# The polynomial ax^3–3x^2−11x + 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....

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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When a polynomial P(x) is divided by a factor (x - a), the remainder theorem gives the remainder as P(a).

It is given that P(x) = ax^3-3x^2-11x + b is divisible by x+2

As a result P(-2) = 0

a*(-2)^3 - 3*(-2)^2 - 11*(-2) + b = 0

-8a - 12 + 22 + b = 0

-8a + b + 10 = 0

Also, when P(x) is divided by x+1 the remainder is 12.

P(-1) = 12

= a*(-1)^3 - 3*(-1)^2 - 11*(-1) + b = 12

= -a - 3 + 11 + b = 12

= -a + 8 + b = 12

= -a + b = 4

From -8a + b + 10 = 0, b = 8a - 10

Substitute b = 8a - 10 in -a + b = 4

-a + 8a - 10 = 4

7a = 14

a = 2

b = 8a - 10 = 16 - 10 = 6

The required values of a and b are 2 and 6 respectively.

`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**