If `a x^3+b x^2+x-6` has `x+2` as a factor and leaves a remainder `4` when divided by `(x-2),` find the value of `a` and `b`.

Explanation

Let `p(x)=ax^3+bx^2+x−6`
Since `(x+2)` is a factor of `p(x)`, then by Factor theorem `p(−2)=0`
`⇒a(−2)^3+b(−2)^2+(−2)−6=0`
`⇒−8a+4b−8=0`
`⇒−2a+b=2` ...(i)
Also when `p(x)` is divided by `(x-2)` the remainder is `4`, therefore by Remainder theorem `p(2)=4`
`⇒a(2)^3+b(2)^2+2−6=4`
`⇒8a+4b+2−6=4`
`⇒8a+4b=8`
`⇒2a+b=2` ...(ii)
Adding equation (i) and equation (ii), we get:
`(−2a+b)+(2a+b)=2+2`
`⇒2b=4`
`⇒b=2`Putting `b=2` in equation (i), we get:
`−2a+2=2`
`⇒−2a=0⇒a=0`
`therefore a=0` and `b=2`