If f(1) = 0, what are all the roots of the function f(x)=x^3+3x^2-x-3? Use the Remainder Theorem.

2 answers:

8 0

There's no if about it,

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

has a zero $f(1)=0$ so $x-1$ is a factor. That's the special case of the Remainder Theorem; since $f(1)=0$ we'll get a remainder of zero when we divide $f(x)$ by $x-1.$

At this point we can just divide or we can try more little numbers in the function. It doesn't take too long to discover $f(-1)=0$ too, so $x+1$ is a factor too by the remainder theorem. I can find the third zero as well; but let's say that's out of range for most folks.

So far we have

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

where $r$ is the zero we haven't guessed yet. Again we could divide $f(x)$ by $(x-1)(x+1)=x^2-1$ but just looking at the constant term we must have

$-3 = -1 (1)(-r) = r$

so

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

We check $f(-3)=(-3)^3+3(-3)^2 -(-3)-3 = 0 \quad\checkmark$

We usually talk about the zeros of a function and the roots of an equation; here we have a function $f(x)$ whose zeros are

$x=1, x=-1, x=-3$