Question #48cc8

Jul 7, 2015

How you find the factors depends a lot on what tools you have in your mathematical toolbox.

Explanation:

Note on terminology, although there is a "quartet" of terms, the word "quartic" refers to the degree being 4. Not to the number of terms. We could call it a quadrinomial, but I don't hear that or see it in textbooks.

To factor a cubic polynomial, First check for common factors and check for sum or difference of two cubes. (Not relevant in this problem.)

I would then try factoring by grouping . It won't always work, but it doesn't take long to try it:

$y = {x}^{3} - 9 {x}^{2} + 27 x - 27$

$\left({x}^{3} - 9 {x}^{2}\right) + \left(27 x - 27\right)$

${x}^{2} \left(x - 9\right) + 27 \left(x - 1\right)$ -- No, factoring by grouping won't work.

You may have noticed that both the first and last terms are perfect cubes. If so, you might try using the guess ${\left(x - 3\right)}^{3}$
Or, you might use the rational zero theorem to learn that $3$ is a zero, so $x - 3$ is a factor. Then do the division to get:

${x}^{3} - 9 {x}^{2} + 27 x - 27 = \left(x - 3\right) \left({x}^{2} - 6 x + 9\right)$

Which can be further factored to get:

${x}^{3} - 9 {x}^{2} + 27 x - 27 = {\left(x - 3\right)}^{3}$