#### Question

The expression 4x^{3} – bx^{2} + x – c leaves remainders 0 and 30 when divided by x + 1 and 2x – 3 respectively. Calculate the values of b and c. Hence, factorise the expression completely.

#### Solution

Let f (x)=4x^3-bx^2+x-c

it is given that when f(x) is dividend by (x+1), the remainder is 0

∴ `f(-1)=0`

`4(-1)^3-b(1)^2+(-1)-c=0`

`-4-b-1-c=0`

`b+c+5=0 `........(1)

It is given that when f(x) is divided by (2x-3) the remainder is 30.

∴` f(3/2)=30 `

`4(3/2)^3 -b(3/2)^2+(3/2)-c =30 `

`27/2-(9b)/4+3/2-c=30 `

`54-9b+6-4c-120=0 `

`9b+4c+60=0 ` .............(2)

Multiplying (1) by 4 and subtracting it from (2), we get,

`5b+40=0 `

`b=-8 `

Substituting the value of b in (1), we get,

`c=-5+8=3 `

Therefore, f(x) =4x^3+8x^x-3

Now, for x-1, wr get,

`f(x)=f(-1)=4(-1)^3+8(-1)^2+(-1)-3=-4+8-1-3=0`

Hence, (x+1) is a factor of f (x)

∴ `4x^3+8x^2+x-3=(x+1)(4x^2+4x-3) `

=`(x+1)(4x^2+6x-2x-3) `

=`(x+1)[2x(2x+3)-(2x+3)]`

=` (x+1)(2x+3)(2x-1) `