Factoring Cubic Equations

I’ve been trying to figure out how to factor cubic equations by studying a few worksheets online such as the one here and was wondering is there any generalized way of factoring these types of equations or do we just need to remember a bunch of different cases.

For example, how would you factor the following:

$x^3 – 7x^2 +7x + 15$

I'm have trouble factoring equations such as these without obvious factors.

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$x=-1$ is an obvious root ! – Yves Daoust May 22 '15 at 14:25

Your example has integer coefficients. We deal mainly with that case, or the essentially equivalent case of rational coefficients, since we can always multiply through by the lcm of the denominators and get integer coefficients,

The following result, called the Rational Roots Theorem, is sometimes useful.

Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0$ be a polynomial with integer coefficients $a_i$, and suppose $a_0\ne 0$. Then any rational root of $P(x)=0$ must have shape $\frac{c}{d}$, where $c$ divides $a_0$ and $d$ divides $a_n$,

In our case, $c$ must divide $15$ and $d$ must divide $1$. That gives the candidates $\pm 1$, $\pm 3$, $\pm 5$, and $\pm 15$.

Test them all. We get very lucky, $3$ works. To find the other roots, we can divide our polynomial by $x-3$, getting a quadratic, which we solve using the Quadratic Formula.

This procedure often works when we are dealing with a "made up" problem. In particular, if the cubic is in a school exercise, almost certainly it has a rational root. But the procedure certainly does not always work. Most cubics with integer coefficients do not have a rational root. And the procedure only applies to polynomials with rational coefficients.

If there are no rational roots, we are in trouble. True, there is a formula for the roots of the general cubic, usually called the Cardano Formula. It dates back to the sixteenth century. One can also find formulas that use trigonometric or hyperbolic functions. But overall the formulas are not very useful.

There are good methods for quickly approximating the roots to high accuracy,

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Thanks! This post along with the following videos: 1, 2, 3, made this concept crystal clear. I appreciate the reference to the rational roots theorem which I see will be useful in the future – Amateur Math Guy Feb 4 '13 at 8:23

In general, there is the cubic equation, but you certainly are not expected to know it, and I am sure you will not be asked a question that would require you to use it.

In fact, I think there is a (somewhat) obvious factor in your question :)

Hint: $$1+7+7=15$$

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The general method involves three important rules for polynomials with integer coefficients:

1. If $a+\sqrt b$ is a root, so is $a-\sqrt b$.
2. If $a+ib$ is a root, so is $a-ib$.
3. If a root is of the form $p/q$, then p is an integer factor of the constant term, and q is an integer factor of the leading coefficient.

Since this is cubic, there is at least one real rational solution (using rules 1 & 2). Let's find it using rule 3: $$q = 1,\ p \in \{\pm1,\pm3,\pm5,\pm15\}$$ Brute force to find that $-1$ is a solution. So we can now write the polynomial as: $$(x+1)(x-a)(x-b)=x^3 - 7x^2 + 7x +15$$ Now use: $$a+b=8$$ $$ab-a-b=7$$ Solve to get: $$a(8-a)-a-(8-a)=7 \to a^2-8a+15 = 0$$ And get the two remaining solutions.

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$x^3-7x^2+7x+15=(x+1)(x-3)(x-5)$

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