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Question
Given that x – 2 and x + 1 are factors of f(x) = x3 + 3x2 + ax + b; calculate the values of a and b. Hence, find all the factors of f(x).
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Solution
f(x)= x3 + 3x2 + ax + b
Since, (x – 2) is a factor of f(x), f(2) = 0
`\implies` (2)3 + 3(2)2 + a(2) + b = 0
`\implies` 8 + 12 + 2a + b = 0
`\implies` 2a + b + 20 = 0 ...(i)
Since, (x + 1) is a factor of f(x), f(–1) = 0
`\implies` (–1)3 + 3(–1)2 + a(–1) + b = 0
`\implies` –1 + 3 – a + b = 0
`\implies` –a + b + 2 = 0 ...(ii)
Subtracting (ii) from (i), we get,
3a + 18 = 0
⇒ a = – 6
Substituting the value of a in (ii), we get,
b = a – 2
= – 6 – 2
= – 8
∴ f(x) = x3 + 3x2 – 6x – 8
Now, for x = –1,
f(x) = f(–1)
= (–1)3 + 3(–1)2 – 6(–1) – 8
= –1 + 3 + 6 – 8
= 0
Hence, (x + 1) is a factor of f(x).
x2 + 2x – 8
`x + 1")"overline(x^3 + 3x^2 - 6x - 8)`
x3 + x2
2x2 – 6x
2x2 + 2x
– 8x – 8
– 8x – 8
0
∴ x3 + 3x2 – 6x – 8 = (x + 1)(x2 + 2x – 8)
= (x + 1)(x2 + 4x – 2x – 8)
= (x + 1)[x(x + 4) – 2(x + 4)]
= (x + 1)(x + 4)(x – 2)
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