Question

Find the value of p for which the polynomial 3x^3 -x^2 + px +1 is exactly divisible by x-1, hence factorise the polynomial

Answer 1

`p = - 3`

The completely factored form is

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

Step-by-step explanation

The given polynomial expression is

`3 {x}^(3) - {x}^(2) + px + 1`

Let

`f(x) = 3 {x}^(3) - {x}^(2) + px + 1`

According to the Remainder Theorem, if f(x) is exactly divisible by x-1, then the remainder is zero.

This implies that:

`f(1) = 0`

`3 {(1)}^(3) - {(1)}^(2) + p(1)+ 1 = 0`

`3 - 1 + p + 1 = 0`

`3 + p = 0`

`p = - 3`

When we substitute the value of p back into the function, f(x) we get:

`f(x) = 3 {x}^(3) - {x}^(2) - 3x + 1`

We now perform long division as shown in the attachment.

We can factor the function to get:

`f(x) =(x - 1)( 3 {x}^(2) + 2x - 1)`

We now split the middle term of the quadratic term and factor it completely to obtain:

`f(x) =(x - 1)( 3 {x}^(2) + 3x - x - 1)`

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

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

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