Question

f 2x³ + ax² – 11x + b leaves remainder 0 and 42 when divided by (x – 2) and (x – 3) respectively, find the values of a and b. With these values of a and b, factorize the given expression.

Answer 1

f(x) = 2x³ + ax² – 11 x + b
Let x – 2 = 0, then x = 2,
Substituting the vaue of x in f(x)
f(2) = 2(2)³ + a(2)² – 11(2) + b
∵ Remainder = 0,
∴ 4a + b – 6 = 0
⇒ 4a + b = 6       ...(i)
Again let x – 3 = 0,
then x = 3
Substituting the value of x is f(x)
f(3) = 2(3)3 + a(3)2 – 11 x 3 + b
= 2 x 27 + 9a – 33 + b
= 54 + 9a – 33 + b
⇒ 9a + b + 21
∵ Remainder = 42
∴ 9a + b + 21 = 42
⇒ 9a + b = 42 – 21
⇒ 9a + b = 21        ...(ii)
Subtracting (i) from (ii)
5a = 15
⇒ `(15)/(5)` = 3
Substituting the value of a is (i)
4(3) + b = 6
⇒ 12 + b = 6
⇒ b = 6 – 12
⇒ b = –6
∴ f(x) will be 2x³ + 3x² – 11 x – 6
∵ x – 2 is a factor (as remainder = 0)
∴ Dividing f(x) by x – 2, we get

`x - 2")"overline(2x^3 + 3x^2 - 11x - 6)("2x^2 + 7x + 3`
           2x3 – 4x2
        –       +                           
                 7x2 – 11x
                 7x2 – 14x
                –      +                    
                        3x – 6
                        3x – 6
                        –   +       
                            x         
∴ 2x³ + 3x² – 11 x –6
= (x – 2)(2x² + 7x + 3)
= (x – 2)[2x² + 6x + x + 3]
= (x – 2)[2x(x + 3) + 1(x + 3)]
= (x – 2)(x + 3)(2x + 1).