Question

Question #c16ad

Answer 1

`x^3+3x^2-x-3=(x-1)^2(x-3)`
`x=1,` and `x=3`

Explanation:

Hope the question is to investigate for the roots of the polynomial
`x^3+3x^2-x-3=0`
By inspection,
When `x=1`
`x^3+3x^2-x-3=1^3+3(1)^2-1-3=1+3-3-1=0`
Hence, `1` is a root
`x=1` is a root
`x-1=0`
`x-1` is a factor
Dividing the polynomial
`x^3+3x^2-x-3` by `x-1`, we have

`(x^3+3x^2-x-3)/(x-1)`

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

Thus, the quotient is `x^2-4x+3`
Simplifying
`x^2-4x+3=x^2-3x-x+3`
`=x(x-3)-1(x-3)`
`x^2-4x+3=(x-3)(x-1)`
Now,
`x^3+3x^2-x-3=(x-1)(x-3)(x-1)`

Rearranging,

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