If (x+2) and (x-4) are the factors of x³ +tx²+ ux +64. Find the values of t and u
Answers
Given :
- (x+2) and (x-4) are the factors of x³ +tx²+ ux +64.
To find :
- Value of t & u
Solution :
★ If (x+2) and (x-4) are the factors of x³ +tx²+ ux +64 then, x³ +tx²+ ux +64 must be equal to zero
°•° (x + 2) is the factor of x³ + tx²+ ux +64
→ x + 2 = 0
→ x = - 2
Substitute the value
→ x³ +tx²+ ux + 64 = 0
→ (-2)³ + t × (-2)² + u × (-2) + 64 = 0
→ - 8 + 4t - 2u + 64 = 0
→ 4t - 2u + 64 - 8 = 0
→ 4t - 2u = - 56 -----(i)
°•° (x - 4) is the factor of x³ + tx²+ ux +64
→ x - 4 = 0
→ x = 4
Substitute the value
→ x³ +tx²+ ux +64 = 0
→ (4)³ + t × (4)² + u × 4 + 64 = 0
→ 64 + 16t + 4u + 64 = 0
→ 16t + 4u + 64 + 64 = 0
→ 16t + 4u + 128 = 0
→ 16t + 4u = - 128 ----(ii)
Multiply (i) by 2 and (ii) by 1
- 8t - 4u = - 112
- 16t + 4u = - 128
Add both the equations
→ 8t - 4u + 16t + 4u = - 112 + (-128)
→ 24t = - 112 - 128
→ 24t = - 240
→ t = - 240/24
→ t = - 10
Put the value of t in eqⁿ (i)
→ 4t - 2u = - 56
→ 4 × (-10) - 2u = - 56
→ - 40 - 2u = - 56
→ - 2u = - 56 + 40
→ - 2u = - 16
→ u = 16/2
→ u = 8
- Value of t = - 10
- Value of u = 8
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(x + 2) is the factor of x³ + tx²+ ux +64
→ x + 2 = 0
→ x = - 2
Substitute the value
→ x³ +tx²+ ux + 64 = 0
→ (-2)³ + t × (-2)² + u × (-2) + 64 = 0
→ - 8 + 4t - 2u + 64 = 0
→ 4t - 2u + 64 - 8 = 0
→ 4t - 2u = - 56 -----(i)
°•° (x - 4) is the factor of x³ + tx²+ ux +64
→ x - 4 = 0
→ x = 4
Substitute the value
→ x³ +tx²+ ux +64 = 0
→ (4)³ + t × (4)² + u × 4 + 64 = 0
→ 64 + 16t + 4u + 64 = 0
→ 16t + 4u + 64 + 64 = 0
→ 16t + 4u + 128 = 0
→ 16t + 4u = - 128 ----(ii)
Multiply (i) by 2 and (ii) by 1
8t - 4u = - 112
16t + 4u = - 128
Add both the equations
→ 8t - 4u + 16t + 4u = - 112 + (-128)
→ 24t = - 112 - 128
→ 24t = - 240
→ t = - 240/24
→ t = - 10
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Put the value of t in eqⁿ (i)
→ 4t - 2u = - 56
→ 4 × (-10) - 2u = - 56
→ - 40 - 2u = - 56
→ - 2u = - 56 + 40
→ - 2u = - 16
→ u = 16/2
→ u = 8
___________________
Value of t = - 10
Value of t = - 10Value of u = 8