The polynomial *ax^3*–3*x^2*−11*x *+ *b*, where *a *and* b *are constants, is denoted by p(*x*). It is given that (*x *+ 2) is a factor of p(*x*), and that when p(*x*) is divided by (*x *+ 1) the remainder is 12.

Find the values of *a *and *b*.

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`P(x)=ax^3-3x^2-11x+b`

`(x+2) is ` factor of P(x)

`=>P(-2)=0`

`-8a-12+22+b=0`

-8a+10+b=0 (i)

Also when (x+1) divides P(x) ,remainder is 12

This implies

P(-1)=12

-a-3+11+b=12

-a-4+b=0 (ii)

subtract (ii) from (i)

-7a+14=0

-7a=-14

a=2

substitute a in (ii)

-2-4+b=0

b=6

**Thus**

**a=2 and b=6**

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