Find all possible roots/zeros of the function \(f(x) = x^3 - 4x^2 + x + 6\)
Answer
We substitute each possible rational root into the function until we find one that makes the function equal to zero. By doing this, we find that \(-1\), \(2\), and \(3\) are the roots of the function.
Step 1 ：First, let's use the Rational Root Theorem (RRT). The RRT states that any rational root, say \( \frac{p}{q}\), of the polynomial \(f(x) = a_n x^n + a_{n-1}x^{n-1} + ... + a_1x + a_0\) must have \(p\) as a factor of the constant term \(a_0\) and \(q\) as a factor of the leading coefficient \(a_n\). In this case, the constant term is 6 and the leading coefficient is 1.
Step 2 ：The factors of 6 are \(\pm1\), \(\pm2\), \(\pm3\), and \(\pm6\). Since the leading coefficient is 1, the possible rational roots of the polynomial are \(\pm1\), \(\pm2\), \(\pm3\), and \(\pm6\).
Step 3 ：We substitute each possible rational root into the function until we find one that makes the function equal to zero. By doing this, we find that \(-1\), \(2\), and \(3\) are the roots of the function.
