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Exercise - 2.4

Polynomials

**Question-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:

(i) 2x³ + x² - 5x + 2; 1/2, 1, -2

(ii) x³ – 4x² + 5x - 2; 2, 1, 1

(i) p(x) = 2x³ + x² - 5x + 2; 1/2, 1, -2 p(1/2) = 2 x (1/2)³ + (1/2)² - 5 x (1/2) + 2 = 2 x 1/8 + 1/4 - 5/2 + 2 = 1/4 + 1/4 - 5/2 + 2 = (1 + 1 - 10 + 8)/4 = 0/4 = 0 p(1) = 2 x (1)³ + (1)² - 5 x 1 + 2 = 2 + 1 - 5 + 2 = 5 - 5 = 0 p(-2) = 2 x (-2)³ + (-2)² - 5 x (-2) + 2 = 2 x (-8) + 4 + 10 + 2 = -16 + 6 + 10 = -16 + 16 = 0 So, 1/2, 1, -2 are the zeroes of the cubic polynomial p(x) = 2x³ + x² - 5x + 2. In general, cubic polynomial is ax³ + bx² + cx + d p(x) = 2x³ + x² - 5x + 2 By compairing, a = 2, b = 1, c = -5, d = 2. Relationship between the zeroes and the coefficients : Given that α = 1/2, β = 1, γ = -2 α + β + γ = 1/2 + 1 - 2 = 3/2 - 2 = -1/2 -b/a = -(1)/2 = -1/2 αβ + βγ + γα = 1/2 x 1 + 1 x (-2) + (-2) x 1/2 = 1/2 - 2 - 1 = (1 - 4 - 2)/2 = -5/2 c/a = -5/2 αβγ = 1/2 x 1 x (-2) = -1 -d/a = -(2)/2 = -1

(ii) p(x) = x³ – 4x² + 5x - 2; 2, 1, 1 p(2) = (2)³ – 4 x (2)² + 5 x 2 - 2 = 8 - 16 + 10 - 2 = 18 - 18 = 0 p(1) = (1)³ – 4 x (1)² + 5 x 1 - 2 = 1 - 4 + 5 - 2 = 6 - 6 = 0 So, 2, 1, 1 are the zeroes of the cubic polynomial p(x) = x³ – 4x² + 5x - 2. In general, cubic polynomial is ax³ + bx² + cx + d p(x) = x³ – 4x² + 5x - 2 By compairing, a = 1, b = -4, c = 5, d = -2. Relationship between the zeroes and the coefficients : Given that α = 2, β = 1, γ = 1 α + β + γ = 2 + 1 + 1 = 4 -b/a = -(-4)/1 = 4 αβ + βγ + γα = 2 x 1 + 1 x 1 + 1 x 2 = 2 + 1 + 2 = 5 c/a = 5/1 = 5 αβγ = 2 x 1 x 1 = 2 -d/a = -(-2)/1 = 2

**Question-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.

In general, the cubic polynomial is ax³ + bx² + cx + d. Now, Given that α = 2, β = -7, γ = -14 α + β + γ = 2 = -b/a = -(-2)/1 αβ + βγ + γα = -7 = c/a = -(7)/1 αβγ = -14 = -d/a = -(14)/1 All coefficients a = 1, b = -2, c = 7, d = 14. Therefore, cubic polynomial is x³ - 2x² + 7x + 14.

**Question-3 :-** Obtain all other zeroes of If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b, find a and b.

p(x) = x³ – 3x² + x + 1 Zeroes of polynomial are a - b, a, a + b. In general, the cubic polynomial is px³ + qx² + rx + s. So, p = 1, q = -3, r = 1, s = 1 Sum of zeroes = a - b + a + a + b = 3a -q/p = -(-3)/1 = 3 3a = 3 a = 1 Product of zeroes = (a - b) x a x (a + b) = a x (a² - b²) = 1 x (1 - b²) = 1 - b² -s/p = -(1)/1 = -1 1 - b² = -1 - b² = -1 - 1 - b² = -2 b² = 2 b = ±√2 All other zeroes are 1, ±√2.

**Question-4 :-** If two zeroes of the polynomial x⁴ – 6x³ – 26x² + 138x – 35 are 2 ± √3 , find other zeroes.

p(x) = x⁴ – 6x³ – 26x² + 138x – 35 Given that two zeroes are 2 + √3 and 2 - √3. Now, g(x) = (x - 2 - √3)(x - 2 + √3) = x² + 4 - 4x – 3 = x² - 4x + 1 q(x) = x² - 2x - 35 = x² - 7x + 5x -35 = x(x - 7) + 5(x - 7) = (x + 5)(x - 7) x + 5 = 0 x = -5 x - 7 = 0 x = 7 All zeroes of the polynomial are 2 + √3, 2 - √3, -5 and 7.

**Question-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.

p(x) = x⁴ – 6x³ + 16x² – 25x + 10 g(x) = x² – 2x + k r(x) = x + a p(x) = g(x) x q(x) + r(x) p(x) - r(x) = g(x) x q(x) x⁴ – 6x³ + 16x² – 25x + 10 - x - a = (x² – 2x + k) x q(x) x⁴ – 6x³ + 16x² – 26x + 10 - a = (x² – 2x + k) x q(x) q(x) = x² – 4x + (8 - k) r(x) = (-10 + 2k)x + (10 - a - 8k + k²) -10 + 2k = 0 2k = 10 k = 5 10 - a - 8k + k² = 0 10 - a - 8 x 5 + 5² = 0 10 - a - 40 + 25 = 0 -a - 5 = 0 a = -5 a = -5, k = 5

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