# What are the factors for a^6 + 7a^3 + 6?

May 15, 2015

$\left({a}^{3} + 6\right) \left({a}^{3} + 1\right)$ is the obvious solution
${a}^{3} + 1 = \left(a + 1\right) \left({a}^{2} - a + 1\right)$ No further real factors all others involve imaginary numbers

May 15, 2015

A simple manipulation of your variables will give us the solution.

Let's kind of factor your function as ${a}^{3} \cdot {a}^{3} + 7 {a}^{3} + 6$, so now you have a common exponential element.

Let's name $b = {a}^{3}$.

Now, we can rewrite it as ${b}^{2} + 7 b + 6$.

Solving it, we will find the roots: ${b}_{1} = 1$ and ${b}_{2} = 6$.

These are our new factors. Now, we can factor the newfound equation using them:

$\left(b + 1\right) \left(b + 6\right) = 0$.

Note that when you expand these factors you will go back to the original ${b}^{2} + 7 b + 6$.

Just replacing $b = {a}^{3}$, we get:

$\left({a}^{3} + 1\right) \left({a}^{3} + 6\right) = 0$.

Note, again, that if you expand it you will get exactly ${a}^{6} + 7 {a}^{3} + 6$.

May 18, 2015

For the sake of a more full answer:

"Factoring completely" without context is ambiguous.

What we need to remove the ambiguity is a statement of what kinds of coefficients we are using.

The integers are the "positive and negative whole numbers and $0$.
The rational numbers are the ratios (fractions) of integers.
The real numbers include all numbers that are not imaginary (that do not involve $\sqrt{- 1}$.
The complex numbers are all two-part numbers $a + b i$ where $i = \sqrt{- 1}$

${a}^{6} + 7 {a}^{3} + 6$

To factor over the integers (or over the rational numbers) means that we are going to use integer (or rationals) This is what beginning and intermediate algebra classes usually mean by "factor completely"..
As others have shown:

${a}^{6} + 7 {a}^{3} + 6 = \left({a}^{3} + 1\right) \left({a}^{3} + 6\right)$

Now $1$ is a perfect cube (the cube of an integer), so we can factor using the sum of cubes:

${a}^{6} + 7 {a}^{3} + 6 = \left(a + 1\right) \left({a}^{2} - a + 1\right) \left({a}^{3} + 6\right)$

These polynomials cannot be factored using integer (or rational) coefficients.
We are finished if we are factoring over the integers (or the rationals)

If we are factoring over the real numbers, (This is what some more advanced algebra and precalculus classes mean by "factor completely".)
We observe that $6 = {\left(\sqrt[3]{6}\right)}^{3}$. Note that $\sqrt[3]{6}$ is an irrational number. Therefore, ${a}^{3} + 6$ can be factored using the sum of cubes and irrational numbers.

${a}^{2} + 6 = {a}^{3} + {\left(\sqrt[3]{6}\right)}^{3}$

$\textcolor{w h i t e}{\text{ssssssssss}}$ $= \left(a + \sqrt[3]{6}\right) \left({a}^{2} - \sqrt[3]{6} x + {\left(\sqrt[3]{6}\right)}^{2}\right)$

$\textcolor{w h i t e}{\text{ssssssssss}}$ $= \left(a + \sqrt[3]{6}\right) \left({a}^{2} - \sqrt[3]{6} x + \sqrt[3]{36}\right)$

${a}^{6} + 7 {a}^{3} + 6 = \left(a + 1\right) \left({a}^{2} - a + 1\right) \left(a + \sqrt[3]{6}\right) \left({a}^{2} - \sqrt[3]{6} x + \sqrt[3]{36}\right)$

The two quadratics have no factors over the real numbers.

Finally, we may be factoring over the complex numbers, as some parts of advanced algebra or precalculus classes sometimes do.we continue. (Often this is asked by including the phrase "over the complex numers" of "find all linear factors of"

We use the quadratic formula (or other methods) to find the zeros of the two quadratics. This gives us the linear factors over the complex numbers.

${a}^{2} - a + 1 = \left(a - \left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right)\right) \left(a - \left(\frac{1}{2} - \frac{\sqrt{3}}{2} i\right)\right)$

${a}^{2} - \sqrt[3]{6} a + \sqrt[3]{36} = \left(a - \sqrt[3]{6} \left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right)\right) \left(a - \sqrt[3]{6} \left(\frac{1}{2} - \frac{\sqrt{3}}{2} i\right)\right)$

If you need to do this, let me know and we'll go through it.