Polynomials

Welcome dear students! Today we are going to learn about Polynomials from Class 10 Maths. In Class 9, 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 x of degree 3 and 7u⁶ – 4u⁵ + 3u⁴ + 8u² – 2 is a polynomial in the variable u of degree 6. Expressions like 1/(x – 1), √x + 2, 2x + 1/(3x²) + 3, etc., are not polynomials.

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A polynomial of degree 1 is called a linear polynomial. For example, 2x – 3, 3x + 5, 2y + √2, 2x/11 – 1, 3z + 4, 2/3 u + 1, etc., are all linear polynomials. Polynomials 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’ has been derived from the word ‘quadrate’, which means ‘square’. x² + 2x – 2, y² – 2, 2x² – 3x + 2, 2u² – 2u + 1, 5v² – 5v + 2/3, 4z² – 3z + 7 are some examples of 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.

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A polynomial of degree 3 is called a cubic polynomial. Some examples of a cubic polynomial are 2 – x³, x³, 3x³/2, 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).

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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 9, 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 2k + 3 = 0, i.e., k = –3/2. In general, if k is a zero of p(x) = ax + b, then p(k) = ak + b = 0, i.e., k = –b/a. So, the zero of the linear polynomial ax + b is –(Constant term)/(Coefficient of x), which is –b/a. 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?

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Section 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 9 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 Figure 2.1, you can see that the graph of y = 2x + 3 intersects the x-axis mid-way between x = –1 and x = –2, that is, at the point (–3/2, 0). You also know that the zero of 2x + 3 is –3/2. Thus, the zero of the polynomial 2x + 3 is the x-coordinate of the point where the graph 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 straight line which intersects the x-axis at exactly one point, namely, (–b/a, 0). 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.

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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. When x is –2, y is 6. When x is –1, y is 0. When x is 0, y is –4. When x is 1, y is –6. When x is 2, y is –6. When x is 3, y is –4. When x is 4, y is 0. When x is 5, y is 6. Note that plotting of graphs of quadratic or cubic polynomials is not meant to be done by the students, nor is to be evaluated. If we locate the points listed above on a graph paper and draw the graph, it will actually look like the one given in Figure 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 or open downwards depending on whether a > 0 or a < 0. (These curves are called parabolas.)

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You can see from Table 2.1 that –1 and 4 are zeroes of the quadratic polynomial. Also note from Figure 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 y = 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 Figure 2.3).

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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 Figure 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 the x-axis or completely below the x-axis. So, it does not cut the x-axis at any point (see Figure 2.5). So, the quadratic polynomial ax² + bx + c has no zero in this case. So, you can see geometrically that a quadratic polynomial can have either two distinct zeroes or two equal zeroes (i.e., one zero), or no zero. This also means that a polynomial of degree 2 has at most two zeroes.

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Now, what do you expect the geometrical meaning of the zeroes of a cubic polynomial to be? Let us find out. Consider the cubic polynomial x³ – 4x. To see what the graph of y = x³ – 4x looks like, let us list a few values of y corresponding to a few values for x as shown in Table 2.2. When x is –2, y is 0. When x is –1, y is 3. When x is 0, y is 0. When x is 1, y is –3. When x is 2, y is 0. 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 Figure 2.6.

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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 Figure 2.7 and Figure 2.8 respectively. Note that 0 is the only zero of the polynomial x³. Also, from Figure 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 Figure 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.

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Remark: In general, given a polynomial p(x) of degree n, the graph of y = p(x) intersects the x-axis at at most n points. Therefore, a polynomial p(x) of degree n has at most n zeroes. Example 1: Look at the graphs in Figure 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). 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?)

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Exercise 2.1: Question 1. The graphs of y = p(x) are given in Figure 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case. Solution: In figure (i), the graph does not intersect the x-axis, so zeroes = 0. In figure (ii), it intersects at one point, zeroes = 1. In figure (iii), it intersects at two points, zeroes = 2. In figure (iv), it intersects at three points, zeroes = 3. In figure (v), it intersects at four points, zeroes = 4. In figure (vi), it intersects at three points, zeroes = 3.

Section 2.3: Relationship between Zeroes and Coefficients of a Polynomial. You have already seen that zero of a linear polynomial ax + b is –b/a. We will now try to 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 9, 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.

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Observe that: Sum of its zeroes = 1 + 3 = 4 = –(–8)/2 = –(Coefficient of x) / (Coefficient of x²). Product of its zeroes = 1 × 3 = 3 = 6/2 = Constant term / Coefficient of x². 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., when x = 1/3 or x = –2. So, the zeroes of 3x² + 5x – 2 are 1/3 and –2. Observe that: Sum of its zeroes = 1/3 + (–2) = –5/3 = –(5)/3 = –(Coefficient of x) / (Coefficient of x²). Product of its zeroes = 1/3 × (–2) = –2/3 = –2/3 = Constant term / Coefficient of x².

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In 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 + αβ] = kx² – k(α + β)x + kαβ. Comparing the coefficients of x², x and constant terms on both the sides, we get a = k, b = –k(α + β) and c = kαβ. This gives α + β = –b/a, αβ = c/a. i.e., sum of zeroes = –(Coefficient of x) / (Coefficient of x²), product of zeroes = Constant term / (Coefficient of x²). (*Note: α, β are Greek letters pronounced as ‘alpha’ and ‘beta’ respectively. We will use later one more letter ‘γ’ pronounced as ‘gamma’.) Let us consider some examples.

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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, sum of zeroes = –2 + (–5) = –7 = –(7)/1 = –(Coefficient of x) / (Coefficient of x²). Product of zeroes = (–2) × (–5) = 10 = 10/1 = Constant term / Coefficient of x². 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, sum of zeroes = √3 + (–√3) = 0 = –(0)/1 = –(Coefficient of x) / (Coefficient of x²). Product of zeroes = √3 × (–√3) = –3 = –3/1 = Constant term / Coefficient of x².

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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 α + β = –3 = –b/a, and αβ = 2 = c/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. You can check that p(x) = 0 for x = 4, –2, 1/2. Since p(x) can have at most three zeroes, these are the zeroes of 2x³ – 5x² – 14x + 8. Now, let us verify the relationships completely. Sum of the zeroes = 4 + (–2) + 1/2 = 2 + 1/2 = 5/2. This equals –(–5)/2, which is –(Coefficient of x²) / (Coefficient of x³). Product of the zeroes = 4 × (–2) × 1/2 = –8 × 1/2 = –4. This equals –8/2, which is –(Constant term) / (Coefficient of x³).

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However, there is one more relationship here. Consider the sum of the products of the zeroes taken two at a time. We have {4 × (–2) + (–2) × 1/2 + 1/2 × 4} = –8 + (–1) + 2 = –7. This equals –14/2, which is (Coefficient of x) / (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 α + β + γ = –b/a, αβ + βγ + γα = c/a, αβγ = –d/a. Let us consider an example. Example 5*: Verify that 3, –1, –1/3 are the zeroes of the cubic polynomial p(x) = 3x³ – 5x² – 11x – 3, and then verify the relationship between the zeroes and the coefficients. (*Not from the examination point of view.) 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. p(–1/3) = 3 × (–1/3)³ – 5 × (–1/3)² – 11 × (–1/3) – 3 = –1/9 – 5/9 + 11/3 – 3 = –6/9 + 33/9 – 27/9 = 0. Therefore, 3, –1 and –1/3 are the zeroes of 3x³ – 5x² – 11x – 3. So, we take α = 3, β = –1 and γ = –1/3.

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Now, let us verify the three relationships clearly: First, α + β + γ = 3 + (–1) + (–1/3) = 2 – 1/3 = 5/3. This matches –b/a = –(–5)/3 = 5/3. Second, αβ + βγ + γα = 3 × (–1) + (–1) × (–1/3) + (–1/3) × 3 = –3 + 1/3 – 1 = –4 + 1/3 = –11/3. This matches c/a = –11/3. Third, αβγ = 3 × (–1) × (–1/3) = 3 × 1/3 = 1. This matches –d/a = –(–3)/3 = 1. All three relationships are successfully verified.

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Exercise 2.2: Question 1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. (i) x² – 2x – 8. We factorise this as (x – 4)(x + 2). Zeroes are 4 and –2. Sum is 4 + (–2) = 2 = –(–2)/1. Product is 4 × (–2) = –8 = –8/1. (ii) 4s² – 4s + 1. This factors as (2s – 1)². Zeroes are 1/2 and 1/2. Sum is 1/2 + 1/2 = 1 = –(–4)/4. Product is 1/2 × 1/2 = 1/4 = 1/4. (iii) 6x² – 3 – 7x. Rearranging gives 6x² – 7x – 3. Splitting the middle term: 6x² – 9x + 2x – 3. Factoring gives 3x(2x – 3) + 1(2x – 3) = (3x + 1)(2x – 3). Zeroes are –1/3 and 3/2. Sum is –1/3 + 3/2 = 7/6 = –(–7)/6. Product is –1/3 × 3/2 = –1/2 = –3/6. (iv) 4u² + 8u. Factoring gives 4u(u + 2). Zeroes are 0 and –2. Sum is 0 + (–2) = –2 = –8/4. Product is 0 × (–2) = 0 = 0/4. (v) t² – 15. Using difference of squares, this is (t – √15)(t + √15). Zeroes are √15 and –√15. Sum is 0 = –0/1. Product is –15 = –15/1. (vi) 3x² – x – 4. Splitting middle term: 3x² – 4x + 3x – 4. Factoring gives x(3x – 4) + 1(3x – 4) = (x + 1)(3x – 4). Zeroes are –1 and 4/3. Sum is –1 + 4/3 = 1/3 = –(–1)/3. Product is –1 × 4/3 = –4/3 = –4/3.

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Question 2: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. (i) Sum = 1/4, Product = –1. Polynomial is x² – (1/4)x – 1. Multiplying by 4 gives 4x² – x – 4. (ii) Sum = √2, Product = 1/3. Polynomial is x² – √2x + 1/3. Multiplying by 3 gives 3x² – 3√2x + 1. (iii) Sum = 0, Product = √5. Polynomial is x² – 0x + √5, which is x² + √5. (iv) Sum = 1, Product = 1. Polynomial is x² – 1x + 1, which is x² – x + 1. (v) Sum = –1/4, Product = 1/4. Polynomial is x² – (–1/4)x + 1/4, which is x² + 1/4x + 1/4. Multiplying by 4 gives 4x² + x + 1. (vi) Sum = 4, Product = 1. Polynomial is x² – 4x + 1.

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Section 2.4: 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 x with 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 α + β = –b/a, αβ = c/a. 6. If α, β, γ are the zeroes of the cubic polynomial ax³ + bx² + cx + d, then α + β + γ = –b/a, αβ + βγ + γα = c/a, and αβγ = –d/a.

Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]