We are already familiar with the content of the **Factor Theorem**, without explicitly having encountered its name till now. This theorem says that given a polynomial \(a\left( x \right)\), if its value at *x* equal to some real number *k* is equal to 0, that is, if \(a\left( k \right) = 0,\)then the linear polynomial \(\left( {x - k} \right)\) is a factor of \(a\left( x \right)\), and we can write

\[a\left( x \right) = \left( {x - k} \right)b\left( x \right),\]

where \(b\left( x \right)\) is a polynomial of degree one less than \(a\left( x \right)\). Let’s take an example. Suppose that

\[a\left( x \right) = {x^3} - 6{x^2} + 11x - 6\]

We make the following observation: \(a\left( 1 \right) = 0\). Thus,

\(a\left( x \right) = \left( {x - 1} \right)b\left( x \right),\)

where \(b\left( x \right)\) is a quadratic polynomial. We make further observations (verify them):

\(a\left( 2 \right) = 0,\,\,a\left( 3 \right) = 0\)

Thus, \(\left( {x - 2} \right),\,\,\left( {x - 3} \right)\) will also be factors of \(a\left( x \right)\). Since a cubic polynomial can have *at the most three linear polynomial factors*, we can immediately conclude that

\[a\left( x \right) = \lambda \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\]

where \(\lambda \) is some real constant. But a simple comparison of coefficients of \(a\left( x \right)\)and the right side above tells us that \(\lambda \) must be equal to 1, and so

\[a\left( x \right) = \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\]

We have thus succeeded in factorizing\(a\left( x \right)\) by inferring the factors using the Factor Theorem.