TOPICS

Unit-2(Examples)

Polynomials

**Example-1 :-** Find the degree of each of the polynomials given below:

(i) x⁵ – x⁴ + 3, (ii) 2 – y² – y³ + 2y⁸, (iii) 2 .

(i) x⁵ – x⁴ + 3, The highest power of the variable is 5. So, the degree of the polynomial is 5. (ii) 2 – y² – y³ + 2y⁸, The highest power of the variable is 8. So, the degree of the polynomial is 8. (iii) 2 . There is no any variable with 2. Therefore, the degree of the polynomial is 0.

**Example-2 :-** Find the value of each of the following polynomials at the indicated value of variables:

(i) p(x) = 5x² – 3x + 7 at x = 1.

(ii) q(y) = 3y³ – 4y + √11 at y = 2.

(iii) p(t) = 4t⁴ + 5t³ – t² + 6 at t = a.

(i) p(x) = 5x² – 3x + 7 at x = 1. p(1) = 5 x (1)² - 3 x 1 + 7 = 5 - 3 + 7 = 12 - 3 = 9 (ii) q(y) = 3y³ – 4y + √11 at y = 2. q(2) = 3 x (2)³ – 4 x 2 + √11 = 3 x 8 - 8 + √11 = 24 - 8 + √11 = 16 - √11 (iii) p(t) = 4t⁴ + 5t³ – t² + 6 at t = a. p(a) = 4 x (a)⁴ + 5 x (a)³ – (a)² + 6 = 4a⁴ + 5a³ – a² + 6

**Example-3 :-** : Check whether –2 and 2 are zeroes of the polynomial x + 2.

Let p(x) = x + 2 at x = -2 p(-2) = -2 + 2 = 0 Now, p(x) = x + 2 at x = 2 p(2) = 2 + 2 = 4 So, -2 is zeroe of polynomial x + 2, but 2 is not.

**Example-4 :-** Find a zero of the polynomial p(x) = 2x + 1.

p(x) = 2x + 1. Finding for zero p(x) = 0 0 = 2x + 1 -1 = 2x -1/2 = x x = -1/2 The zero of polynomial p(x) = 2x + 1 is -1/2.

**Example-5 :-** Verify whether 2 and 0 are zeroes of the polynomial x² – 2x.

Let p(x) = x² – 2x at x = 2 p(2) = (2)² + 2 = 4 + 2 = 6 Now, p(x) = x² – 2x at x = 0 p(2) = 0² + 2 x 0 = 0 + 0 = 0 So, 0 is zero of polynomial x² – 2x, but 2 is not.

**Example-6 :-** : Divide p(x) by g(x), where p(x) = x + 3x² – 1 and g(x) = 1 + x.

p(x) = x + 3x² – 1 and g(x) = 1 + x Here, q(x) = 3x - 2 and r(x) = 1 Verification : g(x) = 0 1 + x = 0 x = -1 Now, x = -1 put in p(x) = x + 3x² – 1 p(-1) = -1 + 3 x (-1)² - 1 = -1 + 3 x 1 - 1 = -1 + 3 - 1 = 3 - 2 = 1 = r(x)

**Example-7 :-** Divide the polynomial 3x⁴ – 4x³ – 3x – 1 by x – 1.

p(x) = 3x⁴ – 4x³ – 3x – 1 and g(x) = x – 1 Here, q(x) = 3x³ – x² – x – 4 and r(x) = -5 Verification : g(x) = 0 x - 1 = 0 x = 1 Now, x = 1 put in p(x) = 3x⁴ – 4x³ – 3x – 1 p(1) = 3x⁴ – 4x³ – 3x – 1 = 3 x (1)⁴ - 4 x (1)³ - 3 x 1 - 1 = 3 - 4 - 3 - 1 = 3 - 8 = -5 = r(x)

**Example-8 :-** Find the remainder obtained on dividing p(x) = x³ + 1 by x + 1.

p(x) = x³ + 1 and g(x) = x + 1 Here, q(x) = x² – x + 1 and r(x) = 0 Verification : g(x) = 0 x + 1 = 0 x = -1 Now, x = -1 put in p(x) = x³ + 1 p(1) = x³ + 1 = (-1)³ + 1 = -1 + 1 = 0 = r(x)

**Example-9 :-** : Find the remainder when x⁴ + x³ – 2x² + x + 1 is divided by x – 1.

p(x) = x⁴ + x³ – 2x² + x + 1 and g(x) = x - 1 Now, g(x) = 0 x - 1 = 0 x = 1 Now, x = 1 put in p(x) = x⁴ + x³ – 2x² + x + 1 p(1) = (1)⁴ + (1)³ – 2(1)² + 1 + 1 = 1 + 1 - 2 + 1 + 1 = 4 - 2 = 2 = r(x) So, by the Remainder Theorem, 2 is the remainder when x⁴ + x³ – 2x² + x + 1 is divided by x – 1.

**Example-10 :-** Check whether the polynomial q(t) = 4t³ + 4t² – t – 1 is a multiple of 2t + 1.

q(t) = 4t³ + 4t² – t – 1 g(t) = 2t + 1 Now, g(t) = 0 2t + 1 = 0 2t = -1 t = -1/2 Now, t = -1/2 put in q(t) = 4t³ + 4t² – t – 1 q(-1/2) = 4(-1/2)³ + 4(-1/2)² – (-1/2) – 1 = 4 x (-1/8) + 4 x 1/4 + 1/2 - 1 = -1/2 + 1 + 1/2 - 1 = 0 = r(t) So the remainder obtained on dividing q(t) by 2t + 1 is 0. So, 2t + 1 is a factor of the given polynomial q(t), i.e., q(t) is a multiple of 2t + 1.

**Example-11 :-** Examine whether x + 2 is a factor of x³ + 3x² + 5x + 6 and of 2x + 4.

p(x) = x³ + 3x² + 5x + 6 and g(x) = x + 2 Now, we take g(x) = 0 x + 2 = 0 x = -2 Now, x = -2 put in p(x) = x³ + 3x² + 5x + 6 p(1) = (-2)³ + 3(-2)² + 5 x (-2) + 6 = -8 + 12 - 10 + 6 = 18 - 18 = 0 So, by the Factor Theorem, x + 2 is a factor of x³ + 3x² + 5x + 6. Again, s(–2) = 2(–2) + 4 = 0. So, x + 2 is a factor of 2x + 4.

**Example-12 :-** : Find the value of k, if x – 1 is a factor of 4x³ + 3x² – 4x + k.

p(x) = 4x³ + 3x² – 4x + k and g(x) = x - 1 Now, g(x) = 0 x - 1 = 0 x = 1 Now, x = 1 put in p(x) = 4x³ + 3x² – 4x + k and p(1) = 0 p(1) = 4(1)³ + 3(1)² – 4(1) + k 0 = 4 + 3 - 4 + k 0 = 3 + k k = -3

**Example-13 :-** Factorise 6x² + 17x + 5 by splitting the middle term, and by using the Factor Theorem.

By splitting method : If we can find two numbers a and b such that a + b = 17 and ab = 6 × 5 = 30, then we can get the factors. So, 6x² + 17x + 5 = 6x² + (2 + 15)x + 5 = 6 x² + 2x + 15x + 5 = 2 (3x + 1) + 5 (3x + 1) = (3x + 1) (2x + 5) By Factor method : 6x² + 17x + 5 = 6(x² + 17x/6 + 5/6) = 6 p(x) If a and b are the zeroes of p(x), then So, 6x² + 17x + 5 = 6(x – a)(x – b). So, ab = 5/6 and factors of 5 = 5 x 1, 6 = 2 x 3 x 1 Let us look at some possibilities for a and b. They could be ±1/2, ±1/3, ±5/2, ±5/3, ±1. Now, p(1/2) = 6[(1/2)² + 17/12 + 5/6] ≠ 0. p(-1/3) = 0 and p(-5/2) = 0. So, (x + 1/3) and (x + 5/2) are factors of p(x). Therefore, = 6(x + 1/3)(x + 5/2) = 6 x (3x + 1)/3 x (2x + 5)/2 = 2(3x + 1)(2x + 5)/2 = (3x + 1)(2x + 5)

**Example-14 :-** Factorise y² – 5y + 6 by using the Factor Theorem.

By Factor method : p(x) = y² – 5y + 6 Factor of 6 = 2 x 3 x 1 Some possibilities are ±1, ±2, ±3. Now, p(1) = (1)² – 5 x (1) + 6 = 1 - 5 + 6 ≠ 0. p(2) = (2)² - 5 x 2 + 6 = 4 - 10 + 6 = 10 - 10 = 0 p(3) = (3)² - 5 x 3 + 6 = 9 - 15 + 6 = 15 - 15 = 0 There are two factors of p(x). i.e., (y - 2) and (y - 3). p(x) = y² – 5y + 6 = (y - 2)(y - 3)

**Example-15 :-** : Factorise x³ – 23x² + 142x – 120.

p(x) = x³ – 23x² + 142x – 120 All factors of -120 is ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±60. Now, p(1) = 0. So x – 1 is a factor of p(x). p(x) = x³ – 23x² + 142x – 120 is divide by x - 1 Now, we can take x³ – 23x² + 142x – 120 = (x - 1)(x² – 22x + 120) p(x) = (x - 1)(x² – 22x + 120) = (x - 1)(x² – 12x – 10x + 120) = (x - 1)[x(x – 12) – 10(x – 12)] = (x - 1)(x – 12)(x – 10) So, x³ – 23x² + 142x – 120 = (x – 1)(x – 10)(x – 12)

**Example-16 :-** Find the following products using appropriate identities:

(i) (x + 3) (x + 3)

(ii) (x – 3) (x + 5)

(i) (x + 3) (x + 3) = (x + 3)² [(a + b)² = a² + b² + 2ab] = x² + 3² + 2 x X x 3 = x² + 9 + 6x (ii) (x – 3) (x + 5) = x² + (-3 + 5)x + (-3)(5) [(x + a)(x + b) = x² + (a + b)x + ab] = x² + 2x - 15

**Example-17 :-** Evaluate 105 × 106 without multiplying directly.

105 × 106 = (100 + 5)(100 + 6) = (100)² + (5 + 6)100 + 5 x 6 [(x + a)(x + b) = x² + (a + b)x + ab] = 10000 + 11 x 100 + 30 = 10000 + 1100 + 30 = 11130

**Example-18 :-** : Factorise:

(i) 49a² + 70ab + 25b²

(ii) 25x²/4 - y²/9

(i) 49a² + 70ab + 25b² = (7a)² + 2 x 7a x 5b + (5b)² [(a + b)² = a² + b² + 2ab] = (7a + 5b)² = (7a + 5b)(7a + 5b) (ii) 25x²/4 - y²/9 = (5x/2)² - (y/3)² [a² - b² = (a + b)(a - b)] = (5x/2 + y/3)(5x/2 - y/3)

**Example-19 :-** Write (3a + 4b + 5c)² in expanded form.

(3a + 4b + 5c)² By using [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac] = (3a)² + (4b)² + (5c)² + 2 x 3a x 4b + 2 x 4b x 5c + 2 x 3a x 5c = 9a² + 16b² + 25c² + 24ab + 40bc + 30ac

**Example-20 :-** Expand (4a – 2b – 3c)².

(4a – 2b – 3c)² By using [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac] = (4a)² + (-2b)² + (-3c)² + 2 x 4a x (-2b) + 2 x (-2b) x (-3c) + 2 x 4a x (-3c) = 16a² + 4b² + 9c² - 16ab + 12bc - 24ac

**Example-21 :-** : Factorise 4x² + y² + z² – 4xy – 2yz + 4xz.

4x² + y² + z² – 4xy – 2yz + 4xz By using [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac] = (2x)² + (-y)² + z² + 2 x (2x) x (-y) + 2 x (-y) x z + 2 x (2x) x z = (2x - y + z)² = (2x - y + z)(2x - y + z)

**Example-22 :-** Write the following cubes in the expanded form:

(i) (3a + 4b)³

(ii) (5p – 3q)³

(i) (3a + 4b)³ By Using [(a + b)³ = a³ + b³ + 3a²b + 3ab²] = (3a)³ + (4b)³ + 3 x (3a)² x 4b + 3 x 3a x (4b)² = 27a³ + 64b³ + 3 x 9a² x 4b + 3 x 3a x 14b² = 27a³ + 64b³ + 108a²b + 126ab² (ii) (5p – 3q)³ By Using [(a - b)³ = a³ - b³ - 3a²b + 3ab²] = (5p)³ - (3q)³ - 3 x (5p)² x 3q + 3 x 5p x (3q)² = 125p³ - 27q³ - 3 x 25p² x 3q + 3 x 5p x 9q² = 125p³ - 27q³ - 225p² + 135pq²

**Example-23 :-** Evaluate each of the following using suitable identities:

(i) (104)³, (ii) (999)³

(i) (104)³ = (100 + 4)³ [(a + b)³ = a³ + b³ + 3a²b + 3ab²] = (100)³ + (4)³ + 3 x (100)² x 4 + 3 x 100 x (4)² = 1000000 + 64 + 3 x 10000 x 4 + 3 x 100 x 16 = 1000000 + 64 + 120000 + 4800 = 1124864 (ii) (999)³ = (1000 - 1)³ [(a - b)³ = a³ - b³ - 3a²b + 3ab²] = (1000)³ - (1)³ - 3 x (1000)² x 1 + 3 x 1000 x (1)² = 1000000000 - 1 - 3 x 1000000 x 1 + 3 x 1000 x 1 = 1000000000 - 1 - 3000000 + 3000 = 1000003000 - 3000001 = 997002999

**Example-24 :-** : Factorise 8x³ + 27y³ + 36x²y + 54xy².

8x³ + 27y³ + 36x²y + 54xy² By using [(a + b)³ = a³ + b³ + 3a²b + 3ab²] = (2x)³ + (3y)³ + 3 x (2x)² x 3y + 3 x 2x x (3y)² = (2x + 3y)³ = (2x + 3y)(2x + 3y)(2x + 3y)

**Example-25 :-** Factorise : 8x³ + y³ + 27z³ – 18xyz

8x³ + y³ + 27z³ – 18xyz By using [x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx) ] = (2x)³ + y³ + (3z)³ – 3 x (2x) x (y) x (3z) = (2 x + y + 3z)[(2x)² + y² + (3z)² – (2x) x (y) – (y) x (3z) – (2x) x (3z)] = (2 x + y + 3z) (4x² + y² + 9z² – 2xy – 3yz – 6xz)

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