Given that x-2 and x + 1 are factors of f(x) = x + 3x + ax + b, calculate the values
of a and b.
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Answer:
Let f(x)=x
3
+3x
2
+ax+b
As, (x–2) is a factor of f(x), so f(2)=0
(2)
3
+3(2)
2
+a(2)+b=0
8+12+2a+b=0
2a+b+20=0…(1)
And as, (x+1) is a factor of f(x), so f(−1)=0
(−1)
3
+3(−1)
2
+a(−1)+b=0
−1+3–a+b=0
−a+b+2=0…(2)
Subtracting (2) from (1), we have
3a+18=0
a=−6
On substituting the value of a in (ii), we have
b=a–2=−6–2=−8
Thus, f(x)=x
3
+3x
2
–6x–8
Now, for x=−1
f(−1)=(−1)
3
+3(−1)
2
–6(−1)–8=−1+3+6–8=0
Therefore, (x+1) is a factor of f(x).
Now, performing long division we have
Hence, f(x)=(x+1)(x
2
+2x–8)
=(x+1)(x
2
+4x–2x–8)
=(x+1)[x(x+4)–2(x+4)]
=(x+1)(x+4)(x–2)
Step-by-step explanation:
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