Polynomials – Exercise 8.4 – Class X

Previous Exercise – Polynomials – Exercise 8.3 – Class X

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Polynomials – Exercise 8.4

1. In each of the following cases, use factor theorem to find whether g(x) is a factor of the polynomials p(x) or not.

(i) p(x) = x³ – 3x² + 6x – 20 ; g(x) = x – 2

(ii) p(x) = 2x⁴ + x³ + 4x² – x – 7 ; g(x) = x + 2

(iii) p(x) = 3x⁴ + 3x² – 2x² – 9x – 12 ; g(x) = x – ¹/₂

(iv) p(x) = 3x³ + x² – 20x + 12 ; g(x) = 3x – 2

(v) p(x) = 2x⁴ + 3x³ – 2x² – 9x – 12 ; g(x) = x² – 3

2. Find the valule of a if (x – 5) is a factor of (x³ – 3x² + ax – 10)
3. If (x³ + ax – bx + 10) is divisible by x² – 3x + 2 , find the values of a and b
4. If both (x – 2) and (x – ¹/₂)are factors of (ax² + 5x + b), show that a = b

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Polynomials – Exercise 8.4 –Solutions:

1. In each of the following cases, use factor theorem to find whether g(x) is a factor of the polynomials p(x) or not.

(i) p(x) = x³ – 3x² + 6x – 20 ; g(x) = x – 2

Solution:

Let p(x) = x³ – 3x² + 6x – 20

By factor theorem, (x – 2) is a factor of p(x) if p(2) = 0

p(2) = 2³ – 3(2)² + 6(2) – 20

= 8 – 12 + 12 – 20

= – 12

⸫ (x – 2) is not a factor of p(x) = x³ – 3x² + 6x – 20

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(ii) p(x) = 2x⁴ + x³ + 4x² – x – 7 ; g(x) = x + 2

Solution:

Let p(x) = 2x⁴ + x³ + 4x² – x – 7

By factor theorem, (x + 2) is a factor of p(x) if p(-2) = 0

p(-2) = 2(-2)⁴ + (-2)³ + 4(-2)² – (-2) – 7

= 32 – 8 + 16 + 2 – 7

= 35

⸫ (x + 2) is not a factor of p(x) = 2x⁴ + x³ + 4x² – x – 7

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 (iii) p(x) = 3x⁴ + 3x³ – 2x² – 9x – 12 ; g(x) = x – ¹/₂

Solution:

Let p(x) = 3x⁴ + 3x³ – 2x² – 9x – 12

By factor theorem, (x – ¹/₂) is a factor of p(x) if p(¹/₂) = 0

p(¹/₂) = 3(¹/₂)⁴ + 3(¹/₂)³ – 2(¹/₂)² – 9(¹/₂) – 12

= 3(¹/₁₆) + 3(¹/₈) – 2(¹/₄) – 9(¹/₂) – 12

= – ²⁶³/₁₆

⸫ (x – ¹/₂) is not a factor of p(x) = 3x⁴ + 3x² – 2x² – 9x – 12

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 (iv) p(x) = 3x³ + x² – 20x + 12 ; g(x) = 3x – 2

Solution:

Let p(x) = 3x³ + x² – 20x + 12

By factor theorem, (3x – 2) is a factor of p(x) if p(²/₃) = 0

p(²/₃) = 3(²/₃)³ + (²/₃)² – 20(²/₃) + 12

= 0

⸫ (3x – 2) is a factor of p(x) = 3x³ + x² – 20x + 12

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 (v) p(x) = 2x⁴ + 3x³ – 2x² – 9x – 12 ; g(x) = x² – 3

Solution:

Let p(x) = 2x⁴ + 3x³ – 2x² – 9x – 12

By factor theorem, (x² – 3) is a factor of p(x) if p(√3) = 0

p (√3) = 2 (√3)⁴ + 3 (√3)³ – 2 (√3)² – 9 (√3) – 12

= 2x3x3 + 3×3√3 – 2×3 – 9√3 – 12

= 18 + 9√3 – 6 – 9√3 – 12

= 0

⸫ (x² – 3) is a factor of p(x) = 2x⁴ + 3x³ – 2x² – 9x – 12

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2. Find the value of a if (x – 5) is a factor of (x³ – 3x² + ax – 10)

Solution:

Let p(x) = x³ – 3x² + ax – 10

By factor theorem, (x – 5) is a factor of p(x) if p(5) = 0

p(x) = x³ – 3x² + ax – 10 = 0

5³ – 3(5²) + a(5) – 10 = 0

125 – 75 + 5a – 10 = 0

40 + 5a = 0

5a = – 40

a = –⁴⁰/₅ = – 8

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3. If (x³ + ax² – bx + 10) is divisible by x² – 3x + 2 , find the values of a and b

Solution:

p(x) = x³ + ax² – bx + 10

g(x) = x² – 3x + 2

g(x) = x² –x – 2x + 2

g(x) = x(x – 1)-2(x – 1)

g(x) = (x – 1)(x – 2)

p(x) = (x³ + ax² – bx + 10) is divisible by g(x) = (x – 1)(x – 2) , then we have p(2) = 0 and p(1) = 0

⇒ p(2) = p(1)

We have to find the values of a and b.

p(x) = x³ + ax² – bx + 10

p(2) = (2)³ + a(2)² – b(2) + 10 = 8 + 4a – 2b + 10 = 4a – 2b + 18 = 2a – b + 9 = 0

p(1) = (1)³ + a(1)² – b(1) + 10 = 1 + a – b + 10 = a – b + 11 = 0

Since p(2) = p(1)

2a – b + 9 = a – b + 11

2a – a = 11 – 9

a = 2

Substitute the value of a in p(2) = 0

p(2) = a – b + 11 = 0

⇒ 2 – b + 11 = 0

-b = -11 – 2 = – 13

b = 13

Therefore, a = 2 and b = 13

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4. If both (x – 2) and (x – ¹/₂)are factors of (ax² + 5x + b), show that a = b

Solution:

p(x) = ax² + 5x + b

By factor theorem, (x – 2) and (x – ¹/₂) is a factor of p(x) then p(2) = p(¹/₂) = 0

p(2) = a(2)² + 5(2) + b = 4a + b + 10 = 0

p(¹/₂) = a(¹/₂)² + 5(¹/₂) + b = ^a/₄ + ⁵/₂ + b = 0

Since p(2) = p(¹/₂)

4a + b + 10 = ^a/₄ + ⁵/₂ + b

4a + 10 = ^a/₄ + ⁵/₂

4a – ^a/₄ = ⁵/₂ – 10

^{16a – a}/₄ = ^5-20/₂

^15a/₄ = ¹⁵/₂

^a/₂ = 1

a = 2

Substitute the value of p(2) = 4a + b + 10 = 0

4a + b + 10 = 0

8 + b + 10 = 0

b = -18

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Next exercise – Polynomials – Exercise 8.5 – Class X

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