This equation is a great candidate for factoring by grouping. Why? Factoring by grouping is a method usually done on polynomials with four or more terms — usually with an even number. Also, factoring by grouping works well when there is no common factor for all the terms in the polynomial, but there

*are* common factors in pairs of the terms.

To factor by grouping, the first step is to rewrite the polynomial in groups:

x^{3} – 3x^{2} – x + 3 = 0 (x^{3} – 3x^{2}) – (x – 3) = 0

There is a common factor of x^{2} in the first pair, so factor it out:

x^{2}(x – 3) – (x – 3) = 0

You can see that each pair has a common factor of (x – 3). After you group, if you *don't* have a common factor in each pair, try rearranging the terms in another way. If you still don't end up with a common factor in each pair, it may be that the equation can't factored (or you've made a mistake — be sure to double check your work!)

Since there is a common factor, factor (x – 3) out of the two groups:

(x – 3)(x^{2} – 1) = 0

Now set each binomial equal to 0 and solve:

x – 3 = 0 x^{2} – 1 = 0 x = 3 (x – 1)(x + 1) = 0 x = 3 OR x = 1 OR x = –1

Check these three possible solutions by substituting the values for x back into the original equation. You should find that all three solutions are valid!