Question

If f(1) = 0 what are all the roots of the function f(x)=x^3+3x^2-x-3 use the remainder theorem.

Answer 1

Solution:

As we are given that f(1) = 0 .

It mean that $(x-1)$ is one of the factor of the given equation.

Remainder theorem can be applied as below:

$\frac{(x^3+3x^2-x-3)}{(x-1)}=\frac{x^3-x^2+4x^2-4x+3x-3}{(x-1)}\\ \\\frac{x^3-x^2+4x^2-4x+3x-3}{(x-1)}=\frac{x^2(x-1)+4x(x-1)+3(x-1)}{(x-1)} \\\\\frac{x^2(x-1)+4x(x-1)+3(x-1)}{(x-1)}=\frac{(x^2+4x+3)(x-1)}{(x-1)} \\\\\frac{(x^2+4x+3)(x-1)}{(x-1)} =\frac{(x^2+3x+x+3)(x-1)}{(x-1)} \\\\\frac{(x^2+3x+x+3)(x-1)}{(x-1)} =\frac{(x-1)(x+3)(x+1)}{(x-1)}$

Hence the factors are (x-1),(x+3) and (x+1).

Hence the correct option is B.

Answer 2

The answer is B quizzlet