If 2x³ + ax² +bx-6 has x-1 as a factor and leaves a remainder 2 when divided by
x - 2, find the value of 'a' and 'b'.
Answers
Answer:
The value of a and b for 2x³ + ax² + bx – 6 is – 8 and 12 respectively.
Step-by-step explanation:
Given that, (x – 1) is a factor of 2x³ + ax² + bx – 6.
From factor theorem, we get the value of p(1) :
➟ 2x³ + ax² + bx – 6 = 0
➟ 2 (1)³ + a (1)² + b (1) – 6 = 0
➟ 2 + a + b – 6 = 0
➟ a + b – 4 = 0
➟ a + b = 4 . . . . . . ⑴
It is also given that, when the polynomial has (x – 2) as its factor, it leaves 2 as remainder. So p(2) :
➟ 2x³ + ax² + bx – 6 = 2
➟ 2 (2)³ + a (2)² + b (2) – 6 = 2
➟ 2 (8) + a (4) + 2b – 6 = 2
➟ 16 + 4a + 2b – 6 = 2
➟ 4a + 2b + 10 = 2
➟ 4a + 2b = 2 – 10
➟ 4a + 2b = – 8 . . . . . ⑵
Now, getting the value of a from ⑴ :
➟ a + b = 4
➟ a = 4 – b . . . . . ⑶
Now, substituting this value of a in ⑵ to get b :
➟ 4a + 2b = – 8
➟ 4 (4 – b) + 2b = – 8
➟ 16 – 4b + 2b = – 8
➟ 16 – 2b = – 8
➟ – 2b = – 8 – 16
➟ – 2b = – 24
➟ b = – 24/– 2
➟ b = 12
Substituting this value of b in ⑴ to get a :
➟ a + b = – 4
➟ a + 12 = – 4
➟ a = – 4 – 12
➟ a = – 8
Therefore the values of a and b are – 8 and 12.