Question

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

Answer 1

We can Split the Middle Term of this expression to factorise it
In this technique, if we have to factorise an expression like `ax^2 + bx + c`, we need to think of 2 numbers such that:
`N_1*N_2 = a*c = 6*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*2 = 12`, and `6+2= 8`

`6x^2 + 8x + 2 = 6x^2 + 6x + 2x + 2`

`= 6x(x+1) + 2(x+1)`

`(x+1)` is a common factor to each of the terms

`= (x+1)(6x+2)`

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

Answer 2

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

`y = 2(3x^2 + 4x + 1).`

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(x) = 2(x + 1)(x + 1/3) = 2(x + 1)(3x + 1)`

Check by developing: `f(x) = 2(3x^2 + x + 3x + 1)` .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.