Question

If the polynomial 2x³ + 3x² + px + q is exactly divisible by 2x² + 9x + 9, find the value of p and q. Hence, factorise the expression completely.

Answer 1

Let f(x) = 2x³ + 3x² + px + q and g(x) = 2x² + 9x + 9

Step 1: Factorise the divisor

The divisor polynomial is 2x² + 9x + 9, which can be factored into its linear factors:

2x² + 9x + 9 = 0

2x² + 6x + 3x + 9 = 0

2x (x + 3) + 3(x + 3) = 0

(2x + 3) (x + 3) = 0

x = −3

x = `−3/2`

Step 2: Apply the factor theorem

We substitute x = −3 and x = `−3/2` into f(x) to form two equations:

f(−3) = 0

⇒ 2(−3)³ + 3(−3)² + p(−3) + q = 0

⇒ 2(−27) + 3(9) − 3p + q = 0

⇒ −54 + 27 − 3p + q = 0

⇒ −27 − 3p + q = 0

⇒ q − 3p = 27

q = 27 + 3p   ...(i)

`f(−3/2)` = 0

⇒ `2(-3/2)^3 + 3(-3/2)^2 + p(-3/2) + q` = 0

⇒ `-27/4 + 27/4 - (3p)/2 + q` = 0

⇒ `-(3p)/2 + q` = 0

⇒ q = `(3p)/2`   ...(ii)

Step 3: Solve equations (i) and (ii)

27 + 3p = `(3p)/2`

Multiply by 2:

54 + 6p = 3p

54 = −3p

p = −18

Now find q:

q = 27 + 3p

= 27 + 3(−18)

= 27 − 54

= −27

Now f(x),

f(x) = 2x³ + 3x² −18x − 27

Step 4: Division by (x + 3)

            2x² − 3x − 9
`x + 3")"overline(2x^3 + 3x^2 - 18x - 27)`
           2x³ + 6x²
         −      −                            
                   − 3x² − 18x
                   − 3x2 − 9x
                   +        +                
                             − 9x − 27
                             − 9x − 27
                             +      +        
                                      x       

2x³ + 3x² −18x − 27 = (x + 3) (2x² − 3x − 9)

= (x + 3) (2x² − 6x + 3x − 9)

= (x + 3) [2x(x − 3) + 3(x − 3)]

= (x + 3) (x − 3) (2x + 3)