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Question
Find the value of the constants a and b, if (x – 2) and (x + 3) are both factors of the expression x3 + ax2 + bx – 12.
Solution
Let x – 2 = 0, then x = 0
Substituting value of x in f(x)
f(x) = x3 + ax2 + bx – 12
f(2) = (2)3 + a(2)2 + b(2) – 12
= 8 + 4a + 2b – 12
= 4a + 2b – 4
∵ x – 2 is a factor
∴ 4a + 2b – 4 = 0
⇒ 4a + 2b = 4
⇒ 2a + b = 2
Again let x + 3 = 0,
then x = –3
Substituting the value of x in f(x)
f(x) = x3 + ax2 + bx – 12
= (–3)3 + a(–3)2 + b(–3) – 12
= –27 + 9a – 3b – 12
= –39 + 9a – 3b
∵ x + 3 is a factor of f(x)
∴ –39 + 9a – 3b = 0
⇒ 9a – 3b = 39
⇒ 3a – b = 13
Adding (i) and (ii)
5a = 15
⇒ a = 3
Substituting the value of a in (i)
2(3) + b = 2
⇒ 6 + b = 2
⇒ b = 2 – 6
∴ b = –4
Hence a = 3, b = –4.
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