Question

Given f(x) = ax² + bx + 2 and g(x) = bx² + ax + 1. If x – 2 is a factor of f(x) but leaves the remainder – 15 when it divides g(x), find the values of a and b. With these values of a and b, factorise the expression. f(x) + g(x) + 4x² + 7x.

Answer 1

f(x) = ax² + bx + 2
g(x) = bx² + ax + 1
x – 2 is a factor of f(x)
Let x – 2 = 0
⇒ x = 2
∴ f(2) = a(2)² + b x 2 + 2 = 4a + 2b + 2
∴ 4a + 2b + 2 = 0    ...(∵ x – 2 s its factor)
⇒ 2a + b + 1 = 0   ....(i) (Dividing by 2)
Dividing g (x) by x – 2, remainder = –15
Let x – 2 = 0
⇒ x = 2
∴ g(2) = b(2)² + a x 2  + 1
= 4b + 2a + 1
∵ Remainder is –15
∴ 4b + 2a + 1 = –15
⇒ 4b + 2a + 1 + 15 = 0
⇒ 4b + 2a + 16 = 0
⇒ 2b + a + 8 = 0  ...(Dividing by 2)
⇒ a + 2b + 8 = 0
Multiplying (i) by 2 and (ii) by 1
4a + 2b + 2 = 0
  a + 2b + 8 = 0
  –      –      –       
3a         –6    = 0

⇒ 3a = 6
⇒ a = `(6)/(3)`
∴ a = 2
Substituting the value of a in (i)
2 x 2 + b + 1 = 0
⇒ 4 + b + 1 = 0
⇒ b + 5 = 0
⇒ b = –5
Hence a = 2, b = –5
Now f(x) + g(x) = 4x2 + 7x
= 2x² – 5x + 2 + (–5x² + 2x + 1) + 4x² + 7x
= 2x² – 5x + 2 – 5x² + 2x + 1 + 4x² + 7x
= 6x² – 5x² – 5x + 2x + 7x + 2 + 1
= x² + 2x + 3
= x² + x  + 3x + 3
= x(x + 1) + 3(x + 1)
= (x + 1)(x + 3).