TOPICS

Exercise - 2.4

Polynomials

**Question-1 :-** Determine which of the following polynomials has (x + 1) a factor :

(i) x³ + x² + x + 1, (ii) x⁴ + x³ + x² + x + 1, (iii) x⁴ + 3x³ + 3x² + x + 1, (iv) x³ - x² - (2 + √2)x + √2.

(i) x³ + x² + x + 1 Let x + 1 is a factor of given polynomial. Now, x + 1 = 0 x = -1 p(-1) = (-1)³ + (-1)² + (-1) + 1 = -1 + 1 - 1 + 1 = 2 - 2 = 0 Therefore, It is confirmed that x + 1 is a factor of x³ + x² + x + 1.

(ii) x⁴ + x³ + x² + x + 1 Let x + 1 is a factor of given polynomial. Now, x + 1 = 0 x = -1 p(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1) + 1 = 1 - 1 + 1 - 1 + 1 = 3 - 2 = 1 Therefore, x + 1 is not a factor of x⁴ + x³ + x² + x + 1.

(iii) x⁴ + 3x³ + 3x² + x + 1 Let x + 1 is a factor of given polynomial. Now, x + 1 = 0 x = -1 p(-1) = (-1)⁴ + 3 x (-1)³ + 3 x (-1)² + (-1) + 1 = 1 - 3 + 3 - 1 + 1 = 5 - 4 = 1 Therefore, x + 1 is not a factor of x⁴ + 3x³ + 3x² + x + 1.

(iv) x³ - x² - (2 + √2)x + √2 Let x + 1 is a factor of given polynomial. Now, x + 1 = 0 x = -1 p(-1) = (-1)³ - (-1)² - (2 + √2) x (-1) + √2 = -1 - 1 + 2 + √2 + √2 = -2 + 2 + 2√2 = 2√2 Therefore, x + 1 is not a factor of x³ - x² - (2 + √2)x + √2.

**Question-2 :-** Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

(i) p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1,

(ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2,

(iii) p(x) = x³ – 4x² + x + 6, g(x) = x – 3 .

(i) p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1, By factor theorem (x + 1) is a factor of 2x³ + x² – 2x – 1. If x + 1 = 0 x = -1 p(-1) = 2(-1)³ + (-1)² – 2(-1) – 1 = -2 + 1 + 2 - 1 = 3 - 3 = 0 Therefore, g(x) = x + 1 is a factor of p(x) = 2x³ + x² – 2x – 1.

(ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2 By factor theorem (x + 2) is a factor of x³ + 3x² + 3x + 1. If x + 2 = 0 x = -2 p(-1) = 2 x (-2)³ + 3 x (-2)² + 3 x (-2) + 1 = 2 x (-8) + 3 x 4 - 6 + 1 = -16 + 12 - 6 + 1 = -22 + 13 = -9 Therefore, g(x) = x + 2 is not a factor of p(x) = x³ + 3x² + 3x + 1.

(iii) p(x) = x³ – 4x² + x + 6, g(x) = x – 3 By factor theorem (x – 3) is a factor of x³ – 4x² + x + 6. If x – 3 = 0 x = 3 p(3) = (3)³ – 4 x (3)² + 3 + 6 = 27 - 36 + 9 = 36 - 36 = 0 Therefore, g(x) = x – 3 is a factor of p(x) = x³ – 4x² + x + 6.

**Question-3 :-** Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:

(i) p(x) = x² + x + k,
(ii) p(x) = 2x² + kx + √2,
(iii) p(x) = kx² – √2x + 1,
(iv) p(x) = kx² – 3x + k.

(i) p(x) = x² + x + k , k = ? Given that x - 1 is a factor of p(x). So, x - 1 = 0 x = 1 And p(1) = 0. p(1) = (1)² + 1 + k 0 = 1 + 1 + k 0 = 2 + k k = -2

(ii) p(x) = 2x² + kx + √2, k = ? Given that x - 1 is a factor of p(x). So, x - 1 = 0 x = 1 And p(1) = 0. p(1) = 2 x (1)² + k x 1 + √2 0 = 2 + k + √2 0 = 2 + √2 + k k = -2 - √2 k = -(2 + √2)

(iii) p(x) = kx² – √2x + 1, k = ? Given that x - 1 is a factor of p(x). So, x - 1 = 0 x = 1 And p(1) = 0. p(1) = k x (1)² – √2 x 1 + 1 0 = k - √2 + 1 0 = -√2 + 1 + k k = √2 - 1

(iv) p(x) = kx² – 3x + k, k = ? Given that x - 1 is a factor of p(x). So, x - 1 = 0 x = 1 And p(1) = 0. p(1) = k x (1)² - 3 + k 0 = k - 3 + k 0 = -3 + 2k 2k = 3 k = 3/2

**Question-4 :-** Factorise :

(i) 12x² – 7x + 1,

(ii) 2x² + 7x + 3,

(iii) 6x² + 5x – 6,

(iv) 3x² – x – 4.

(i) 12x² – 7x + 1 By Spiliting method : [x² + (a + b)x + ab] a + b = -7 , ab = 12 x 1 = 12 = 12x² – 4x - 3x + 1 = 4x(3x - 1) - 1(3x - 1) = (3x - 1)(4x - 1)

(ii) 2x² + 7x + 3 By Spiliting method : [x² + (a + b)x + ab] a + b = 7 , ab = 2 x 3 = 6 = 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1)

(iii) 6x² + 5x – 6 By Spiliting method : [x² + (a + b)x + ab] a + b = 5 , ab = 6 x (-6) = -36 = 6x² + 9x - 4x - 6 = 3x(2x + 3) - 2(2x + 3) = (2x + 3)(3x - 2)

(iv) 3x² – x – 4 By Spiliting method : [x² + (a + b)x + ab] a + b = -1 , ab = 3 x (-4) = -12 = 3x² – 4x + 3x - 4 = x(3x - 4) + 1(3x - 4) = (3x - 4)(x + 1)

**Question-5 :-** Factorise :

(i) x³ – 2x² – x + 2,

(ii) x³ – 3x² – 9x – 5,

(iii) x³ + 13x² + 32x + 20,

(iv) 2y³ + y² – 2y – 1

(i) x³ – 2x² – x + 2 Firstly, the factor of 2 is ±1, ±2. p(1) = (1)³ – 2(1)² – 1 + 2 = 1 - 2 - 1 + 2 = 2 - 2 = 0 Then, x = 1 and x - 1 is a factor of x³ – 2x² – x + 2. Now, x³ – 2x² – x + 2 is divide by x - 1 p(x) = x³ – 2x² – x + 2 = (x - 1)(x² – x - 2) = (x - 1)(x² – 2x + x - 2) = (x - 1)[x(x - 2) + 1(x - 2)] = (x - 1)(x - 2)(x + 1)

(ii) x³ – 3x² – 9x – 5 Firstly, the factor of 5 is ±1, ±5. p(1) = (1)³ – 3(1)² – 9 x 1 - 5 = 1 - 3 - 9 - 5 = 6 - 17 = -11 ≠ 0 p(-1)= (-1)³ – 3 x (-1)² – 9 x (-1) - 5 = -1 - 3 + 9 - 5 = 9 - 9 = 0 Then, x = -1 and x + 1 is a factor of x³ – 3x² – 9x - 5. Now, x³ – 3x² – 9x - 5 is divide by x + 1 p(x) = x³ – 3x² – 9x - 5 = (x + 1)(x² – 4x - 5) = (x + 1)(x² – 5x + x - 5) = (x + 1)[x(x - 5) + 1(x - 5)] = (x + 1)(x - 5)(x + 1)

(iii) x³ + 13x² + 32x + 20 Firstly, the factor of 20 is ±1, ±2, ±4, ±5. p(-1)= (-1)³ + 13 x (-1)² + 32 x (-1) + 20 = -1 + 13 - 32 + 20 = 33 - 33 = 0 Then, x = -1 and x + 1 is a factor of x³ + 13x² + 32x + 20. Now, x³ + 13x² + 32x + 20 is divide by x + 1 p(x) = x³ + 13x² + 32x + 20 = (x + 1)(x² + 12x + 20) = (x + 1)(x² + 10x + 2x + 20) = (x + 1)[x(x + 10) + 2(x + 10)] = (x + 1)(x + 10)(x + 2)

(iv) 2y³ + y² – 2y – 1 Firstly, the factor of 1 is ±1. p(1) = 2(1)³ + (1)² – 2 x 1 - 1 = 2 + 1 - 2 - 1 = 3 - 3 = 0 Then, y = 1 and y - 1 is a factor of 2y³ + y² – 2y – 1. Now, 2y³ + y² – 2y – 1 is divide by y - 1 p(y) = 2y³ + y² – 2y – 1 = (y - 1)(2y² + 3y + 1) = (y - 1)(2y² + 2y + y + 1) = (y - 1)[2y(y + 1) + 1(y + 1)] = (y - 1)(2y + 1)(y + 1)

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