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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1. 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:

   Hope it helps you dear friend

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