Chap-2 (8th Nov.)
MATHEMATICS2isPOLYNOMIALShed20bl2.1 Introduction no NCtt Eo Rbe TrepuIn Class IX, you have studied polynomials in one variable and their degrees. Recallthat if p(x) is a polynomial in x, the highest power of x in p(x) is called the degree ofthe polynomial p(x). For example, 4x 2 is a polynomial in the variable x ofdegree 1, 2y2 – 3y 4 is a polynomial in the variable y of degree 2, 5x3 – 4x2 x –is a polynomial in the variable x of degree 3 and 7u6 –in the variable u of degree 6. Expressions like23 4u 4 u 2 u 8 is a polynomial21,x 1x 2,1etc., arex 2x 32not polynomials.A polynomial of degree 1 is called a linear polynomial. For example, 2x – 3,223 x 5, y 2 , x , 3z 4, u 1 , etc., are all linear polynomials. Polynomials311such as 2x 5 – x2, x3 1, etc., are not linear polynomials.A polynomial of degree 2 is called a quadratic polynomial. The name ‘quadratic’2has been derived from the word ‘quadrate’, which means ‘square’. 2 x2 3 x ,5u21 2u2 5, 5v2 v, 4z 2 are some examples of337quadratic polynomials (whose coefficients are real numbers). More generally, anyquadratic polynomial in x is of the form ax 2 bx c, where a, b, c are real numbersand a 0. A polynomial of degree 3 is called a cubic polynomial. Some examples ofy2 – 2, 2 x2 3 x,File Name : C:\Computer Station\Class - X (Maths)/Final/Chap-2/Chap–2(8th Nov).pmd
P OLYNOMIALS21a cubic polynomial are 2 – x3, x3,2 x 3 , 3 – x2 x3 , 3x3 – 2x2 x – 1. In fact, the mostgeneral form of a cubic polynomial isdax3 bx 2 cx d,where, a, b, c, d are real numbers and a 0.heNow consider the polynomial p(x) x2 – 3x – 4. Then, putting x 2 in thepolynomial, we get p(2) 22 – 3 2 – 4 – 6. The value ‘– 6’, obtained by replacingx by 2 in x2 – 3x – 4, is the value of x2 – 3x – 4 at x 2. Similarly, p(0) is the value ofp(x) at x 0, which is – 4.isIf p(x) is a polynomial in x, and if k is any real number, then the value obtained byreplacing x by k in p(x), is called the value of p(x) at x k, and is denoted by p(k).p(–1) (–1) 2 –{3 (–1)} – 4 0Also, note thatp(4) 4 2 – (3 4) – 4 0.blWhat is the value of p(x) x2 –3x – 4 at x –1? We have : no NCtt Eo Rbe TrepuAs p(–1) 0 and p(4) 0, –1 and 4 are called the zeroes of the quadraticpolynomial x2 – 3x – 4. More generally, a real number k is said to be a zero of apolynomial p(x), if p(k) 0.You have already studied in Class IX, how to find the zeroes of a linearpolynomial. For example, if k is a zero of p(x) 2x 3, then p(k) 0 gives us32k 3 0, i.e., k 2 b In general, if k is a zero of p(x) ax b, then p(k) ak b 0, i.e., k a b (Constant term)So, the zero of the linear polynomial ax b is. aCoefficient of xThus, the zero of a linear polynomial is related to its coefficients. Does thishappen in the case of other polynomials too? For example, are the zeroes of a quadraticpolynomial also related to its coefficients?In this chapter, we will try to answer these questions. We will also study thedivision algorithm for polynomials.2.2 Geometrical Meaning of the Zeroes of a PolynomialYou know that a real number k is a zero of the polynomial p(x) if p(k) 0. But whyare the zeroes of a polynomial so important? To answer this, first we will see thegeometrical representations of linear and quadratic polynomials and the geometricalmeaning of their zeroes.File Name : C:\Computer Station\Class - X (Maths)/Final/Chap-2/Chap–2(8th Nov).pmd
22MATHEMATICS–22y 2x 3–17hexdConsider first a linear polynomial ax b, a 0. You have studied in Class IX that thegraph of y ax b is a straight line. For example, the graph of y 2x 3 is a straightline passing through the points (– 2, –1) and (2, 7).From Fig. 2.1, you can seeisthat the graph of y 2x 3intersects the x - axis mid-way3that is, at the point , 0 . 2 no NCtt Eo Rbe TrepuYou also know that the zero ofblbetween x –1 and x – 2,32x 3 is . Thus, the zero of2the polynomial 2x 3 is thex-coordinate of the point where thegraph of y 2x 3 intersects theFig. 2.1x-axis.In general, for a linear polynomial ax b, a 0, the graph of y ax b is a b straight line which intersects the x-axis at exactly one point, namely, , 0 . a Therefore, the linear polynomial ax b, a 0, has exactly one zero, namely, thex-coordinate of the point where the graph of y ax b intersects the x-axis.Now, let us look for the geometrical meaning of a zero of a quadratic polynomial.Consider the quadratic polynomial x2 – 3x – 4. Let us see what the graph* ofy x2 – 3x – 4 looks like. Let us list a few values of y x2 – 3x – 4 corresponding toa few values for x as given in Table 2.1.* Plotting of graphs of quadratic or cubic polynomials is not meant to be done by the students,nor is to be evaluated.File Name : C:\Computer Station\Class - X (Maths)/Final/Chap-2/Chap–2(8th Nov).pmd
P OLYNOMIALS23xy x2 – 3x – 4–2–101234560–4–6–6–406isbl no NCtt Eo Rbe TrepuIn fact, for any quadraticpolynomial ax2 bx c, a 0, thegraph of the correspondingequation y ax 2 bx c has oneof the two shapes either openupwards likeor openheIf we locate the points listedabove on a graph paper and drawthe graph, it will actually look likethe one given in Fig. 2.2.dTable 2.1downwards likedepending onwhether a 0 or a 0. (Thesecurves are called parabolas.)You can see from Table 2.1that –1 and 4 are zeroes of thequadratic polynomial. Alsonote from Fig. 2.2 that –1 and 4are the x-coordinates of the pointswhere the graph of y x2 – 3x – 4intersects the x - axis. Thus, thezeroes of the quadratic polynomialx2 – 3x – 4 are x -coordinates ofthe points where the graph ofy x 2 – 3x – 4 intersects thex -axis.Fig. 2.2This fact is true for any quadratic polynomial, i.e., the zeroes of a quadraticpolynomial ax2 bx c, a 0, are precisely the x -coordinates of the points where theparabola representing y ax 2 bx c intersects the x-axis.From our observation earlier about the shape of the graph of y ax 2 bx c, thefollowing three cases can happen:File Name : C:\Computer Station\Class - X (Maths)/Final/Chap-2/Chap–2(8th Nov).pmd
24MATHEMATICSCase (i) : Here, the graph cuts x-axis at two distinct points A and A′.The x-coordinates of A and A′ are the two zeroes of the quadratic polynomialax bx c in this case (see Fig. 2.3). no NCtt Eo Rbe Trepublished2Fig. 2.3Case (ii) : Here, the graph cuts the x-axis at exactly one point, i.e., at two coincidentpoints. So, the two points A and A′ of Case (i) coincide here to become one point A(see Fig. 2.4).Fig. 2.4The x -coordinate of A is the only zero for the quadratic polynomial ax 2 bx cin this case.File Name : C:\Computer Station\Class - X (Maths)/Final/Chap-2/Chap–2(8th Nov).pmd
P OLYNOMIALS25 no NCtt Eo Rbe TrepublishedCase (iii) : Here, the graph is either completely above the x -axis or completely belowthe x -axis. So, it does not cut the x - axis at any point (see Fig. 2.5).Fig. 2.5So, the quadratic polynomial ax 2 bx c has no zero in this case.So, you can see geometrically that a quadratic polynomial can have either twodistinct zeroes or two equal zeroes (i.e., one zero), or no zero. This also means that apolynomial of degree 2 has atmost two zeroes.Now, what do you expect the geometrical meaning of the zeroes of a cubicpolynomial to be? Let us find out. Consider the cubic polynomial x3 – 4x. To see whatthe graph of y x3 – 4x looks like, let us list a few values of y corresponding to a fewvalues for x as shown in Table 2.2.Table 2.2xy x3 – 4x–2–1012030–30Locating the points of the table on a graph paper and drawing the graph, we seethat the graph of y x3 – 4x actually looks like the one given in Fig. 2.6.File Name : C:\Computer Station\Class - X (Maths)/Final/Chap-2/Chap–2(8th Nov).pmd
26MATHEMATICSheisbl no NCtt Eo Rbe TrepuLet us take a few moreexamples. Consider the cubicpolynomials x3 and x3 – x2. We drawthe graphs of y x3 and y x3 – x2in Fig. 2.7 and Fig. 2.8 respectively.dWe see from the table abovethat – 2, 0 and 2 are zeroes of thecubic polynomial x3 – 4x. Observethat – 2, 0 and 2 are, in fact, thex -coordinates of the only pointswhere the graph of y x3 – 4xintersects the x -axis. Since the curvemeets the x - axis in only these 3points, their x -coordinates are theonly zeroes of the polynomial.Fig. 2.6Fig. 2.7File Name : C:\Computer Station\Class - X (Maths)/Final/Chap-2/Chap–2(8th Nov).pmdFig. 2.8
P OLYNOMIALS27dNote that 0 is the only zero of the polynomial x3. Also, from Fig. 2.7, you can seethat 0 is the x - coordinate of the only point where the graph of y x3 intersects thex -axis. Similarly, since x3 – x2 x2 (x – 1), 0 and 1 are the only zeroes of the polynomialx3 – x2 . Also, from Fig. 2.8, these values are the x - coordinates of the only pointswhere the graph of y x3 – x2 intersects the x-axis.heFrom the examples above, we see that there are at most 3 zeroes for any cubicpolynomial. In other words, any polynomial of degree 3 can have at most three zeroes.isRemark : In general, given a polynomial p(x) of degree n, the graph of y p(x)intersects the x- axis at atmost n points. Therefore, a polynomial p(x) of degree n hasat most n zeroes. no NCtt Eo Rbe TrepublExample 1 : Look at the graphs in Fig. 2.9 given below. Each is the graph of y p(x),where p(x) is a polynomial. For each of the graphs, find the number of zeroes of p(x).Fig. 2.9Solution :(i) The number of zeroes is 1 as the graph intersects the x- axis at one point only.(ii) The number of zeroes is 2 as the graph intersects the x -axis at two points.(iii) The number of zeroes is 3. (Why?)File Name : C:\Computer Station\Class - X (Maths)/Final/Chap-2/Chap–2(8th Nov).pmd
28MATHEMATICS(iv) The number of zeroes is 1. (Why?)(v) The number of zeroes is 1. (Why?)d(vi) The number of zeroes is 4. (Why?)heEXERCISE 2.1 no NCtt Eo Rbe Trepublis1. The graphs of y p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find thenumber of zeroes of p(x), in each case.Fig. 2.102.3 Relationship between Zeroes and Coefficients of a PolynomialYou have already seen that zero of a linear polynomial ax b is b . We will now tryato answer the question raised in Section 2.1 regarding the relationship between zeroesand coefficients of a quadratic polynomial. For this, let us take a quadratic polynomial,say p(x) 2x2 – 8x 6. In Class IX, you have learnt how to factorise quadraticpolynomials by splitting the middle term. So, here we need to split the middle term‘– 8x’ as a sum of two terms, whose product is 6 2x2 12x2. So, we write2x2 – 8x 6 2x2 – 6x – 2x 6 2x(x – 3) – 2(x – 3) (2x – 2)(x – 3) 2(x – 1)(x – 3)File Name : C:\Computer Station\Class - X (Maths)/Final/Chap-2/Chap–2(8th Nov).pmd
P OLYNOMIALS29 ( 8) (Coefficient of x) 2Coefficient of x 26Constant termProduct of its zeroes 1 3 3 2 Coefficient of x 2 1 3 4 heSum of its zeroesdSo, the value of p(x) 2x2 – 8x 6 is zero when x – 1 0 or x – 3 0, i.e., whenx 1 or x 3. So, the zeroes of 2x2 – 8x 6 are 1 and 3. Observe that : (3x – 1)(x 2)isLet us take one more quadratic polynomial, say, p(x) 3x2 5x – 2. By themethod of splitting the middle term,3x2 5x – 2 3x2 6x – x – 2 3x(x 2) –1(x 2)11or x –2. So, the zeroes of 3x2 5x – 2 are and – 2. Observe that :33 no NCtt Eo Rbe Trepuwhen x blHence, the value of 3x2 5x – 2 is zero when either 3x – 1 0 or x 2 0, i.e.,Sum of its zeroes Product of its zeroes 1 5 (Coefficient of x) ( 2) 33Coefficient of x 21 2Constant term ( 2) 33Coefficient of x 2In general, if α* and β* are the zeroes of the quadratic polynomial p(x) ax2 bx c,a 0, then you know that x – α and x – β are the factors of p(x). Therefore,ax2 bx c k(x – α) (x – β), where k is a constant k[x2 – (α β)x α β] kx2 – k(α β)x k α βComparing the coefficients of x2, x and constant terms on both the sides, we geta k, b – k(α β) and c kαβ.This givesα β αβ –b,aca* α,β are Greek letters pronounced as ‘alpha’ and ‘beta’ respectively. We will use later onemore letter ‘γ’ pronounced as ‘gamma’.File Name : C:\Computer Station\Class - X (Maths)/Final/Chap-2/Chap–2(8th Nov).pmd
30product of zeroes αβ b (Coefficient of x) ,aCoefficient of x2dsum of zeroes α β cConstant term .aCoefficient of x2hei.e.,MATHEMATICSLet us consider some examples.Solution : We haveblx2 7x 10 (x 2)(x 5)isExample 2 : Find the zeroes of the quadratic polynomial x2 7x 10, and verify therelationship between the zeroes and the coefficients.So, the value of x2 7x 10 is zero when x 2 0 or x 5 0, i.e., when x – 2 orx –5. Therefore, the zeroes of x2 7x 10 are – 2 and – 5. Now, (7) – (Coefficient of x ), 1Coefficient of x 2 no NCtt Eo Rbe Trepusum of zeroes – 2 (– 5) – (7) product of zeroes ( 2) ( 5) 10 10Constant term 1 Coefficient of x2Example 3 : Find the zeroes of the polynomial x2 – 3 and verify the relationshipbetween the zeroes and the coefficients.Solution : Recall the identity a 2 – b 2 (a – b)(a b). Using it, we can write:()(x2 – 3 x 3 x 3So, the value of x2 – 3 is zero when x Therefore, the zeroes of x2 – 3 are)3 or x – 3 3 and 3 Now,sum of zeroes product of zeroes 3 3 0 (Coefficient of x) ,Coefficient of x 2Constant term( 3 )( 3) – 3 13 Coefficientof xFile Name : C:\Computer Station\Class - X (Maths)/Final/Chap-2/Chap–2(8th Nov).pmd2
P OLYNOMIALS31Example 4 : Find a quadratic polynomial, the sum and product of whose zeroes are– 3 and 2, respectively.c.aisαβ 2 and b ,aheα β –3 dSolution : Let the quadratic polynomial be ax2 bx c, and its zeroes be α and β.We haveIf a 1, then b 3 and c 2.blSo, one quadratic polynomial which fits the given conditions is x2 3x 2.You can check that any other quadratic polynomial that fits these conditions willbe of the form k(x2 3x 2), where k is real. no NCtt Eo Rbe TrepuLet us now look at cubic polynomials. Do you think a similar relation holdsbetween the zeroes of a cubic polynomial and its coefficients?Let us consider p(x) 2x3 – 5x2 – 14x 8.1 Since p(x) can have atmost three2zeroes, these are the zeores of 2x3 – 5x2 – 14x 8. Now,You can check that p(x) 0 for x 4, – 2,sum of the zeroes 4 ( 2) product of the zeroes 4 ( 2) 1 5 ( 5) (Coefficient of x2 ) ,2 22Coefficient of x 31 8– Constant term. 4 22 Coefficient of x3However, there is one more relationship here. Consider the sum of the productsof the zeroes taken two at a time. We have{4 ( 2)} ( 2) 1 1 4 2 2 – 8 1 2 7 14Coefficient of x .2Coefficient of x3In general, it can be proved that if α, β, γ are the zeroes of the cubic polynomialax bx2 cx d, then3File Name : C:\Computer Station\Class - X (Maths)/Final/Chap-2/Chap–2(8th Nov).pmd
32MATHEMATICS–b,acαβ βγ γα ,a–dα β γ .aLet us consider an example.hedα β γ 1are the zeroes of the cubic polynomial3p(x) 3x3 – 5x2 – 11x – 3, and then verify the relationship between the zeroes and thecoefficients.blisExample 5* : Verify that 3, –1, Solution : Comparing the given polynomial with ax3 bx 2 cx d, we get no NCtt Eo Rbe Trepua 3, b – 5, c –11, d – 3. Furtherp(3) 3 33 – (5 32) – (11 3) – 3 81 – 45 – 33 – 3 0,p(–1) 3 (–1)3 – 5 (–1)2 – 11 (–1) – 3 –3 – 5 11 – 3 0,32 1 1 1 1 p 3 5 11 3 , 3 3 3 3 –1 5 112 2 3 – 09 933 31are the zeroes of 3x3 – 5x2 – 11x – 3.31So, we take α 3, β –1 and γ 3Now,1 5 ( 5) b , 1 α β γ 3 ( 1) 2 33 33a Therefore, 3, –1 and αβ β γ γ α 3 ( 1) ( 1) 1 1 1 11 c , 3 3 1 3 3 33a ( 3) d . 1 αβγ 3 ( 1) 1 3a 3 * Not from the examination point of view.File Name : C:\Computer Station\Class - X (Maths)/Final/Chap-2/Chap–2(8th Nov).pmd
P OLYNOMIALS33EXERCISE 2.2(i) x2 – 2x – 8(iii) 6x 2 – 3 – 7x(v) t2 – 15(vi) 3x 2 – x – 4he(iv) 4u2 8u(ii) 4s2 – 4s 1d1. Find the zeroes of the following quadratic polynomials and verify the relationship betweenthe zeroes and the coefficients.(v) 1, 14 42.4 Division Algorithm for Polynomials(vi) 4, 1bl(iv) 1, 1is2. Find a quadratic polynomial each with the given numbers as the sum and product of itszeroes respectively.1,1 1(i)(ii) 2 ,(iii) 0, 543 no NCtt Eo Rbe TrepuYou know that a cubic polynomial has at most three zeroes. However, if you are givenonly one zero, can you find the other two? For this, let us consider the cubic polynomialx3 – 3x2 – x 3. If we tell you that one of its zeroes is 1, then you know that x – 1 isa factor of x3 – 3x2 – x 3. So, you can divide x3 – 3x2 – x 3 by x – 1, as you havelearnt in Class IX, to get the quotient x2 – 2x – 3.Next, you could get the factors of x2 – 2x – 3, by splitting the middle term, as(x 1)(x – 3). This would give youx3 – 3x2 – x 3 (x – 1)(x2 – 2x – 3) (x – 1)(x 1)(x – 3)So, all the three zeroes of the cubic polynomial are now known to you as1, – 1, 3.Let us discuss the method of dividing one polynomial by another in some detail.Before noting the steps formally, consider an example.Example 6 : Divide 2x2 3x 1 by x 2.Solution : Note that we stop the division process wheneither the remainder is zero or its degree is less than thedegree of the divisor. So, here the quotient is 2x – 1 andthe remainder is 3. Also,x 22x – 122x 3x 122x 4x(2x – 1)(x 2) 3 2x2 3x – 2 3 2x2 3x 1i.e.,2x2 3x 1 (x 2)(2x – 1) 3Therefore, Dividend Divisor Quotient RemainderLet us now extend this process to divide a polynomial by a quadratic polynomial.File Name : C:\Computer Station\Class - X (Maths)/Final/Chap-2/Chap–2(8th Nov).pmd
34MATHEMATICSExample 7 : Divide 3x3 x2 2x 5 by 1 2x x2.x 2x 123x 6x 3 x––2–5 x – x 5–5x2 – 10 x – 5 9x 10d3–heSolution : We first arrange the terms of thedividend and the divisor in the decreasing orderof their degrees. Recall that arranging the termsin this order is called writing the polynomials instandard form. In this example, the dividend isalready in standard form, and the divisor, instandard form, is x2 2x 1.3x – 52blisStep 1 : To obtain the first term of the quotient, divide the highest degree term of thedividend (i.e., 3x3) by the highest degree term of the divisor (i.e., x2). This is 3x. Thencarry out the division process. What remains is – 5x2 – x 5. no NCtt Eo Rbe TrepuStep 2 : Now, to obtain the second term of the quotient, divide the highest degree termof the new dividend (i.e., –5x2) by the highest degree term of the divisor (i.e., x2). Thisgives –5. Again carry out the division process with – 5x2 – x 5.Step 3 : What remains is 9x 10. Now, the degree of 9x 10 is less than the degreeof the divisor x2 2x 1. So, we cannot continue the division any further.So, the quotient is 3x – 5 and the remainder is 9x 10. Also,(x2 2x 1) (3x – 5) (9x 10) 3x3 6x2 3x – 5x2 – 10x – 5 9x 10 3x3 x2 2x 5Here again, we see thatDividend Divisor Quotient RemainderWhat we are applying here is an algorithm which is similar to Euclid’s divisionalgorithm that you studied in Chapter 1.This says thatIf p(x) and g(x) are any two polynomials with g(x) 0, then we can findpolynomials q(x) and r(x) such thatp(x) g(x) q(x) r(x),where r(x) 0 or degree of r(x) degree of g(x).This result is known as the Division Algorithm for polynomials.Let us now take some examples to illustrate its use.Example 8 : Divide 3x2 – x3 – 3x 5 by x – 1 – x2, and verify the division algorithm.File Name : C:\Computer Station\Class - X (Maths)/Final/Chap-2/Chap–2(8th Nov).pmd
P OLYNOMIALS35x–22he–x x – 1 – x3 3 x2 – 3 x 532–x x – x – 2So, dividend –x3 3x 2 – 3x 5 and2 x – 2x 52divisor –x2 x – 1.2x – 2 x 2– –Division process is shown on the
20 MATHEMATICS File Name : C:\Computer Station\Class - X (Maths)/Final/Chap-2/Chap–2(8th Nov).pmd 2 2.1 Introduction In
Preparing the living compartment (inside)P.4 Chap. III CArrYInG PASSEnGErS P.5 Chap. IV LoADS AnD LoADED wEIGHt P.6 Chap. V wInDowS AnD SKYLIGHtS P.8 Chap. VI DrIVInG P. 10 Chap. VII tYrES P.11 Chap. VIII SIGnALInG, lights P.12 Chap. IX PArKInG mAnEuVErS P. 14 Chap. X tHE CAmPEr DrIVEr'S CHArtEr P. 16 Chap. XI oPErAtIon oF APPLIAnCES P.17
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