Question

If ax³ + 3x² + bx – 3 has a factor (2x + 3) and leaves remainder – 3 when divided by (x + 2), find the values of a and b. With these values of a and b, factorise the given expression.

Answer 1

Let,

p(x) = ax³ + 3x² + bx − 3

g(x) = 2x + 3 = 0 ⇒ x = `−3/2​`

f(x) = x + 2 = 0 ⇒ x = − 2

 

Given, g(x) is a factor of f(x)

∴ By factor theorem,

 

`p(−3/2​) = 0`

 

⇒ `a(−3/2​)^3 + 3(−3/2)^2+b(−3/2​)−3=0`

 

⇒ `a(−27/8​)+3(9/4​)+b(−3/2)−3=0`

 

⇒ `−(27a)/8 ​+ 27/4 ​− (3b)/2 − 3 = 0`

 

⇒ `−(27a + 54 - 12b)/8 = 3`

 

⇒ −27a + 54 − 12b = 24

 

⇒ − 3(9a + 4b) = 24 − 54 = −30

 

⇒ 9a + 4b = 10             ...(i)

 

Also, p(x) when divided by f(x) leaves a remainder −3

∴ By remainder theorem,

p(−2) = −3

⇒ a(−2)³ + 3(−2)² + b(−2) − 3 = −3

⇒ −8a + 12 − 2b = 0

⇒ 8a + 2b = 12

⇒ 4a + b = 6            ...(ii)

 

Solving (i) and  (ii), we get

a = 2 and  b = −2

 

Hence, p(x) = 2x³ + 3x² − 2x − 3

= x²(2x + 3) − (2x + 3)

= (2x + 3)(x² − 1)

= (2x + 3)(x + 1)(x − 1)