Question

If (1)=0, what are all the roots of the function f(x)=x3+3x2-x-3?use the remainder theorem

Answer 1

If $f(1)=0$, then by the remainder theorem $\dfrac{f(x)}{x-1}$ has a remainder of 0, which means $f(x)$ is divisible by $x-1$. In fact,

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

So we know 1 is a root, and the other two roots are -1 and -3.