20 MATHEMATICS POLYNOMIALS


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1 0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of the polynomil p(x). For exmple, 4x + is polynomil in the vrible x of degree 1, y y + 4 is polynomil in the vrible y of degree, 5x 4x + x is polynomil in the vrible x of degree nd 7u u + u + u is polynomil 1 in the vrible u of degree 6. Expressions like x 1, x +, 1 etc., re x + x + not polynomils. A polynomil of degree 1 is clled liner polynomil. For exmple, x, x + 5, y +, x, z + 4, u + 1, etc., re ll liner polynomils. Polynomils 11 such s x + 5 x, x + 1, etc., re not liner polynomils. A polynomil of degree is clled qudrtic polynomil. The nme qudrtic hs been derived from the word qudrte, which mens squre. x + x, 5 y u 1, x + x, u + 5, 5 v v, 4z + re some exmples of 7 qudrtic polynomils (whose coefficients re rel numbers). More generlly, ny qudrtic polynomil in x is of the form x + bx + c, where, b, c re rel numbers nd 0. A polynomil of degree is clled cubic polynomil. Some exmples of
2 POLYNOMIALS 1 cubic polynomil re x, x, x, x + x, x x + x 1. In fct, the most generl form of cubic polynomil is x + bx + cx + d, where,, b, c, d re rel numbers nd 0. Now consider the polynomil p(x) = x x 4. Then, putting x = in the polynomil, we get p() = 4 = 6. The vlue 6, obtined by replcing x by in x x 4, is the vlue of x x 4 t x =. Similrly, p(0) is the vlue of p(x) t x = 0, which is 4. If p(x) is polynomil in x, nd if k is ny rel number, then the vlue obtined by replcing x by k in p(x), is clled the vlue of p(x) t x = k, nd is denoted by p(k). Wht is the vlue of p(x) = x x 4 t x = 1? We hve : p( 1) = ( 1) { ( 1)} 4 = 0 Also, note tht p(4) = 4 ( 4) 4 = 0. As p( 1) = 0 nd p(4) = 0, 1 nd 4 re clled the zeroes of the qudrtic polynomil x x 4. More generlly, rel number k is sid to be zero of polynomil p(x), if p(k) = 0. You hve lredy studied in Clss IX, how to find the zeroes of liner polynomil. For exmple, if k is zero of p(x) = x +, then p(k) = 0 gives us k + = 0, i.e., k = b In generl, if k is zero of p(x) = x + b, then p(k) = k + b = 0, i.e., k = b (Constnt term) So, the zero of the liner polynomil x + b is =. Coefficient of x Thus, the zero of liner polynomil is relted to its coefficients. Does this hppen in the cse of other polynomils too? For exmple, re the zeroes of qudrtic polynomil lso relted to its coefficients? In this chpter, we will try to nswer these questions. We will lso study the division lgorithm for polynomils.. Geometricl Mening of the Zeroes of Polynomil You know tht rel number k is zero of the polynomil p(x) if p(k) = 0. But why re the zeroes of polynomil so importnt? To nswer this, first we will see the geometricl representtions of liner nd qudrtic polynomils nd the geometricl mening of their zeroes.
3 MATHEMATICS Consider first liner polynomil x + b, 0. You hve studied in Clss IX tht the grph of y = x + b is stright line. For exmple, the grph of y = x + is stright line pssing through the points (, 1) nd (, 7). x y = x From Fig..1, you cn see tht the grph of y = x + intersects the x xis midwy between x = 1 nd x =, tht is, t the point, 0. You lso know tht the zero of x + is. Thus, the zero of the polynomil x + is the xcoordinte of the point where the grph of y = x + intersects the Fig..1 xxis. In generl, for liner polynomil x + b, 0, the grph of y = x + b is b stright line which intersects the xxis t exctly one point, nmely,, 0. Therefore, the liner polynomil x + b, 0, hs exctly one zero, nmely, the xcoordinte of the point where the grph of y = x + b intersects the xxis. Now, let us look for the geometricl mening of zero of qudrtic polynomil. Consider the qudrtic polynomil x x 4. Let us see wht the grph* of y = x x 4 looks like. Let us list few vlues of y = x x 4 corresponding to few vlues for x s given in Tble.1. * Plotting of grphs of qudrtic or cubic polynomils is not ment to be done by the students, nor is to be evluted.
4 POLYNOMIALS Tble.1 x y = x x If we locte the points listed bove on grph pper nd drw the grph, it will ctully look like the one given in Fig... In fct, for ny qudrtic polynomil x + bx + c, 0, the grph of the corresponding eqution y = x + bx + c hs one of the two shpes either open upwrds like or open downwrds like depending on whether > 0 or < 0. (These curves re clled prbols.) You cn see from Tble.1 tht 1 nd 4 re zeroes of the qudrtic polynomil. Also note from Fig.. tht 1 nd 4 re the xcoordintes of the points where the grph of y = x x 4 intersects the xxis. Thus, the zeroes of the qudrtic polynomil x x 4 re xcoordintes of the points where the grph of y = x x 4 intersects the xxis. Fig.. This fct is true for ny qudrtic polynomil, i.e., the zeroes of qudrtic polynomil x + bx + c, 0, re precisely the xcoordintes of the points where the prbol representing y = x + bx + c intersects the xxis. From our observtion erlier bout the shpe of the grph of y = x + bx + c, the following three cses cn hppen:
5 4 MATHEMATICS Cse (i) : Here, the grph cuts xxis t two distinct points A nd A. The xcoordintes of A nd A re the two zeroes of the qudrtic polynomil x + bx + c in this cse (see Fig..). Fig.. Cse (ii) : Here, the grph cuts the xxis t exctly one point, i.e., t two coincident points. So, the two points A nd A of Cse (i) coincide here to become one point A (see Fig..4). Fig..4 The xcoordinte of A is the only zero for the qudrtic polynomil x + bx + c in this cse.
6 POLYNOMIALS 5 Cse (iii) : Here, the grph is either completely bove the xxis or completely below the xxis. So, it does not cut the xxis t ny point (see Fig..5). Fig..5 So, the qudrtic polynomil x + bx + c hs no zero in this cse. So, you cn see geometriclly tht qudrtic polynomil cn hve either two distinct zeroes or two equl zeroes (i.e., one zero), or no zero. This lso mens tht polynomil of degree hs tmost two zeroes. Now, wht do you expect the geometricl mening of the zeroes of cubic polynomil to be? Let us find out. Consider the cubic polynomil x 4x. To see wht the grph of y = x 4x looks like, let us list few vlues of y corresponding to few vlues for x s shown in Tble.. Tble. x y = x 4x Locting the points of the tble on grph pper nd drwing the grph, we see tht the grph of y = x 4x ctully looks like the one given in Fig..6.
7 6 MATHEMATICS We see from the tble bove tht, 0 nd re zeroes of the cubic polynomil x 4x. Observe tht, 0 nd re, in fct, the xcoordintes of the only points where the grph of y = x 4x intersects the xxis. Since the curve meets the xxis in only these points, their xcoordintes re the only zeroes of the polynomil. Let us tke few more exmples. Consider the cubic polynomils x nd x x. We drw the grphs of y = x nd y = x x in Fig..7 nd Fig..8 respectively. Fig..6 Fig..7 Fig..8
8 POLYNOMIALS 7 Note tht 0 is the only zero of the polynomil x. Also, from Fig..7, you cn see tht 0 is the xcoordinte of the only point where the grph of y = x intersects the xxis. Similrly, since x x = x (x 1), 0 nd 1 re the only zeroes of the polynomil x x. Also, from Fig..8, these vlues re the xcoordintes of the only points where the grph of y = x x intersects the xxis. From the exmples bove, we see tht there re t most zeroes for ny cubic polynomil. In other words, ny polynomil of degree cn hve t most three zeroes. Remrk : In generl, given polynomil p(x) of degree n, the grph of y = p(x) intersects the xxis t tmost n points. Therefore, polynomil p(x) of degree n hs t most n zeroes. Exmple 1 : Look t the grphs in Fig..9 given below. Ech is the grph of y = p(x), where p(x) is polynomil. For ech of the grphs, find the number of zeroes of p(x). Fig..9 Solution : (i) The number of zeroes is 1 s the grph intersects the xxis t one point only. (ii) The number of zeroes is s the grph intersects the xxis t two points. (iii) The number of zeroes is. (Why?)
9 8 MATHEMATICS (iv) The number of zeroes is 1. (Why?) (v) The number of zeroes is 1. (Why?) (vi) The number of zeroes is 4. (Why?) EXERCISE.1 1. The grphs of y = p(x) re given in Fig..10 below, for some polynomils p(x). Find the number of zeroes of p(x), in ech cse. Fig..10. Reltionship between Zeroes nd Coefficients of Polynomil You hve lredy seen tht zero of liner polynomil x + b is b. We will now try to nswer the question rised in Section.1 regrding the reltionship between zeroes nd coefficients of qudrtic polynomil. For this, let us tke qudrtic polynomil, sy p(x) = x 8x + 6. In Clss IX, you hve lernt how to fctorise qudrtic polynomils by splitting the middle term. So, here we need to split the middle term 8x s sum of two terms, whose product is 6 x = 1x. So, we write x 8x + 6 = x 6x x + 6 = x(x ) (x ) =(x )(x ) = (x 1)(x )
10 POLYNOMIALS 9 So, the vlue of p(x) = x 8x + 6 is zero when x 1 = 0 or x = 0, i.e., when x = 1 or x =. So, the zeroes of x 8x + 6 re 1 nd. Observe tht : ( 8) (Coefficient of x) Sum of its zeroes = 1+ = 4 = = Coefficient of x Product of its zeroes = 1 = = 6 = Constnt term Coefficient of x Let us tke one more qudrtic polynomil, sy, p(x) = x + 5x. By the method of splitting the middle term, x + 5x = x + 6x x = x(x + ) 1(x + ) =(x 1)(x + ) Hence, the vlue of x + 5x is zero when either x 1 = 0 or x + = 0, i.e., when x = 1 or x =. So, the zeroes of x + 5x re 1 nd. Observe tht : 1 5 (Coefficient of x) Sum of its zeroes = + ( ) = = Coefficient of x 1 Constnt term Product of its zeroes = ( ) = = Coefficient of x In generl, if α* nd β* re the zeroes of the qudrtic polynomil p(x) = x + bx + c, 0, then you know tht x α nd x β re the fctors of p(x). Therefore, x + bx + c = k(x α) (x β), where k is constnt = k[x (α + β)x + α β] = kx k(α + β)x + k α β Compring the coefficients of x, x nd constnt terms on both the sides, we get = k, b = k(α + β) nd c = kαβ. This gives α + β = b, αβ = c * α,β re Greek letters pronounced s lph nd bet respectively. We will use lter one more letter γ pronounced s gmm.
11 0 MATHEMATICS b (Coefficient of x) i.e., sum of zeroes = α + β = =, Coefficient of x product of zeroes = αβ = Let us consider some exmples. c Constnt term =. Coefficient of x Exmple : Find the zeroes of the qudrtic polynomil x + 7x + 10, nd verify the reltionship between the zeroes nd the coefficients. Solution : We hve x + 7x + 10 = (x + )(x + 5) So, the vlue of x + 7x + 10 is zero when x + = 0 or x + 5 = 0, i.e., when x = or x = 5. Therefore, the zeroes of x + 7x + 10 re nd 5. Now, (7) (Coefficient of x) sum of zeroes = + ( 5) = (7) = =, 1 Coefficient of x 10 Constnt term product of zeroes = ( ) ( 5) = 10 = = 1 Coefficient of x Exmple : Find the zeroes of the polynomil x nd verify the reltionship between the zeroes nd the coefficients. Solution : Recll the identity b = ( b)( + b). Using it, we cn write: x = ( x )( x + ) So, the vlue of x is zero when x = or x = Therefore, the zeroes of x re nd Now, sum of zeroes = = 0 = (Coefficient of x), Coefficient of x Constnt term = = = 1 Coefficient of x product of zeroes = ( )( )
12 POLYNOMIALS 1 Exmple 4 : Find qudrtic polynomil, the sum nd product of whose zeroes re nd, respectively. Solution : Let the qudrtic polynomil be x + bx + c, nd its zeroes be α nd β. We hve α + β = = b, c nd αβ = =. If = 1, then b = nd c =. So, one qudrtic polynomil which fits the given conditions is x + x +. You cn check tht ny other qudrtic polynomil tht fits these conditions will be of the form k(x + x + ), where k is rel. Let us now look t cubic polynomils. Do you think similr reltion holds between the zeroes of cubic polynomil nd its coefficients? Let us consider p(x) = x 5x 14x + 8. You cn check tht p(x) = 0 for x = 4,, 1 Since p(x) cn hve tmost three zeroes, these re the zeores of x 5x 14x + 8. Now, sum of the zeroes = product of the zeroes = 1 5 ( 5) (Coefficient of x ) 4 + ( ) + = = =, Coefficient of x 1 8 Constnt term 4 ( ) = 4 = =. Coefficient of x However, there is one more reltionship here. Consider the sum of the products of the zeroes tken two t time. We hve ( ) + ( ) + 4 { } = 14 Coefficient of x 8 1+ = 7= = Coefficient of x. In generl, it cn be proved tht if α, β, γ re the zeroes of the cubic polynomil x + bx + cx + d, then
13 MATHEMATICS α + β + γ = b, αβ + βγ + γα = c, α β γ = d. Let us consider n exmple. 1 Exmple 5* : Verify tht, 1, re the zeroes of the cubic polynomil p(x) = x 5x 11x, nd then verify the reltionship between the zeroes nd the coefficients. Solution : Compring the given polynomil with x + bx + cx + d, we get =, b = 5, c = 11, d =. Further p() = (5 ) (11 ) = = 0, p( 1) = ( 1) 5 ( 1) 11 ( 1) = = 0, p = 5 11, = = + = Therefore,, 1 nd re the zeroes of x 5x 11x. 1 So, we tke α =, β = 1 nd γ = Now, ( 5) b α+β+γ= + ( 1) + = = = =, αβ+ βγ +γα = ( 1) + ( 1) + = + 1= = c, 1 ( ) αβγ = ( 1) = 1= = * Not from the exmintion point of view. d.
14 POLYNOMIALS EXERCISE. 1. Find the zeroes of the following qudrtic polynomils nd verify the reltionship between the zeroes nd the coefficients. (i) x x 8 (ii) 4s 4s + 1 (iii) 6x 7x (iv) 4u + 8u (v) t 15 (vi) x x 4. Find qudrtic polynomil ech with the given numbers s the sum nd product of its zeroes respectively. (i) 1, 1 4 (ii) 1, (iii) 0, 5 (iv) 1, 1 (v).4 Division Algorithm for Polynomils 1, 1 (vi) 4, You know tht cubic polynomil hs t most three zeroes. However, if you re given only one zero, cn you find the other two? For this, let us consider the cubic polynomil x x x +. If we tell you tht one of its zeroes is 1, then you know tht x 1 is fctor of x x x +. So, you cn divide x x x + by x 1, s you hve lernt in Clss IX, to get the quotient x x. Next, you could get the fctors of x x, by splitting the middle term, s (x + 1)(x ). This would give you x x x + = (x 1)(x x ) =(x 1)(x + 1)(x ) So, ll the three zeroes of the cubic polynomil re now known to you s 1, 1,. Let us discuss the method of dividing one polynomil by nother in some detil. Before noting the steps formlly, consider n exmple. Exmple 6 : Divide x + x + 1 by x +. x 1 Solution : Note tht we stop the division process when x + x + x + 1 either the reminder is zero or its degree is less thn the x + 4x degree of the divisor. So, here the quotient is x 1 nd the reminder is. Also, (x 1)(x + ) + = x + x + = x + x + 1 i.e., x + x + 1 = (x + )(x 1) + Therefore, Dividend = Divisor Quotient + Reminder Let us now extend this process to divide polynomil by qudrtic polynomil.
15 4 MATHEMATICS Exmple 7 : Divide x + x + x + 5 by 1 + x + x. Solution : We first rrnge the terms of the dividend nd the divisor in the decresing order of their degrees. Recll tht rrnging the terms in this order is clled writing the polynomils in stndrd form. In this exmple, the dividend is lredy in stndrd form, nd the divisor, in stndrd form, is x + x + 1. x + x + 1 Step 1 : To obtin the first term of the quotient, divide the highest degree term of the dividend (i.e., x ) by the highest degree term of the divisor (i.e., x ). This is x. Then crry out the division process. Wht remins is 5x x + 5. Step : Now, to obtin the second term of the quotient, divide the highest degree term of the new dividend (i.e., 5x ) by the highest degree term of the divisor (i.e., x ). This gives 5. Agin crry out the division process with 5x x + 5. Step : Wht remins is 9x Now, the degree of 9x + 10 is less thn the degree of the divisor x + x + 1. So, we cnnot continue the division ny further. So, the quotient is x 5 nd the reminder is 9x Also, (x + x + 1) (x 5) + (9x + 10) = x + 6x + x 5x 10x 5 + 9x + 10 =x + x + x + 5 Here gin, we see tht Dividend = Divisor Quotient + Reminder Wht we re pplying here is n lgorithm which is similr to Euclid s division lgorithm tht you studied in Chpter 1. This sys tht If p(x) nd g(x) re ny two polynomils with g(x) 0, then we cn find polynomils q(x) nd r(x) such tht p(x) = g(x) q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of g(x). This result is known s the Division Algorithm for polynomils. Let us now tke some exmples to illustrte its use. x 5 x + 6x +x 5 x x x 10x x + 10 Exmple 8 : Divide x x x + 5 by x 1 x, nd verify the division lgorithm.
16 POLYNOMIALS 5 Solution : Note tht the given polynomils re not in stndrd form. To crry out division, we first write both the dividend nd divisor in decresing orders of their degrees. So, dividend = x + x x + 5 nd divisor = x + x 1. Division process is shown on the right side. x x + x 1 x + x x + 5 x + x x + + x x+ 5 x x+ + We stop here since degree () = 0 < = degree ( x + x 1). So, quotient = x, reminder =. Now, Divisor Quotient + Reminder =( x + x 1) (x ) + = x + x x + x x + + = x + x x + 5 = Dividend In this wy, the division lgorithm is verified. Exmple 9 : Find ll the zeroes of x 4 x x + 6x, if you know tht two of its zeroes re nd. Solution : Since two zeroes re nd x x + = x is fctor of the given polynomil. Now, we divide the given polynomil by x., ( )( ) x x + 1 x x 4 x x +6x x 4 4x + First term of quotient is x x 4 = x x + x + 6 x x + 6x + x x + 0 Second term of quotient is x x Third term of quotient is x = x 1 = x
17 6 MATHEMATICS So, x 4 x x + 6x = (x )(x x + 1). Now, by splitting x, we fctorise x x + 1 s (x 1)(x 1). So, its zeroes re given by x = 1 nd x = 1. Therefore, the zeroes of the given polynomil re 1,,, nd 1. EXERCISE. 1. Divide the polynomil p(x) by the polynomil g(x) nd find the quotient nd reminder in ech of the following : (i) p(x) = x x + 5x, g(x) = x (ii) p(x) = x 4 x + 4x + 5, g(x) = x + 1 x (iii) p(x) = x 4 5x + 6, g(x) = x. Check whether the first polynomil is fctor of the second polynomil by dividing the second polynomil by the first polynomil: (i) t, t 4 + t t 9t 1 (ii) x + x + 1, x 4 + 5x 7x + x + (iii) x x + 1, x 5 4x + x + x Obtin ll other zeroes of x 4 + 6x x 10x 5, if two of its zeroes re nd 4. On dividing x x + x + by polynomil g(x), the quotient nd reminder were x nd x + 4, respectively. Find g(x). 5. Give exmples of polynomils p(x), g(x), q(x) nd r(x), which stisfy the division lgorithm nd (i) deg p(x) = deg q(x) (ii) deg q(x) = deg r(x) (iii) deg r(x) = 0 EXERCISE.4 (Optionl)* 1. Verify tht the numbers given longside of the cubic polynomils below re their zeroes. Also verify the reltionship between the zeroes nd the coefficients in ech cse: 1 (i) x + x 5x + ;, 1, (ii) x 4x + 5x ;, 1, 1. Find cubic polynomil with the sum, sum of the product of its zeroes tken two t time, nd the product of its zeroes s, 7, 14 respectively. *These exercises re not from the exmintion point of view.
18 POLYNOMIALS 7. If the zeroes of the polynomil x x + x + 1 re b,, + b, find nd b. 4. If two zeroes of the polynomil x 4 6x 6x + 18x 5 re ±, find other zeroes. 5. If the polynomil x 4 6x + 16x 5x + 10 is divided by nother polynomil x x + k, the reminder comes out to be x +, find k nd..5 Summry In this chpter, you hve studied the following points: 1. Polynomils of degrees 1, nd re clled liner, qudrtic nd cubic polynomils respectively.. A qudrtic polynomil in x with rel coefficients is of the form x + bx + c, where, b, c re rel numbers with 0.. The zeroes of polynomil p(x) re precisely the xcoordintes of the points, where the grph of y = p(x) intersects the x xis. 4. A qudrtic polynomil cn hve t most zeroes nd cubic polynomil cn hve t most zeroes. 5. If α nd β re the zeroes of the qudrtic polynomil x + bx + c, then b α+β=, c αβ =. 6. If α, β, γ re the zeroes of the cubic polynomil x + bx + cx + d = 0, then b α+β+γ=, c αβ+ βγ+ γα =, d nd αβγ =. 7. The division lgorithm sttes tht given ny polynomil p(x) nd ny nonzero polynomil g(x), there re polynomils q(x) nd r(x) such tht where p(x) = g(x) q(x) + r(x), r(x) = 0 or degree r(x) < degree g(x).
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