Factor Theorem
2026-02-28 10:08 Diff
https://brightchamps.com/en-us/math/algebra/factor-theorem

To factorize a cubic polynomial:
Find zero using the trail- and-error method. Then, using synthetic division method, divide the given polynomial f(x) by the given binomial \((x-a)\), 

After division, if the remainder is not zero, then \((x-a)\) is not a factor of \(f(x)\).

If the remainder is zero, use the division algorithm and write the given polynomial as a product of \((x-a)\) and quadratic quotient \(q(x)\); \(f(x) = (x-a)q(y) + r\)

If possible, factor the quadratic quotient further

Then, represent the polynomial in factored form.

Let’s factor \(f(x) = x^3 - 6x^2 + 11x - 6\) using the aforementioned procedure 

The first step is to find a zero using the hit and try method and dividing the given polynomial
Try \(x = 1\),
            \(f(1) = 1^3 - 6(1)^2 + 11(1) - 6\\ = 1 - 6 + 11 - 6\\ = 0 \)

So, x = 1 is a zero, a (x-1) is a factor.

Now, we will use synthetic division to divide f(x) by (x-1)
            \(\begin{array}{r|rrrr} & 1 & -6 & 11 & -6 \\ 1\, & & 1 & -5 & 6 \\ \hline & 1 & -5 & 6 & 0 \\ \end{array} \)

  The quotient is: \(x^2 - 5x = 6 \)

Since the remainder is zero, 

                 \(x^2 - 5x + 6 = (x-2) (x-3)\)


The polynomial, in its final factored form, is 


                    \(f(x) = (x-1)(x-2)(x-3)\)

Thus, the zeroes of the given polynomial \(x^3 - 6x^2 + 11x - 6 \)are 1, 2, and 3