# How do you factor 6x^2 + 8x + 2?

May 19, 2015

We can Split the Middle Term of this expression to factorise it
In this technique, if we have to factorise an expression like $a {x}^{2} + b x + c$, we need to think of 2 numbers such that:
${N}_{1} \cdot {N}_{2} = a \cdot c = 6 \cdot 2 = 12$
AND
${N}_{1} + {N}_{2} = b = 8$
After trying out a few numbers we get ${N}_{1} = 6$ and ${N}_{2} = 2$
$6 \cdot 2 = 12$, and $6 + 2 = 8$

$6 {x}^{2} + 8 x + 2 = 6 {x}^{2} + 6 x + 2 x + 2$

$= 6 x \left(x + 1\right) + 2 \left(x + 1\right)$

$\left(x + 1\right)$ is a common factor to each of the terms

$= \left(x + 1\right) \left(6 x + 2\right)$

=color(green)(2(x+1)(3x+1)

May 19, 2015

There is another shortcut that avoids the lengthy factoring by grouping.

$y = 2 \left(3 {x}^{2} + 4 x + 1\right) .$

Since a + b + c = 0, the trinomial in parentheses has one factor (x + 1) and another (x + c/a) = (x + 1/3)

Factored form: $f \left(x\right) = 2 \left(x + 1\right) \left(x + \frac{1}{3}\right) = 2 \left(x + 1\right) \left(3 x + 1\right)$

Check by developing: $f \left(x\right) = 2 \left(3 {x}^{2} + x + 3 x + 1\right)$ .OK

Reminder of the shortcut for f(x) = ax^2 + bx + c = 0.
1. When a + b + c = 0, one real roots is (1) and the other is (c/a)
2. When a - b + c = 0, one real root is (-1) and the other is (-c/a)

Remember this TIP. It will save you a lot of time and effort.