If ax^3+bx^2+x-6 has (x+2) as a factor and leaves a remainder 4 when divided by(x-2), find the value of'a' and 'b'
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F(x) = ax3 + bx2 + x – 6
F(- 2) = -8a + 4b – 2 – 6 = -8a + 4b – 8 = 0
So, 2a – b + 2 = 0 ……(i)
F(2) = 8a + 4b + 2 – 6 = 8a + 4b – 4 = 4
So, 2a + b – 2 = 0 ……..(ii)
From (i) and (ii), we get
a = 0 , b = 2,
F(- 2) = -8a + 4b – 2 – 6 = -8a + 4b – 8 = 0
So, 2a – b + 2 = 0 ……(i)
F(2) = 8a + 4b + 2 – 6 = 8a + 4b – 4 = 4
So, 2a + b – 2 = 0 ……..(ii)
From (i) and (ii), we get
a = 0 , b = 2,
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