TOPICS

Unit-2(Examples)

Polynomials

**Example-1 :-** Look at the graphs in Figures 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).

(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 as the graph intersects the x-axis at three points. (iv) The number of zeroes is 1 as the graph intersects the x-axis at one point only. (v) The number of zeroes is 1 as the graph intersects the x-axis at one point only. (vi) The number of zeroes is 4 as the graph intersects the x-axis at four points.

**Example-2 :-** Find the zeroes of the quadratic polynomial x² + 7x + 10, and verify the relationship between the zeroes and the coefficients.

p(x) = x² + 7x + 10 = x² + 5x + 2x + 10 = x(x + 5) + 2(x + 5) = (x + 2)(x + 5) Zeroes of polynomial : x + 2 = 0 x = -2 x + 5 = 0 x = -5 Relationship between the zeroes and the coefficients : p(x) = x² + 7x + 10 a = 1, b = 7, c = 10 Sum of zeroes = α + β = -b/a = -7/1 = -7 Product of zeroes = αβ = c/a = 10/1 = 10

**Example-3 :-** Find the zeroes of the polynomial x² – 3 and verify the relationship between the zeroes and the coefficients.

p(x) = x² – 3 = x² – (√3)² = (x + √3)(x – √3) Zeroes of polynomial : x + √3 = 0 x = –√3 x – √3 = 0 x = √3 Relationship between the zeroes and the coefficients : p(x) = x² – 3 a = 1, b = 0, c = -3 Sum of zeroes = α + β = -b/a = 0/1 = 0 Product of zeroes = αβ = c/a = -3/1 = -3

**Example-4 :-** Find a quadratic polynomial, the sum and product of whose zeroes are –3 and 2, respectively.

Given that : Sum of zeroes = α + β = -b/a = -3/1 Product of zeroes = αβ = c/a = 2/1 Therefore, a = 1, b = -3, c = 2. In general form, p(x) = ax² + bx + c. So, quadratic polynomial is x² - 3x + 2.

**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.

p(x) = 3x³ – 5x² – 11x – 3 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 x (-1/3)³ - 5 x (-1/3)² - 11 x (-1/3) - 3 = 3 x (-1/27) - 5 x (1/9) + 11/3 - 3 = -1/9 - 5/9 + 11/3 - 3 = (-1 - 5 + 33 - 27)/9 = 0/9 = 0 So, 3, –1, -1/3 are the zeroes of the cubic polynomial p(x) = 3x³ – 5x² – 11x – 3. In general, cubic polynomial is ax³ + bx² + cx + d p(x) = 3x³ – 5x² – 11x – 3 By compairing, a = 3, b = -5, c = -11, d = -3. Relationship between the zeroes and the coefficients : Given that α = 3, β = -1, γ = -1/3 α + β + γ = 3 - 1 - 1/3 = (9 - 3 - 1)/3 = 5/3 -b/a = -(-5)/3 = 5/3 αβ + βγ + γα = 3 x (-1) + (-1) x (-1/3) + (-1/3) x 3 = -3 + 1/3 - 1 = (-9 + 1 - 3)/3 = -11/3 c/a = -11/3 αβγ = 3 x (-1) x (-1/3) = 1 -d/a = -(-3)/3 = 1

**Example-6 :-** Divide 2x² + 3x + 1 by x + 2.

p(x) = 2x² + 3x + 1 g(x) = x + 2 q(x) = 2x - 1, r(x) = 3 Verification : p(x) = g(x) x q(x) + r(x) = (x + 2)(2x - 1) + 3 = 2x² - x + 4x - 2 + 3 = 2x² + 3x + 1

**Example-7 :-** Divide 3x³ + x² + 2x + 5 by 1 + 2x + x².

p(x) = 3x³ + x² + 2x + 5 g(x) = 1 + 2x + x² q(x) = 3x - 5, r(x) = 9x + 10 Verification : p(x) = g(x) x q(x) + r(x) = (1 + 2x + x²) x (3x - 5) + (9x + 10) = 3x - 5 + 6x² - 10x + 3x³ - 5x² + 9x + 10 = 3x³ + (6x² - 5x²) + (3x + 9x - 10x) + (10 - 5) = 3x³ + x² + 2x + 5

**Example-8 :-** : Divide 3x² – x³ – 3x + 5 by x – 1 – x², and verify the division algorithm.

p(x) = 3x² – x³ – 3x + 5 g(x) = x – 1 – x² q(x) = x - 2, r(x) = 3 Verification : p(x) = g(x) x q(x) + r(x) = (x – 1 – x²)(x - 2) + 3 = x² - 2x - x³ + 2 - x + 2x² + 3 = 3x² – x³ – 3x + 5

**Example-6 :-** Find all the zeroes of 2x⁴ – 3x³ – 3x² + 6x – 2, if you know that two of its zeroes are √2 and -√2.

p(x) = 2x⁴ – 3x³ – 3x² + 6x – 2 Given that two zeroes are √2 and -√2. So, (x - √2)(x + √2) = x² - 2 [a² - b² = (a + b)(a - b)] q(x) = 2x² - 3x + 1 Now, p(x) = 2x⁴ – 3x³ – 3x² + 6x – 2 = (x - √2)(x + √2)(2x² - 3x + 1) = (x - √2)(x + √2)[2x² - 2x - x + 1] = (x - √2)(x + √2)[2x(x - 1) - 1(x - 1)] = (x - √2)(x + √2)(x - 1)(2x - 1) x - 1 = 0 x = 1 2x - 1 = 0 2x = 1 x = 1/2 Therefore, all zeroes are √2, -√2, 1 and 1/2.

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