POLYNOMIALS 2.1 Introduction In Class IX, you have studied polynomials in one variable and their degrees. Recall that if p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of the polynomial p(x). For example, 4x + 2 is a polynomial in the variable x of degree 1, 2y2 – 3y + 4 is a polynomial in the variable y of degree 2, 5x3 – 4x2 + x – 2 is a polynomial in the variable xof degree 3 and 7u6 – 3 u4 + 4u2 + u −8 is a polynomial2 1 1in the variable u of degree 6. Expressions like , x + 2 , 2 etc., are x − 1 x + 2x +3 not polynomials. A polynomial of degree 1 is called a linear polynomial. For example, 2x – 3, 223x + 5, y+ 2, x − , 3z + 4, u +1 , etc., are all linear polynomials. Polynomials11 3 such as 2x + 5 – x2, x3 + 1, etc., are not linear polynomials. A polynomial of degree 2 is called a quadratic polynomial. The name ‘quadratic’ 22 , 5has been derived from the word ‘quadrate’, which means ‘square’. 2x + 3x − u 22 21 y2 – 2, 2 − x2 + 3, x − 2u2 + 5, 5v − v,4 z + are some examples of3 37 quadratic polynomials (whose coefficients are real numbers). More generally, any quadratic polynomial in x is of the form ax2 + bx + c, where a, b, c are real numbers and a ≠ 0. A polynomial of degree 3 is called a cubic polynomial. Some examples of a cubic polynomial are 2 – x3, x3, 2,x3 3 – x2 + x3, 3x3 – 2x2 + x – 1. In fact, the most general form of a cubic polynomial is ax3 + bx2 + cx + d, where, a, b, c, d are real numbers and a ≠ 0. Now consider the polynomial p(x) = x2 – 3x – 4. Then, putting x = 2 in the polynomial, we get p(2) = 22 – 3 × 2 – 4 = – 6. The value ‘– 6’, obtained by replacing x by 2 in x2 – 3x – 4, is the value of x2 – 3x – 4 at x = 2. Similarly, p(0) is the value of p(x) at x = 0, which is – 4. If p(x) is a polynomial in x, and if k is any real number, then the value obtained by replacing x by k in p(x), is called the value of p(x) at x = k, and is denoted by p(k). What is the value of p(x) = x2 –3x – 4 at x = –1? We have : p(–1) = (–1)2 –{3 × (–1)} – 4 = 0 Also, note that p(4) = 42 – (3 × 4) – 4 = 0. As p(–1) = 0 and p(4) = 0, –1 and 4 are called the zeroes of the quadratic polynomial x2 – 3x – 4. More generally, a real number k is said to be a zero of a polynomial p(x), if p(k) = 0. You have already studied in Class IX, how to find the zeroes of a linear polynomial. For example, if k is a zero of p(x) = 2x + 3, then p(k) = 0 gives us 32k + 3 = 0, i.e., k = −⋅ 2 −bIn 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 = . a Coefficient of x Thus, the zero of a linear polynomial is related to its coefficients. Does this happen in the case of other polynomials too? For example, are the zeroes of a quadratic polynomial also related to its coefficients? In this chapter, we will try to answer these questions. We will also study the division algorithm for polynomials. 2.2 Geometrical Meaning of the Zeroes of a Polynomial You know that a real number k is a zero of the polynomial p(x) if p(k) = 0. But why are the zeroes of a polynomial so important? To answer this, first we will see the geometrical representations of linear and quadratic polynomials and the geometrical meaning of their zeroes. Consider first a linear polynomial ax + b, a ≠ 0. You have studied in Class IX that the graph of y = ax + b is a straight line. For example, the graph of y = 2x + 3 is a straight line passing through the points (– 2, –1) and (2, 7). From Fig. 2.1, you can see that the graph of y = 2x + 3 intersects the x-axis mid-way between x = –1 and x = –2, 3that is, at the point ⎜⎛− ,0 ⎞ .⎟⎝ 2 ⎠ You also know that the zero of 32x + 3 is − . Thus, the zero of 2 the polynomial 2x + 3 is the x-coordinate of the point where the Fig. 2.1graph of y = 2x + 3 intersects the x-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, the x-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* of y = x2 – 3x – 4 looks like. Let us list a few values of y = x2 – 3x – 4 corresponding to a 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. Table 2.1 x – 2 –1 0 1 2 3 4 5 y = x2 – 3x – 4 6 0 – 4 – 6 – 6 – 4 0 6 If we locate the points listed above on a graph paper and draw the graph, it will actually look like the one given in Fig. 2.2. In fact, for any quadratic polynomial ax2 + bx + c, a ≠ 0, the graph of the corresponding equation y = ax2 + bx + c has one of the two shapes either open upwards like or open downwards like depending on whether a > 0 or a < 0. (These curves are called parabolas.) You can see from Table 2.1 that –1 and 4 are zeroes of the quadratic polynomial. Also note from Fig. 2.2 that –1 and 4 are the x-coordinates of the points where the graph of y = x2 – 3x – 4 intersects the x-axis. Thus, the zeroes of the quadratic polynomial x2 – 3x – 4 are x-coordinates of the points where the graph of Fig. 2.2y = x2 – 3x – 4 intersects the x-axis. This fact is true for any quadratic polynomial, i.e., the zeroes of a quadratic polynomial ax2 + bx + c, a ≠ 0, are precisely the x-coordinates of the points where the parabola representing y = ax2 + bx + c intersects the x-axis. From our observation earlier about the shape of the graph of y = ax2 + bx + c, the following three cases can happen: Case (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 polynomial ax2 + bx + c in this case (see Fig. 2.3). Fig. 2.3 Case (ii) : Here, the graph cuts the x-axis at exactly one point, i.e., at two coincident points. So, the two points A and A′ of Case (i) coincide here to become one point A (see Fig. 2.4). Fig. 2.4 The x-coordinate of A is the only zero for the quadratic polynomial ax2 + bx + c in this case. Case (iii) : Here, the graph is either completely above thex-axis or completely below the x-axis. So, it does not cut the x-axis at any point (see Fig. 2.5). Locating the points of the table on a graph paper and drawing the graph, we see that 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 We see from the table above that – 2, 0 and 2 are zeroes of the cubic polynomial x3 – 4x. Observe that – 2, 0 and 2 are, in fact, the x-coordinates of the only points where the graph of y = x3 – 4x intersects the x-axis. Since the curve meets the x-axis in only these 3 points, their x-coordinates are the only zeroes of the polynomial. Let us take a few more examples. Consider the cubic polynomials x3 and x3 – x2. We draw the graphs of y = x3 and y = x3 – x2 in Fig. 2.7 and Fig. 2.8 respectively. Fig. 2.6 Fig. 2.7 Fig. 2.8 Note that 0 is the only zero of the polynomial x3. Also, from Fig. 2.7, you can see that 0 is the x-coordinate of the only point where the graph of y = x3 intersects the x-axis. Similarly, since x3 –x2 = x2 (x– 1), 0 and 1 are the only zeroes of the polynomial x3 – x2. Also, from Fig. 2.8, these values are the x-coordinates of the only points where the graph of y = x3 – x2 intersects the x-axis. From the examples above, we see that there are at most 3 zeroes for any cubic polynomial. In other words, any polynomial of degree 3 can have at most three zeroes. Remark : 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 has at most n zeroes. Example 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. 2.9 Solution : (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?) (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 2.1 1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case. Fig. 2.10 2.3 Relationship between Zeroes and Coefficients of a Polynomial bYou have already seen that zero of a linear polynomial ax + b is − .We will now try ato answer the question raised in Section 2.1 regarding the relationship between zeroes and 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 quadratic polynomials 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 write 2x2 – 8x + 6 =2x2 – 6x – 2x + 6 = 2x(x – 3) – 2(x – 3) =(2x – 2)(x – 3) = 2(x – 1)(x – 3) So, the value of p(x) = 2x2 – 8x + 6 is zero when x – 1 = 0 or x – 3 = 0, i.e., when x = 1 or x = 3. So, the zeroes of 2x2 – 8x + 6 are 1 and 3. Observe that : ( 8) −(Coefficient of x−− )Sum of its zeroes = +=4 =13 = 22 Coefficient of x 6 Constant termProduct of its zeroes = 13×=3 == 22 Coefficient of x Let us take one more quadratic polynomial, say, p(x) = 3x2 + 5x – 2. By the method of splitting the middle term, 3x2 + 5x – 2 = 3x2 + 6x – x – 2 = 3x(x + 2) –1(x + 2) =(3x – 1)(x + 2) Hence, the value of 3x2 + 5x – 2 is zero when either 3x – 1 = 0 or x + 2 = 0, i.e., 11 when x = or x = –2. So, the zeroes of 3x2 + 5x – 2 are and – 2. Observe that :331 −5 −(Coefficient of x)(2) =Sum of its zeroes = +− = 3 3 Coefficient of x2 1 −2 Constant term×− =Product of its zeroes = (2) = 23 3 Coefficient of x In general, ifα* and β* are the zeroes of the quadratic polynomialp(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 β Comparing the coefficients of x2, x and constant terms on both the sides, we get a = k, b =– k(α + β) and c = kαβ. –b This gives α+ β=,a cαβ= a * α,β are Greek letters pronounced as ‘alpha’ and ‘beta’ respectively. We will use later one more letter ‘γ’ pronounced as ‘gamma’. b −(Coefficient of x)i.e., sum of zeroes = α + β = −= 2,a Coefficient of x c Constant term product of zeroes = αβ = = 2. a Coefficient of x Let us consider some examples. Example 2 : Find the zeroes of the quadratic polynomial x2 + 7x + 10, and verify the relationship between the zeroes and the coefficients. Solution : We have x2 + 7x + 10 = (x + 2)(x + 5) So, the value of x2 + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0, i.e., when x = – 2 or x = –5. Therefore, the zeroes of x2 + 7x + 10 are – 2 and – 5. Now, −(7) –(Coefficient of x)sum of zeroes = –2 +(– 5) =– (7) == 2,1 Coefficient of x 10 Constant term product of zeroes = ( 2) −10 = 2−×( 5) == ⋅1 Coefficient of x Example 3 : Find the zeroes of the polynomial x2 – 3 and verify the relationship between the zeroes and the coefficients. Solution : Recall the identity a2 – b2 = (a – b)(a + b). Using it, we can write: x2 – 3 = (x −3 )(x +3 ) So, the value of x2 – 3 is zero when x = 3 or x = – 3⋅ Therefore, the zeroes of x2 – 3 are 3 and 3−⋅ Now, −(Coefficient of x)sum of zeroes = 3 −3 == 2,0 Coefficient of x −3 Constant term 3 )(−3)=–3 == ⋅product of zeroes = (1 Coefficient of x2 Example 4 : Find a quadratic polynomial, the sum and product of whose zeroes are – 3 and 2, respectively. Solution : Let the quadratic polynomial be ax2 + bx + c, and its zeroes be α and β. We have −b ,α + β = – 3 = a c .and αβ =2= a If a = 1, then b = 3 and c = 2. So, one quadratic polynomial which fits the given conditions is x2 + 3x + 2. You can check that any other quadratic polynomial that fits these conditions will be of the form k(x2 + 3x + 2), where k is real. Let us now look at cubic polynomials. Do you think a similar relation holds between the zeroes of a cubic polynomial and its coefficients? Let us consider p(x) = 2x3 – 5x2 – 14x + 8. 1You can check that p(x) = 0 for x = 4, – 2, ⋅Since p(x) can have atmost three2 zeroes, these are the zeores of 2x3 – 5x2 – 14x + 8. Now, 15 −− −(Coefficient of x2(5) )sum of the zeroes = 4( 2) == ,+−+= 2 2 2 Coefficient of x3 1 −8 –Constant term product of the zeroes = 4( 2) ×−× =4.−= = 2 2 Coefficient of x3 However, there is one more relationship here. Consider the sum of the products of the zeroes taken two at a time. We have ⎧ ⎫⎧ ⎫{×− } (2) 1 + 14(2) +−× ×4⎨ ⎬⎨ ⎬⎩ 2 ⎭⎩ 2 ⎭ −14 Coefficient of x =–812 −+ =− = =7.2 Coefficient of x3 In general, it can be proved that if α, β, γ are the zeroes of the cubic polynomial ax3 + bx2 + cx + d, then –bα+ β+ γ=,a c αβ+ βγ+ γα=,a –d α β γ=. a Let us consider an example. 1Example 5* : Verify that 3, –1, − are the zeroes of the cubic polynomial3 p(x) = 3x3 – 5x2 – 11x – 3, and then verify the relationship between the zeroes and the coefficients. Solution : Comparing the given polynomial with ax3 + bx2 + cx + d, we get a = 3, b = – 5, c = –11, d = – 3. Further p(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, ⎛ 1 ⎞⎛ 1 ⎞3 ⎛ 1 ⎞2 ⎛ 1 ⎞ p −=3 ×− −×−5 −11 ×− −3⎜⎟⎜⎟ ⎜⎟ ⎜⎟ ,⎝ 3 ⎠⎝ 3 ⎠⎝ 3 ⎠⎝ 3 ⎠ 1511 22 =– −+ −3–= +=0 993 33 1 Therefore, 3, –1 and − are the zeroes of 3x3 – 5x2 – 11x – 3.3 1So, we take α = 3, β = –1 and γ = −⋅ 3Now, ⎛ 1 ⎞ 15 −−( 5) −b =3(1) +− =2 =α+β+γ+− −== ,⎜⎟⎝ 3 ⎠ 33 3 a ⎛ 1 ⎞⎛ 1 ⎞ 1 −11 c3( 1) (1) +− ×3 =3 −1 == ,αβ+βγ+γα= ×− +− ×− − +⎜⎟⎜ ⎟⎝ 3 ⎠⎝ 3 ⎠ 33 a ⎛ 1 ⎞ −−( 3) −d .αβγ=3(1)×− ×− =1 ==⎜⎟⎝ 3 ⎠ 3 a * Not from the examination point of view. EXERCISE 2.2 1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. (i) x2 – 2x – 8 (ii) 4s2 – 4s + 1 (iii) 6x2 – 3 – 7x (iv) 4u2 + 8 u (v) t2 – 15 (vi) 3x2 – x – 4 2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. 1 ,(i) 1 −1 (ii) 2, (iii) 0, 54 3 11(iv) 1, 1 (v) − , (vi) 4, 1 44 2.4 Division Algorithm forPolynomials You know that a cubic polynomial has at most three zeroes. However, if you are given only one zero, can you find the other two? For this, let us consider the cubic polynomial x3 – 3x2 – x + 3. If we tell you that one of its zeroes is 1, then you know that x – 1 is a factor of x3 – 3x2 – x + 3. So, you can divide x3 – 3x2 – x + 3 by x – 1, as you have learnt 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 you x3 – 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 as 1, – 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. 2x – 1 Solution : Note that we stop the division process when x + 2 either the remainder is zero or its degree is less than the 2 + 4xxdegree of the divisor. So, here the quotient is 2x – 1 and the remainder is 3. Also, (2x – 1)(x + 2) + 3 = 2x2 + 3x – 2 + 3 = 2x2 + 3x + 1 i.e., 2x2 + 3x + 1 = (x + 2)(2x – 1) + 3 Therefore, Dividend = Divisor × Quotient + Remainder Let us now extend this process to divide a polynomial by a quadratic polynomial. Example 7 : Divide 3x3 + x2 + 2x + 5 by 1 + 2x + x2. 3x – 5 Solution : We first arrange the terms of the x 2 + 2x + 1 3dividend and the divisor in the decreasing order 3 + 62 +3xxx –––of their degrees. Recall that arranging the terms 2 –5 –x + 5xin this order is called writing the polynomials in –5x2 – 10x – 5standard form. In this example, the dividend is ++ +already in standard form, and the divisor, in 9x + 10standard form, is x2 +2x + 1. Step 1 : To obtain the first term of the quotient, divide the highest degree term of the dividend (i.e., 3x3) by the highest degree term of the divisor (i.e., x2). This is 3x. Then carry out the division process. What remains is – 5x2 – x + 5. Step 2 : Now, to obtain the second term of the quotient, divide the highest degree term of the new dividend (i.e., –5x2) by the highest degree term of the divisor (i.e., x2). This gives –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 degree of 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 + 5 Here again, we see that Dividend = Divisor × Quotient + Remainder What we are applying here is an algorithm which is similar to Euclid’s division algorithm that you studied in Chapter 1. This says that If p(x) and g(x) are any two polynomials with g(x) ≠0, then we can find polynomials q(x) and r(x) such that 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 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. Solution : Note that the given polynomials x – 2 are not in standard form. To carry out 3–x 2 + x – 1 – x + 3x 2 – 3x + 5division, we first write both the dividend and – x3 + x 2 – xdivisor in decreasing orders of their degrees. + – + So, dividend = –x3 + 3x2 – 3x + 5 and divisor = –x2 + x – 1. 2 – 2x 2 x + 2 – + – Division process is shown on the right side. 3 We stop here since degree (3) = 0 < 2 = degree (–x2 + x – 1). So, quotient = x – 2, remainder = 3. Now, Divisor × Quotient + Remainder =(–x2 + x – 1) (x – 2) + 3 =–x3 + x2 – x + 2x2 – 2x + 2 + 3 =–x3 + 3x2 – 3x + 5 = Dividend In this way, the division algorithm is verified. Example 9 : Find all the zeroes of 2x4 – 3x3 – 3x2 + 6x – 2, if you know that two of its zeroes are 2 and − 2 . Solution : Since two zeroes are 2 and − 2, (x − 2 )(x +2 ) = x2 – 2 is a factor of the given polynomial. Now, we divide the given polynomial by x2 – 2. 2x2– 3x +1 2x42First term of quotient is 2 = 2x x –+ 32 3–3x + x + 6 x– 2 −3x Second term of quotient is 2 =−3x –3x 3 + 6xx +– x 2 – 2 x2 2Third term of quotient is =1 x – 2 2x –+ 0 So, 2x4 – 3x3 – 3x2 + 6x – 2 = (x2 – 2)(2x2 – 3x + 1). Now, by splitting –3x, we factorise 2x2 – 3x + 1 as (2x – 1)(x – 1). So, its zeroes 1are given by x = and x = 1. Therefore, the zeroes of the given polynomial are2 2, − 2, 1 ,2 and 1. EXERCISE 2.3 1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following : (i) p(x) = x3 – 3x2 + 5 x – 3, g(x) = x2 – 2 (ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x (iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2 2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: (i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12 (ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2 (iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1 55and – ⋅3. Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are 33 4. On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Find g(x). 5. Give examples of polynomials p(x), g(x), q(x) andr(x), which satisfy the division algorithm and (i) deg p(x) = deg q(x) (ii) deg q(x) = deg r(x) (iii) deg r(x) = 0 EXERCISE 2.4 (Optional)* 1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case: 1(i) 2x3 + x2 – 5x + 2; ,1,–2 (ii) x3 – 4x2 + 5x – 2; 2, 1, 1 2 2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively. *These exercises are not from the examination point of view. 3. If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, find a and b. 4. If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x– 35 are 2 ±3 , find other zeroes. 5. If the polynomial x4 – 6x3 + 16x2 – 25x+ 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a. 2.5 Summary In this chapter, you have studied the following points: 1. Polynomials of degrees 1, 2 and 3 are called linear, quadratic and cubic polynomials respectively. 2. A quadratic polynomial in xwith real coefficients is of the form ax2 + bx + c, where a, b, c are real numbers with a ≠ 0. 3. The zeroes of a polynomial p(x) are precisely the x-coordinates of the points, where the graph of y = p(x) intersects the x-axis. 4. A quadratic polynomial can have at most 2 zeroes and a cubic polynomial can have at most 3 zeroes. 5. If α and β are the zeroes of the quadratic polynomial ax2 + bx + c, then bcα+β=− , αβ= . aa 6. If α, β, γ are the zeroes of the cubic polynomial ax3 + bx2 + cx + d, then −bα+β+γ= , a c ,αβ+βγ+γα= a −d .αβγ=and a 7. The division algorithm states that given any polynomial p(x) and any non-zero polynomial g(x), there are polynomials q(x) and r(x) such that p(x) = g(x) q(x) + r(x), where r(x) = 0 or degree r(x) < degree g(x).