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Question
If (x + 1) and (x – 2) are factors of x3 + (a + 1)x2 – (b – 2)x – 6, find the values of a and b. And then, factorise the given expression completely.
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Solution
Let f(x) = x3 + (a + 1)x2 – (b – 2)x – 6
Since, (x + 1) is a factor of f(x).
∴ Remainder = f(–1) = 0
(–1)3 + (a + 1)(–1)2 – (b – 2)(–1) – 6 = 0
–1 + (a + 1) + (b – 2) – 6 = 0
a + b – 8 = 0 ...(i)
Since, (x – 2) is a factor of f(x).
∴ Remainder = f(2) = 0
(2)3 + (a + 1)(2)2 – (b – 2)(2) – 6 = 0
8 + 4a + 4 – 2b + 4 – 6 = 0
4a – 2b + 10 = 0
2a – b + 5 = 0 ...(ii)
Adding (i) and (ii), we get,
3a – 3 = 0
a = 1
Substituting the value of a in (i), we get,
1 + b – 8 = 0
b = 7
∴ f(x) = x3 + 2x2 – 5x – 6
Now, (x + 1) and (x – 2) are factors of f(x).
Hence, (x + 1)(x – 2) = x2 – x – 2 is a factor of f(x).
x + 3
`x^2 - x - 2")"overline(x^3 + 2x^2 - 5x - 6)`
x3 – x2 – 2x
3x2 – 3x – 6
3x2 – 3x – 6
0
∴ f(x) = x3 + 2x2 – 5x – 6 = (x + 1)(x – 2)(x + 3)
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