find the values of the contants a and b if (x-2) and (x+3) are both factors of the expression x^3+ax^2+bx-12
Answers
Answer:
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Step-by-step explanation:
Expression x3 + ax2 + bx – 12
(x – 2) is a factor i.e., at x = 2 the remainder will be zero
⇒ (2)3 + a(2)2 + b(2) – 12 = 0
⇒ 8 + 4a + 2b – 12 = 0
⇒ 4a + 2b = 4
⇒ 2a + b = 2 ….(i)
When x + 3 is a factor i.e., at x = - 3 the remainder will be zero.
⇒ (- 3)3 + a(- 3)2 + b(- 3) – 12 = 0
⇒ - 27 + 9a – 3b – 12 = 0
⇒ 9a – 3b = 39
⇒ 3a – b = 13 ......(ii)
Solving (i) and (ii) simultaneously
2a + b = 2
By adding 3a – b = 13
5a = 15
a = 3
Substituting the value of a in the equation (i)
⇒ 2 × 3 + b = 2
⇒ 6 + b = 2
⇒ b = 2 – 6 = - 4
⇒ a = 3, b = - 4
Answer:
Step-by-step explanation:
Expression x3 + ax2 + bx – 12
(x – 2) is a factor i.e., at x = 2 the remainder will be zero
⇒ (2)3 + a(2)2 + b(2) – 12 = 0
⇒ 8 + 4a + 2b – 12 = 0
⇒ 4a + 2b = 4
⇒ 2a + b = 2 ….(i)
When x + 3 is a factor i.e., at x = - 3 the remainder will be zero.
⇒ (- 3)3 + a(- 3)2 + b(- 3) – 12 = 0
⇒ - 27 + 9a – 3b – 12 = 0
⇒ 9a – 3b = 39
⇒ 3a – b = 13 ......(ii)
Solving (i) and (ii) simultaneously
2a + b = 2
By adding 3a – b = 13
5a = 15
a = 3
Substituting the value of a in the equation (i)
⇒ 2 × 3 + b = 2
⇒ 6 + b = 2
⇒ b = 2 – 6 = - 4
⇒ a = 3, b = - 4