Question

If (x + 2) and (x – 3) are factors of x³ + ax + b, find the values of a and b. With these values of a and b, factorise the given expression.

Answer 1

Let x + 2 = 0, then x = –2

Substituting the value of x in f(x),

f(x) = x³ + ax + b

f(–2) = (–2)² + a(–2) + b

= –8 – 2a + b

∵ x + 2 is a factor

∴ Remainder is zero

∴ –8 – 2a + b = 0

⇒ –2a + b = 8

∴ 2a – b = –8              ...(i)

Again let x – 3 = 0, then x = 3

Substituting the value of x in f(x),

f(x) = x³ + ax + b

f(3) = (3)³ + a(3) + b

= 27 + 3a + b

∵ x – 3 is a factor  

∴ Remainder = 0

⇒ 27 + 3a + b = 0

⇒ 3a + b = –27          ...(ii)

Adding (i) and (ii)

5a = –35

⇒ a = `(-35)/(5)`

⇒ a = –7

Substituting value of a in (i)

2(–7) –b = –8

⇒ –14 – b = –8

–b = –8 + 14

⇒ –b = 6

∴ b = –6

Hence, a = –7, b = –6

(x + 2) and (x – 3) are the factors of 

x³ + ax + b

⇒  x³ – 7x – 6

Now dividing x³ – 7x – 6 by (x + 2)

(x – 3) or x² – x – 6, we get

`x^2 - x  –  6")"overline(x^3 - 7x  –  6)("x + 1`
                    x³ –  x² – 6x         
                    –      +     +                     
                           x² –  x – 6
                           x² –  x – 6
                         –      +    +              
                                  x             
∴ Factors are (x + 2), (x – 3) and (x + 1).