How do you factor 3x ^ { 4} + 12x ^ { 3} + 4x + 24?

Jan 25, 2018

A few thoughts...

Explanation:

The quartic as it stands has no rational factors.

If there is a typo in the question and one of the terms is in error, then it may be possible to factor by grouping.

For example:

$3 {x}^{4} + 12 {x}^{3} + 4 x + \textcolor{red}{16} = 3 {x}^{3} \left(x + 4\right) + 4 \left(x + 4\right) = \left(3 {x}^{3} + 4\right) \left(x + 4\right)$

$3 {x}^{4} + 12 {x}^{3} + \textcolor{red}{6} x + 24 = 3 {x}^{3} \left(x + 4\right) + 6 \left(x + 4\right) = 3 \left({x}^{3} + 2\right) \left(x + 4\right)$

$3 {x}^{4} + \textcolor{red}{18} {x}^{3} + 4 x + 24 = 3 {x}^{3} \left(x + 6\right) + 4 \left(x + 6\right) = \left(3 {x}^{3} + 4\right) \left(x + 6\right)$

$\textcolor{red}{2} {x}^{4} + 12 {x}^{3} + 4 x + 24 = 2 {x}^{3} \left(x + 6\right) + 4 \left(x + 6\right) = 2 \left({x}^{3} + 2\right) \left(x + 6\right)$

It is possible to factor each of the above possibilities further using irrational coefficients.

For example:

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

So:

${x}^{3} + 2 = {x}^{3} + {\left(\sqrt[3]{2}\right)}^{3} = \left(x + \sqrt[3]{2}\right) \left({x}^{2} - \sqrt[3]{2} x + \sqrt[3]{4}\right)$

Alternatively, if the question is correct as given, then it is possible to factorise, but the answer gets very messy and complicated. I may write up a separate answer to sketch the process.