<s>[INST] How do I factorise this polynomial Please help factorise this: 
$$6x^2+x+4=0$$
In my attempts, I assumed $a=6$, $b=1$ and $c=4$.
I multiplied $c$ by $a$ and attempted to get the factors that give us the sum of $b$.
The only factor pairs of $24$ are $1\cdot 24$, $2\cdot 12$, $3\cdot 8$ and $4\cdot 6$.
None of the above can give me a sum of $1$.
 [/INST] Let us complete the square:
\begin{align*}
6 x^2 + x + 4 &= 6 \left(x^2 + \frac{1}{6} x + \frac{3}{2}\right) \\
&= 6 \left(x^2 + \frac{1}{6} x + \frac{1}{144} + \frac{3}{2} - \frac{1}{144}\right) \\
&= 6 \left(x + \frac{1}{12}\right)^2 + 6 \left(\frac{3}{2} - \frac{1}{144}\right) \\
&= 6 \left(x + \frac{1}{12}\right)^2 + \left(9 - \frac{1}{24}\right)
\end{align*}
Setting this to zero gives
$$6 \left(x + \frac{1}{12}\right)^2 = - \left(9 - \frac{1}{24}\right) < 0$$
which has no real solutions.
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