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 $1$.
 A: 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.
A: We can use the quadratic formula 
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
As $$ x  = \frac{(-1)\pm \sqrt{-95}}{12},$$
you will not get real roots.
A: There is a simple way to determine whether a given quadratic polynomial will have real roots, and that is by using the discriminant.
For some polynomial $\mathcal{P} = ax^2 + bx + c$ then the discriminant $\Delta$ of $\mathcal{P}$ is defined as
$$\Delta = b^2 - 4ac$$
The discriminant provides the following information:


*

*For $\Delta$ > 0, there are two real roots

*For $\Delta$ = 0, there is exactly one real root

*For $\Delta$ < 0, there are no real roots


In this case we have $\mathcal{P} = 6x^2+x+4=0$, so
$$\Delta = (1)^2 - 4(6)(4) = -95 < 0$$
Therefore there are no real roots. Furthermore, this makes $\mathcal{P}$ an irreducible polynomial of order 2. So, the factored form is simply itself.
A: Use the box method. ac = 24; so I need to find factors of 24 that add up to 1. Thus, I have no factors, and it is irreducible.
