Factor Theorem states that if you have a polynomial, \(f(x)\), and a value, \(c\), then \(c\) is a root of the polynomial \(f(x)\) if and only if \((x - c)\) is a factor of \(f(x)\).
In other words, if you substitute \(x = c\) into the polynomial \(f(x)\) and the result is zero, then \(x - c\) is a factor of \( f(x) \). Conversely, if \(x - c\) is a factor of \(f(x)\), then \(c\) is a root of \( f(x)\).
Mathematically, this can be expressed as:
\[ f(c) = 0 \quad \Leftrightarrow \quad (x - c) \text{ is a factor of } f(x) \]
Determine if \(x -2\) is a root of \( f(x) = x^3 - 4x^2 + x + 6 \).
According to Factor Theorem, we can substitute \(2\) for \(x\) in the function. If \(f(2) = 0\), this indicates that \(x- 2\) is a factor:
\(f(2) = (2)^3 - 4(2)^2 + 2 + 6 \)
\(f(2) = 8 - 4(4) + 8\)
\(f(2) = 16 - 16\)
\(f(2) = 0\)
Since \(f(2) = 0\), we can determine that \(x - 2\) is a factor of \(f(x)\); consequently \(x = 2\) is a root of \(f(x)\).
According to Factor Theorem, if \(x = 1\) is a root of \(f(x)\), this indicates that \(x-1\) must be a factor. We can substitute \(1\) for \(x\) and evaluate the expression:
\(f(1) = (1)^3 - 5(1)^2 + 6(1) - 4\)
\(f(1) = 1 - 5(1) + 6 - 4\)
\(f(1) = 3 - 5\)
\(f(1) = -2\)
Since \(f(1) = -2\) and not \(0\) this means that \(x-1\) isn't a factor of \(f(x)\) and \(x = 1\) isn't a root.
Since \(x = 1\) isn't a root, we need to look for other possible roots.
We can use Synthetic Division to divide \(f(x)\) by a linear factor. In this instance, we will use Synthetic Division with \(x=2\) to determine if its a root:
\(f(2) = (2)^3 - 5(2)^2 + 6(2) - 4 \)
\(f(2) = 8 - 5(4) + 12 - 4 \)
\(f(2) = 16 - 20\)
\(f(2) = -4\)
Since \(f(2) = -4\) and not \(0\), this means that \(x - 2\) is not a factor of \(f(x)\). Therefore, \( x = 2\) is not a root of \(f(x)\).
Select a value for \(x\). This function will determine whether or not this value is a root of \(f(x) = x^5 - 3x^4 + 2x - 1\)