5, , (iv), , 2 x 2 " 5x ! 1, , Let us observe the following tables and fill the blanks., (i) Types of polynomials according to degree :, Degree of a, polynomial, , Name of the, polynomial, , Example, , Not defined, , Zero polynomial, , 0, , Zero, , Constant polynomial, , !12; 5;, , 1, , ......................................, , x ! 12; !7x + 8; ax + b etc., , 2, , Quadratic polynomial, , ......................................, , 3, , Cubic polynomial, , 3x3 ! 2x2 + 5x + 7, , 3, etc, 4, , Usually, a polynomial of degree ‘n’ is called nth degree polynomial., , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 5 :
POLYNOMIALS AND FACTORISATION, , 31, , (ii) Types of polynomials according to number of terms:, Number of non - zero, terms, , Name of the, polynomial, , Example, , Terms, , 1, , Monomial, , !3x, , !3x, , 2, , Binomial, , 3x + 5, , 3x, 5, , 3, , Trinomial, , 2x2 + 5x + 1, , ..........................., , Multinomial, , ..........................., , 3x3, 2x2, !7x, 5, , More than 3, , Note : Every polynomial is a multinomial but every multinomial need not be a polynomial., A linear polynomial with one variable may be a monomial or a binomial., Eg : 3x or 2x ! 5, , THINK, DISCUSS AND WRITE, How many terms a cubic (degree 3) polynomial with one variable can have?, Give examples., If the variable in a polynomial is x, we may denote the polynomial by p(x), q(x) or r(x), etc. So for example, we may write, p(x) = 3x2 + 2x + 1, q(x) = x3 ! 5x2 + x ! 7, r(y) = y4 ! 1, t(z) = z2 + 5z + 3, , TRY THESE, 1. Write a polynomial with, 2 terms in variable x., 2. How can you write a, polynomial with 15 terms, in variable p ?, , A polynomial can have any finite number of terms., So far mostly we have discussed the polynomials in one variable only. We can also have, polynomials in more than one variable. For example x + y, x2 + 2xy + y2, x2 - y2 are, polynomials in two variables x, y. Similarly x2 + y2 + z2, x3 + y3 + z3 are polynomials in three, variables. You will study such polynomials later in detail., , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 6 :
32, , IX-CLASS MATHEMATICS, , EXERCISE - 2.1, 1. Find the degree of each of the polynomials given below, (i), (iv), , x5 ! x4 + 3, , (ii) x2 + x ! 5, , 3x6 + 6y3 ! 7 (v) 4 ! y2, , (iii) 5, (vi) 5t !, , 3, , 2. Which of the following expressions are polynomials in one variable and which are not ?, Give reasons for your answer., (i), , 3x2 ! 2x + 5 (ii) x2 +, , (v), , 5 x"x 5, , 2, , (iii) p2 ! 3p + q, , (iv) y ", , 2, y, , (vi) x100 + y100, , 3. Write the coefficient of x3 in each of the following, (i), , x3 + x + 1, , (v), , x3 " x, , 2, , (ii) 2 ! x3 + x2, (vi), , 2, ! x3, 3, , (iii), , 2 x3 " 5, , (vii) 2x2 + 5, , (iv) 2x3 + 5, (vi) %, , 4. Classify the following as linear, quadratic and cubic polynomials, (i), (v), , 5x2 + x ! 7, 3p, , (ii) x ! x3, (vi), , r, , (iii) x2 + x + 4, , (iv) x ! 1, , 2, , 5. Write whether the following statements are True or False. Justify your answer, (i) A binomial can have at the most two terms, (ii) Every polynomial is a binomial, (iii) A binomial may have degree 3, (iv) Degree of zero polynomial is zero, (v) The degree of x2 + 2xy + y2 is 2, (vi), , r2 is monomial., , 6. Give one example each of a monomial and trinomial of degree 10., , 2.4 ZEROES, l, , OF A POL, POLYNOMIAL, YNOMIAL, , Consider the polynomial p(x) = x2 + 5x + 4., What is the value of p(x) for x = 1., For this we have to replace x by 1 every where in p(x), , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 7 :
POLYNOMIALS AND FACTORISATION, , By doing this, , 33, , p(1) = (1)2 + 5(1) + 4,, , we get, , = 1 + 5 + 4 = 10, , So, we say that the value of p(x) at x = 1 is 10, Similarly find p(x) for x = 0 and x = !1, p(0) = (0)2 + 5(0) + 4, , p(!1) = (!1)2 + 5(!1) + 4, , = 0+0+4, , = 1!5+4, , = 4, , =0, , Can you find the value of p(!4)?, l, , Consider another polynomial, s(y) = 4y4 ! 5y3 ! y2 + 6, s(1) = 4(1)4 ! 5(1)3 ! (1)2 + 6, = 4(1) ! 5(1) ! 1 + 6, = 4!5!1+6, = 10 ! 6, = 4, Can you find s(!1) = ?, , DO THIS, Find the value of each of the following polynomials for the indicated value of, variables:, (i) p(x) = 4x2 ! 3x + 7 at x = 1, (ii) q(y) = 2y3 ! 4y + 11 at y = 1, (iii) r(t) = 4t4 + 3t3 ! t2 + 6 at t = p, t & R, (iv) s(z) = z3 ! 1 at z = 1, (v) p (x) = 3x2 + 5x - 7 at x = 1, (vi) q (z) = 5z3 - 4z +, , l, , 2 at z = 2, , Now consider the polynomial r (t) = t ! 1, What is r(1) ? It is r(1) = 1 ! 1 = 0, As r(1) = 0, we say that 1 is a zero of the polynomial r(t)., In general, we say that a zero of a polynomial p(x) is the value of x, for which p(x) = 0., , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 8 :
34, , IX-CLASS MATHEMATICS, , This value is also called a root of the polynomial p (x), What is the zero of polynomial f(x) = x + 1?, You must have observed that the zero of the polynomial x + 1 is obtained by equating, it to 0. i.e., x + 1 = 0, which gives x = !1. If f(x) is a polynomial in x then f(x) = 0 is called a, polynomial equation in x . We observe that ‘!1’ is the root of the polynomial f(x) in the above, example So we say that ‘!1’ is the zero of the polynomial x + 1, or a root of the polynomial, equation x + 1 = 0., l, , Now, consider the constant polynomial 3., Can you tell what is its zero ? It does not, have a zero. As 3 = 3x0 no real value of x, gives value of 3x0. Thus a constant polynomial, has no zeroes. But zero polynomial is a, constant polynomial having many zeros., , TRY THESE, Find zeroes of the following, polynomials, 1. 2x ! 3, 2. x2 ! 5x + 6, 3. x + 5, , Example-1. p(x) = x + 2. Find p(1), p(2), p(!1) and p(!2). What are zeroes of the, polynomial x + 2?, Solution : Let p(x) = x + 2, replace x by 1, p(1) = 1 + 2 = 3, replace x by 2, p(2) = 2 + 2 = 4, replace x by !1, p(!1) = !1 + 2 = 1, replace x by !2, p(!2) = !2 + 2 = 0, Therefore, ' , 2, !1 are not the zeroes of the polynomial x + 2, but !2 is the zero of the, polynomial., Example-2., , Find a zero of the polynomial p(x) = 3x + 1, , Solution : Finding a zero of p(x), is same as solving the equation, p(x) = 0, i.e., , 3x + 1 = 0, 3x = !1, x = !, , 1, 3, , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 9 :
POLYNOMIALS AND FACTORISATION, , So, !, , 35, , 1, is a zero of the polynomial 3x + 1., 3, , Example-3., , Find a zero of the polynomial 2x ! 1., , Solution : Finding a zero of p(x), is the same as solving the equation p(x) = 0, As 2x ! 1 = 0, x =, , 1, (how ?), 2, , (1), Check it by finding the value of P * +, ,2-, , Now, if p(x) = ax + b, a $ 0, a linear polynomial, how will you find a zero of p(x)?, As we have seen to find zero of a polynomial p(x), we need to solve the polynomial, equation p(x) = 0, Which means ax + b = 0, a $ 0, So ax = !b, i.e., x =, , !b, a, , !b, is the only zero of the polynomial p(x) = ax + b i.e., A linear polynomial, a, in one variable has only one zero., , So, x =, , DO THIS, Fill in the blanks:, Linear, Polynomial, , Zero of the polynomial, , x+a, , !a, , x!a, , -------------, , ax + b, , -------------, , ax ! b, , b, a, , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 10 :
36, , IX-CLASS MATHEMATICS, , Verify whether 2 and 1 are zeroes of the polynomial x2 ! 3x +2, , Example-4., , Solution : Let p(x) = x2 ! 3x + 2, replace x by 2, p(2) = (2)2 ! 3(2) + 2, = 4!6+2=0, also replace x by 1, p(1) = (1)2 ! 3(1) + 2, =1!3+2, =0, Hence, both 2 and 1 are zeroes of the polynomial x2 ! 3x + 2., Is there any other way of checking this?, What is the degree of the polynomial x2 ! 3x + 2 ? Is it a linear polynomial ? No,, it is a quadratic polynomial. Hence, a quadratic polynomial has two zeroes., Example-5. If 3 is a zero of the polynomial x2 + 2x ! a. Find a?, Solution : Let p(x) = x2 + 2x ! a, As the zero of this polynomial is 3, we know that p(3) = 0., x2 + 2x ! a = 0, Put x = 3, (3)2 + 2 (3) ! a = 0, 9+6!a =0, 15 ! a = 0, .!a = !15, a = 15, , or, , THINK AND DISCUSS, 1. x2 + 1 has no zeros. Why?, 2. Can you tell the number of zeroes a polynomial of degree ‘n’ will have?, , EXERCISE - 2.2, 1. Find the value of the polynomial 4x2 ! 5x + 3, when, (i) x = 0, , (ii) x = !1, , (iii) x = 2, , (iv) x =, , 1, 2, , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 11 :
POLYNOMIALS AND FACTORISATION, , 37, , 2. Find p(0), p(1) and p(2) for each of the following polynomials., p(x) = x2! x +1, , (ii) p(y) = 2 + y + 2y2 ! y3, , (iii), , p(z) = z3, , (iv) p(t) = (t ! 1) (t + 1), , (v), , 2, , (i), , p(x) = x ! 3x + 2, , 3. Verify whether the values of x given in each case are the zeroes of the polynomial or not?., (i), , p(x) = 2x + 1; x = !, , 1, 2, , (ii) p(x) = 5x !. ; x =, , !3, 2, , (iii), , p(x) = x2 !.1; x = +1, , (iv) p(x) = (x - 1)(x ". 2); x = !.1, !2, , (v), , p(y) = y2; y = 0, , (vi) p(x) = .ax + b ; x = !, , (vii), , f(x) = 3x2 ! 1; x / !, , 1, 2, ,, 3, 3, , (viii) f (x) = 2x - 1, x =, , b, a, , 1 !1, ,, 2 2, , 4. Find the zero of the polynomial in each of the following cases., (i), , f(x) = x " 2, , (ii) f(x) = x ! 2, , (iii) f(x) = 2x + 3, , (iv), , f(x) = 2x ! 3 (v) f(x) = x2, , (vii), , f(x) = px + q, p $ 0, p q are real numbers., , (vi) f(x) = px, p $ 0, , 5. If 2 is a zero of the polynomial p(x) = 2x2 ! 3x + 7a, find the value of a., 6. If 0 and 1 are the zeroes of the polynomial f (x) = 2x3 ! 3x2 + ax + b, find the values, of a and b., , 2.5 DIVIDING POL, S, YNOMIAL, YNOMIALS, OLYNOMIAL, Observe the following examples, (i) Let us consider two numbers 25 and 3. Divide 25 by 3. We get the quotient 8 and, remainder 1. We write, Dividend = (Divisor # Quotient) + Remainder, So, 25 = (8 # 3) + 1, Similarly divide 20 by 5, we get 20 = (4 # 5) + 0, The remainder here is 0. In this case we say that 5 is a factor of 20 or 20 is a multiple, of 5., As we divide a number by another non-zero number, we can also divide a polynomial by, another polynomial? Let’s see., , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 12 :
38, , IX-CLASS MATHEMATICS, , (ii) Divide the polynomial 3x3 + x2 + x by the monomial x., 3x 3 x 2 x, " ", x, x x, 2, = 3x + x + 1, , We have (3x3 + x2 + x) 0 x =, , In fact x is a common factor to each term of 3x3 + x2 + x So we can write, 3x3 + x2 + x as x(3x2 + x + 1), What are the factors of 3x3 + x2 + x ?, (iii) Consider another example (2x2 + x + 1) 0 x, , 2x " 1, 2, x 2x " x " 1, !2 x 2, x "1, , 2 x2 x 1, " ", Here, (2x + x + 1) 0 x =, x, x x, 2, , = 2x + 1 +, , 1, x, , !x, 1, , Is it a polynomial ?, As one of the term, , 1, 1, has a negative integer exponent (i.e. = x!1), x, x, , 1, is not a polynomial., x, We can however write, , 1 2x + 1 +, , (2x2 + x + 1) = [x # (2x + 1)] + 1, By taking out 1 separately the rest of the polynomial can be written as product of two, polynomials., Here we can say (2x + 1) is the quotient, x is the divisor and 1 is the remainder. We must, keep in mind that since the remainder is not zero, ‘x’ is not a factor of 2x2 + x + 1., , DO THESE, 1. Divide 3y3 + 2y2 + y by ‘y’ and write division fact, 2. Divide 4p2 + 2p + 2 by ‘2p’ and write division fact., Example-6., , Divide 3x2 + x ! 1 by x + 1., , Solution : Consider p(x) = 3x2 + x ! 1 and q(x) = x + 1., Divide p(x) by q(x). Recall the division process you have learnt in earlier classes., 3x2, / 3x , it becomes first term in quotient., Step 1: Divide, x, FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 13 :
POLYNOMIALS AND FACTORISATION, , Step 2 : Multiply (x + 1) 3x = 3x2 + 3x, , 39, , 3x ! 2, , by subtracting 3x2 + 3x from 3x2 + x, we get !2x, , x " 1 3x 2 " x ! 1, , !2 x, / !2 , it becomes the 2nd term in the quotient., Step 3 : Divide, x, Step 4 : Multiply (x + 1)(!2) = !2x ! 2,, Subtract it from !2x ! 1, which gives ‘1’., , Step 5 : We stop here as the remainder is 1, a constant., (Can you tell why a constant is not divided by a polynomial?), , 3 x 2 " 3x, ! !, ! 2x !1, ! 2x ! 2, " ", "1, , This gives us the quotient as (3x ! 2) and remainder (+1)., Note : The division process is said to be complete if we get the remainder 0 or the degree of, the remainder is less than the degree of the divisor., Now, we can write the division fact as, 3x2 + x ! 1 = (x + 1) (3x ! 2) + 1, i.e. Dividend = (Divisor # Quotient) + Remainder., Let us see by replacing x by !1 in p(x), p(x), , = 3x2 + x ! 1, , p(!1) = 3(!1)2 + (!1) ! 1, , It is observed that the value of p(!1), is equal the remainder; i.e. is ‘1’., , = 3(+1) + (!1) ! 1 = 1., So, the remainder obtained on dividing p(x) = 3x2 + x ! 1 by (x + 1) is same as p (-1), where -1 is the zero of x + 1. i.e. x = !1., 2 x3 ! 2 x 2 ! 2 x ! 5, Let us consider some more examples., x ! 1 2 x 4 ! 4 x3 ! 3x ! 1, Example-7. Divide the polynomial 2x4 ! 4x3 ! 3x ! 1, 2 x 4 ! 2 x3, by (x ! 1) and verify the remainder with zero of the divisor., !, ", 4, 3, Solution : Let f(x) = 2x ! 4x ! 3x ! 1, ! 2 x3 ! 3x ! 1, First see how many times 2x4 is of x., 2 x4, / 2 x3, x, Now multiply (x ! 1) (2x3) = 2x4 ! 2x3, Then again see the first term of the remainder that, is !2x3. Now do the same., , FREE DISTRIBUTION BY A.P. GOVERNMENT, , ! 2 x3 " 2 x 2, ", !, ! 2 x 2 ! 3x ! 1, ! 2x2 " 2x, ", !, ! 5x ! 1, ! 5x " 5, " !, !6

Page 14 :
40, , IX-CLASS MATHEMATICS, , Here the quotient is 2x3 ! 2x2 ! 2x ! 5 and the remainder is !6., Now, the zero of the polynomial (x ! 1) is 1., Put x = 1 in f (x), f (x) = 2x4 ! 4x3 ! 3x ! 1, f(1) = 2(1)4 ! 4(1)3 ! 3(1) ! 1, = 2(1) ! 4(1) ! 3(1) ! 1, =2!4!3!1, = !6, Is the remainder same as the value of the polynomial f(x) at zero of (x ! 1) ?, From the above examples we shall now state the fact in the form of the following theorem., It gives a remainder without actual division of a polynomial by a linear polynomial in one, variable., Remainder Theorem :, 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)., Let us now look at the proof of this theorem., Proof : Let p(x) be any polynomial with degree greater than or equal to 1., Further suppose that when p(x) is divided by a linear polynomial g(x) = (x ! a), the, quotient is q(x) and the remainder is r(x). In other words, p(x) and g(x) are two polynomials, such that the degree of p(x) > degree of g(x) and g(x) $ 0 then we can find polynomials q(x), and r(x) such that, where r(x) = 0 or degree of r(x) < degree of g(x)., By division algorithm,, p(x) = g(x) . q(x) + r(x), 1 p(x) = (x ! a) . q(x) + r(x), , g(x) = (x ! a), , Since the degree of (x ! a) is 1 and the degree of r(x) is less than the degree of (x ! a)., 1 Degree of r(x) = 0, implies r(x) is a constant, say K., , So, for every real value of x, r(x) = K., Therefore,, p(x) = (x ! a) q(x) + K, If x = a, then p(a) = (a ! a) q(a) + K, =0+K, =K, , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 15 :
POLYNOMIALS AND FACTORISATION, , 41, , Hence proved., Let us use this result in finding remainders when a polynomial is divided by a linear polynomial, without actual division., Example-8., , Find the remainder when x3 + 1 divided by (x + 1), , Solution : Here p(x) = x3 + 1, The zero of the linear polynomial x + 1, , is .!1 [x + 1 = 0, x = !1], , So replacing x by !1, p(!1) = (!1)3 + 1, = !1 + 1, =0, So, by Remainder Theorem, we know that (x3 + 1) divided by (x + 1) gives 0 as the, remainder., You can also check this by actually dividing x3 + 1 by x + 1., Can you say (x + 1) is a factor of (x3 + 1) ?, Example-9., , Check whether (x ! 2) is a factor of x3 ! 2x2 ! 5x + 4, , Solution : Let p(x) = x3 ! 2x2 ! 5x + 4, To check whether the linear polynomial (x ! 2) is a factor of the given polynomial,, Replace x, by the zero of (x ! 2) i.e. x ! 2 = 0 2 x = 2., p(2) = (2)3 ! 2(2)2 ! 5(2) + 4, = 8 ! 2(4) ! 10 + 4, = 8 ! 8 ! 10 + 4, = ! 6., As the remainder is not equal to zero, the polynomial (x ! 2) is not a factor of the given, polynomial x3 ! 2x2 ! 5x + 4., Example10., , Check whether the polynomial p(y) = 4y3 + 4y2 ! y ! 1 is a multiple of (2y + 1)., , Solution : As you know, p(y) will be a multiple of (2y + 1) only, if (2y + 1) divides p(y) exactly., We shall first find the zero of the divisor , 2y + 1, i.e., y =, , FREE DISTRIBUTION BY A.P. GOVERNMENT, , !1, ,, 2

Page 16 :
42, , IX-CLASS MATHEMATICS, , Replace y by, , !1, in p(y), 2, 3, , 2, , ( !1 ), ( !1 ), ( !1 ) ( !1 ), p * + / 4 * + " 4 * + ! * + !1, , 2 , 2 , 2 - , 2 ( !1), ( 1) 1, / 4* + " 4* + " !1, , 8, 4- 2, /, , !1, 1, "1" !1, 2, 2, , =0, So, (2y + 1) is a factor of p(y). That is p(y) is a multiple of (2y + 1)., Example-11. If the polynomials ax3 + 3x2 ! 13 and 2x3 ! 5x + a are divided by (x ! 2) leave, the same remainder, find the value of a., Solution : Let p(x) = ax3 + 3x2 ! 13 and q(x) = 2x3 ! 5x " a, p(x) and q(x) when divided by x ! 2 leave same remainder., 1 p(2) = q(2), a(2)3 + 3(2)2 ! 13 = 2(2)3 ! 5(2) + a, 8a + 12 ! 13 = 16 ! 10 + a, 8a ! 1 = a + 6, 8a ! a = 6 + 1, 7a = 7, a=1, , EXERCISE - 2.3, 1. Find the remainder when x3 + 3x2 + 3x + 1 is divided by the following, Linear polynomials :, (i), , x+1, , (v), , 5 + 2x, , (ii) x !., , 1, 2, , (iii) x, , (iv) x +, , 2. Find the remainder when x3 ! px2 + 6x ! p is divided by x ! p., , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 17 :
POLYNOMIALS AND FACTORISATION, , 43, , 3. Find the remainder when 2x2 ! 3x + 5 is divided by 2x ! 3. Does it exactly divide the, polynomial ? State reason., 2, 4. Find the remainder when 9x3 ! 3x2 + x ! 5 is divided by x !, 3, 5. If the polynomials 2x3 + ax2 + 3x ! 5 and x3 + x2 ! 4x + a leave the same remainder, when divided by x ! 2, find the value of a., 6. If the polynomials x3 + ax2 " 5 and x3 ! 2x2 + a are divided by (x " 2) leave the same, remainder, find the value of a., 7. Find the remainder when f (x) = x4 ! 3x2 + 4 is divided by g(x) = x ! 2 and verify the, result by actual division., 8. Find the remainder when p(x) = x3 ! 6x2 + 14x ! 3 is divided by g(x) = 1 ! 2x and, verify the result by long division., 9. When a polynomial 2x3 +3x2 + ax + b is divided by (x ! 2) leaves remainder 2, and, (x + 2) leaves remainder !2. Find a and b., , 2.6 FACT, ORISING A POL, YNOMIAL, CTORISING, POLYNOMIAL, As we have already studied that a polynomial q(x) is said to have divided a polynomial, p(x) exactly if the remainder is zero. In this case q(x) is a factor of p(x)., For example. When, is zero (verify), , p(x) = 4x3 + 4x2 ! x ! 1 is divided by g(x) = 2x + 1, if the remainder, , then 4x3 + 4x2 ! x ! 1 = q(x) (2x + 1) + 0, So, , p(x) = q(x) (2x + 1), , Therefore g(x) = 2x + 1 is a factor of p(x)., With the help of Remainder Theorem can you state a theorem that helps find the factors of, a given polynomial ?, Factor Theorem : 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 (ii) and its converse, “if (x ! a) is a factor of a polynomial p(x) then p(a) = 0., Let us see the simple proof of this theorem., Proof : By Remainder Theorem,, p(x) = (x ! a) q(x) + p(a), (i) Consider proposition (i) If p(a) = 0, then p(x) = (x ! a) q(x) + 0., = (x ! a) q(x), , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 18 :
44, , IX-CLASS MATHEMATICS, , Which shows that (x ! a) is a factor of p(x)., Hence proved., (ii) Consider proposition (ii) Since (x ! a) is a factor of p(x), then p(x) = (x ! a) q(x) for, some polynomial q(x), 1 p(a) = (a ! a) q(a), , = 0, 1 Hence p(a) = 0 when (x ! a) is a factor of p(x), Let us consider some examples., Example-12. Examine whether x + 2 is a factor of x3 + 2x2 + 3x + 6, Solution : Let p(x) = x3 + 2x2 + 3x + 6 and g(x) = x + 2, The zero of g(x) is !2, Then, , p(!2) = (!2)3 + 2(!2)2 + 3(!2) + 6, = ! 8 + 2(4) ! 6 + 6, =!8+8!6+6, =0, , So, by the Factor Theorem, x + 2 is a factor of x3 + 2x2 + 3x + 6., Example-13. Find the value of K, if 2x ! 3 is a factor of 2x3 ! 9x2 + x + K., Solution : (2x ! 3) is a factor of p(x) = 2x3 ! 9x2 + x + K,, If (2x ! 3) = 0, x =, , 3, 2, , 1 The zero of (2x ! 3) is, , 3, 2, , (3), If (2x ! 3) is a factor of p(x) then p * + / 0, ,23, 2, p(x) = 2x ! 9x + x + K,, 3, , 2, , 3, ( 3), ( 3), ( 3), 2 p* + / 2* + ! 9* + " " K / 0, , 2, 2, 22, ( 27 ), ( 9) 3, 2 2* + ! 9* + " " K / 0, , 8, 4- 2, , ( 27 81 3, ), 2 * ! " " K = 0+ # 4, , 4, 4 2, , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 19 :
POLYNOMIALS AND FACTORISATION, , 45, , 27 ! 81 + 6 + 4K = 0, !48 + 4K = 0, 4K = 48, So, , K = 12, , Example-14. Show that (x ! 1) is a factor of x10 ! 1 and also of x11 ! 1., Solution : Let p(x) = x10 ! 1 and g(x) = x11 ! 1, To prove (x ! 1) is a factor of both p(x) and g(x), it is sufficient to show that p(1) = 0, and g(1) = 0., Now, p(x) = x10 ! 1, , and, , g(x) = x11 ! 1, , p(1) = (1)10 ! 1, , and, , g(1) = (1)11 ! 1, , =1!1, , =1!1, , =0, , =0, , Thus by Factor Theorem,, (x ! 1) is a factor of both p(x) and g(x)., , TRY THESE, Show that (x - 1) is a, factor of xn - 1., , We shall now try to factorise quadratic polynomial of the type ax2 + bx + c, (where, a $ 0 and a, b, c are constants)., Let its factors be (px + q) and (rx + s)., Then ax2 + bx + c = (px + q) (rx + s), = prx2 + (ps + qr)x + qs, By comparing the coefficients of x2, x and constants, we get that,, a = pr, b = ps + qr, c = qs, This shows that b is the sum of two numbers ps and qr,, whose product is (ps) (qr) = (pr)(qs), = ac, Therefore, to factorise ax2 + bx + c, we have to write b as the sum of two numbers, whose product is ac., , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 20 :
46, , IX-CLASS MATHEMATICS, , Example-15. Factories 3x2 + 11x + 6, Solution : If we can find two numbers p and q such that p + q = 11 and pq = 3 # 6 = 18, then, we can get the factors., So, let us see the pairs of factors of 18., (1, 18), (2, 9), (3, 6) of these pairs, 2 and 9 will satisfy p + q = 11, So, , 3x2 + 11x + 6 = 3x2 + 2x + 9x + 6, = x(3x + 2) + 3(3x + 2), = (3x + 2) (x + 3)., , DO THESE, Factorise the following, 1., , 6x2 + 19x + 15, , 2.10m2 - 31m - 132, , 3. 12x2 + 11x + 2, , Now, consider an example., Example-16. Verify whether 2x4 ! 6x3 + 3x2 + 3x ! 2 is divisible by x2 ! 3x + 2 or not ?, How can you verify using Factor Theorem ?, Solution : The divisor is not a linear polynomial. It is a quadratic polynomial. You have learned, the factorisation of a quadratic polynomial by splitting the middle term as follows., x2 ! 3x + 2 = x2 ! 2x ! x + 2, = x(x ! 2) ! 1(x ! 2), = (x ! 2) (x ! 1)., To show x2 ! 3x + 2 is a factor of polynomial 2x4 ! 6x3 + 3x2 + 3x ! 2, we have to, show (x ! 2) and (x .!1) are the factors of 2x4 ! 6x3 + 3x2 + 3x ! 2., Let, then, , p(x) = 2x4 ! 6x3 + 3x2 + 3x ! 2, p(2) = 2(2)4 ! 6(2)3 + 3(2)2 + 3(2) ! 2, = 2(16) ! 6(8) + 3(4) + 6 ! 2, = 32 ! 48 + 12 + 6 ! 2, = 50 ! 50, =0, , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 21 :
POLYNOMIALS AND FACTORISATION, , 47, , As p(2) = 0, (x ! 2) is a factor of p(x)., Then, , p(1), , = 2(1)4 ! 6(1)3 + 3(1)2 + 3(1) ! 2, = 2(1) ! 6(1) + 3(1) + 3 ! 2, = 2!6+3+3!2, = 8!8, = 0, , As, , p(1) = 0, (x ! 1) is a factor of p(x)., , As both (x ! 2) and (x ! 1) are factors of p(x), the product x2 ! 3x + 2 will also be a, factor of p(x) = 2x4 ! 6x3 + 3x2 + 3x ! 2., Example-17. Factorise x3 ! 23x2 + 142x ! 120, Solution : Let p(x) = x3 ! 23x2 + 142x ! 120, By trial, we find that p(1) = 0. (verify), So (x ! 1) is a factor of p(x), When we divide p (x) by (x ! 1), we get x2 - 22x + 120., Another way of doing this is,, x3 ! 23x2 + 142x ! 120 = x3 ! x2 ! 22x2 + 22x + 120x ! 120, = x2(x ! 1) ! 22x (x ! 1) + 120 (x ! 1) (why?), = (x ! 1) (x2 ! 22x + 120), Now x2 ! 22x + 120 is a quadratic expression that can be factorised by splitting the, middle term. We have, x2 ! 22x + 120 = x2 ! 12x ! 10x + 120, = x(x ! 12) ! 10 (x ! 12), = (x ! 12) (x ! 10), So, x3 ! 23x2 " 142x ! 120 = (x ! 1) (x ! 10) (x ! 12)., , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 22 :
48, , IX-CLASS MATHEMATICS, , EXERCISE - 2.4, 1. Determine which of the following polynomials has (x + 1) as a factor., (i), (iii), , x3 ! x2 ! x + 1, , (ii) x4 ! x3 + x2 ! x + 1, , x4 + 2x3 + 2x2 + x + 1, , (iv) x3 ! x2 !.(3 !, , 3) x +, , 3, , 2. Use the Factor Theorem to determine whether g(x) is factor of f(x) in each of the, following cases :, (i) f(x) = 5x3 + x2 ! 5x ! 1, g(x) = x + 1, (ii) f(x) = x3 + 3x2 " 3x " 1, g(x) = x + 1, (iii) f(x) = x3 ! 4x2 + x " 6, g(x) = x ! 2, (iv) f(x) = 3x3 " x2 ! 20x " 12, g(x) = 3x ! 2, (v) f(x) = 4x3 + 20x2 + 33x + 18, g(x) = 2x + 3, 3. Show that (x ! 2), (x " 3) and (x ! 4) are factors of x3 ! 3x2 ! 10x + 24., 4. Show that (x " 4), (x ! 3) and (x ! 7) are factors of x3 ! 6x2 ! 19x + 84., 1), (, 5. If both (x ! 2) and * x ! + are factors of px2 + 5x + r, show that p = r., 2,, , 6. If (x2 ! 1) is a factor of ax4 + bx3 " cx2 + dx + e, show that a " c +e = b + d = 0, 7. Factorize (i) x3 ! 2x2 ! x + 2, (iii) x3 + 13x2 + 32x + 20, , (ii) x3 ! 3x2 ! 9x ! 5, (iv) y3 + y2 ! y ! 1, , 8. If ax2 + bx + c and bx2 + ax + c have a common factor x + 1 then show that, c = 0 and a = b., 9. If x2 ! x ! 6 and x2 + 3x ! 18 have a common factor (x ! a) then find the value of a., 10. If (y ! 3) is a factor of y3 ! 2y2 ! 9y ! 18 then find the other two factors., , 2.6, , ALGEBRAIC IDENTITIES, Recall that an algebraic Identity is an algebraic equation that is true for all values of the, , variables occurring in it. You have studied the following algebraic identities in earlier classes, Identity I : (x + y)2 3 x2 + 2xy + y2, Identity II : (x ! y)2 3 x2 ! 2xy + y2, , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 23 :
POLYNOMIALS AND FACTORISATION, , 49, , Identity III : (x + y) (x ! y) 3 x2 ! y2, Identity IV : (x + a)(x + b) 3 x2 + (a + b)x + ab., , Step-II, Step-III, , y, , Subtract length y from x., Calculate for (x ! y)2, = x2 ! [(x ! y) y + (x ! y) y + y2], , (x-y) × y, , (x-y), , x, , 2, , (x-y), , = x2 ! xy + y2 ! xy + y2 ! y2, = x2 ! 2xy + y2, , (x-y), x, , (y×y), (x-y) × y, , Geometrical Proof :, For Identity, (x ! y)2, Step-I, Make a square of side x., , y, , TRY THIS, Try to draw the geometrical figures for other identities., (i), , ( x " y ) 2 3 x 2 " 2 xy " y 2, , (ii) ( x " y )( x ! y ) 3 x 2 ! y 2, , (iii), , ( x " a )( x " b) 3 x 2 " ( a " b) x " ab, , DO THESE, Find the following product using appropriate identities, (i) (x + 5) (x + 5), (ii) (p ! 3) (p + 3), (iii) (y ! 1) (y ! 1), (iv) (t + 2) (t + 4), (v) 102 # 98, Identities are useful in factorisation of algebraic expressions. Let us see some examples., Example-18. Factorise, (i) x2 + 5x + 4, (iii) 25a2 + 40ab + 16b2, , (ii) 9x2 ! 25, (iv) 49x2 ! 112xy + 64y2, , Solution :, (i) Here x2 + 5x + 4 = x2 + (4 + 1)x + (4) (1), Comparing with Identity (x + a) (x + b) 3 x2 + (a + b)x + ab, we get (x + 4) (x + 1)., (ii) 9x ! 25 = (3x)2 ! (5)2, 2, , Now comparing it with Identity III, x2 ! y2 3 (x + y) (x ! y), we get, ..1 9x2 ! 25 = (3x + 5) (3x ! 5)., , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 30 :
56, , IX-CLASS MATHEMATICS, , 1, , length = (x + 5), breadth = (2x ! 1), Let, x = 1, l = 6, b = 1, x = 2, l = 7, b = 3, x = 3, l = 8, b = 5, .............................., .............................., Can you find more values ?, , EXERCISE - 2.5, 1. Use suitable identities to find the following products, (i), (iv), , (ii) (x !.5) (x !.5) (iii) (3x ".2)(3x !.2), , (x + 5) (x + 2), , ( 2 1 )( 2 1 ), * x " 2 +* x ! 2 +, x -,, x ,, , (1 + x) (1 + x), , (v), , 2. Evaluate the following products without actual multiplication., (i), , 101 # 99, , (ii) 999 # 999, , (iv), , 501 # 501, , (v) 30.5 # 29.5, , 1, 1, (iii) 50 # 49, 2, 2, , 3. Factorise the following using appropriate identities., (i), , 16x2 + 24xy + 9y2, , (ii) 4y2 ! 4y + 1, , y2, 25, , (iii), , 4 x2 !, , (v), , x2 + 5x + 6, , (iv) 18a2 ! 50, (vi) 3p2 ! 24p + 36, , 4. Expand each of the following, using suitable identities, (i), (iv), , (x + 2y + 4z)2 (ii) (2a ! 3b)3, (a b ), * ! " 1+, ,4 2 -, , 2, , (v) (p + 1), , 3, , (iii) (!2a + 5b ! 3c)2, 2 ), (, (vi) * x ! y +, 3 ,, , 3, , 5. Factorise, (i), , 25x2 + 16y2 + 4z2 ! 40xy + 16yz ! 20xz, , (ii), , 9a2 + 4b2 + 16c2 + 12ab ! 16bc ! 24ca, , FREE DISTRIBUTION BY A.P. GOVERNMENT

Page 32 :
58, , IX-CLASS MATHEMATICS, , WHA, T, HAT, , WE HA, HAVE, VE DISCUSSED, , In this chapter, you have studied the following points., 1. A polynomial p(x) in one variable x is an algebraic expression in x of the form, p(x) = anxn + an!1xn!1 + ....... + a2x2 + a1 x + a0, where a0, a1, a2, .... an are, respectively the coefficients of x0, x1, x2, .... xn and n is called the degree of the, polynomial if an $ 0. Each anxn ; an!1xn!1 ; .... a0, is called a term of the polynomial, p(x)., 2. Polynomials are classified as monomial, binomial, trinomial etc. according to the number, of terms in it., 3. Polynomials are also named as linear polynomial, quadratic polynomial, cubic polynomial, etc. according to the degree of the polynomial., 4. A real number ‘a’ is a zero of a polynomial p(x) if p(a) = 0. In this case, ‘a’ is also, called a root of the polynomial equation p(x) = 0., 5. Every linear polynomial in one variable has a unique zero, a non-zero constant polynomial, has no zero., 6. Remainder Theorem : If p(x) is any polynomial of degree greater than or equal to 1 and, p(x) is divided by the linear polynomial (x ! a), then the remainder is p(a)., 7. Factor Theorem : If x ! a is a factor of the polynomial p(x), then p(a) = 0. Also if, p(a) = 0 then (x ! a) is a factor of p(x)., 8. Some Algebraic Identities are:, (i) (x + y + z)2 3 x2 + y2 + z2 + 2xy + 2yz + 2zx, (ii) (x + y)3 3 x3 + y3 + 3xy(x + y), (iii) (x ! y)3 3 x3 ! y3 ! 3xy(x ! y), (iv) x3 + y3 + z3 ! 3xyz 3 (x + y + z) (x2 + y2 + z2 ! xy ! yz ! zx) also, (v) x3 + y3 3 (x + y) (x2 ! xy + y2), (vi) x3 ! y3 3 (x ! y) (x2 + xy + y2), , Brain teaser, If, , x " x " x " ... / x x x ... then, , what is the value of x., , FREE DISTRIBUTION BY A.P. GOVERNMENT