Question

If (x + 3) and (x – 4) are factors of x³ + ax² – bx + 24, find the values of a and b: With these values of a and b, factorise the given expression.

Answer 1

f(x) = x³ + ax² – bx + 24
Let x + 3 = 0, then x = -3
Substituting the value of x in f(x)
f(–3) = (–3)3 + a(–3)2– b(–3) + 24,
= –27 + 9a + 3b + 24
= 9a + 3b – 3
∵ x + 3 is a factor,
∴ Remainder = 0,
∴ 9a + 3b – 3 = 0
⇒ 3a + b – 1 = 0   ...(Dividing by 3)
⇒ 3a + b = 1                               ...(i)
Again Let x – 4 = 0,
then x = 4
Substituting the value of x in f(x)
f(x) = (4)² + a(4)² – b(4) + 24
= 64 + 16a – 4b + 24
= 16a – 4b + 88
∵ x – 4 is a factor
∴ Remainder = 0
16a – 4b + 88 = 0
⇒ 16a – 4b = – 88    ...(Dividing by 4)
⇒ 4a – b = –22                               ...(ii)
Adding (i) and (ii)
7a = –21,
⇒ a = –3
Substituting the value of a in (i)
3(–3) + b = 1
⇒ –9 + b = 1
⇒ b = 1 + 9 = 10
∴ a = –3, b = 10
Now f(x) will be
f(x) = x³ – 3x² – 10x + 24
∵ x + 3 and x – 4 are factors of f(x)
∴ Dividing f(x) by (x + 3)(x – 4)
or
x² – x – 12

`x^3 - x - 12")"overline(x^3 - 3x^2 - 10x + 24)("x - 2`
                       x³ – x² – 12x
                      –       +       +               
                              –2x² + 2x + 24
                              –2x² + 2x + 24
                              +       –      –        
                                       x                
x³ – 3x² – 10x + 24
= (x² – x – 12)(x – 2)
= (x + 3)(x – 4)(x – 2).