Question

How do you factor the trinomial `18x^2 - 21x + 6`?

Answer 1

`18x^2 -21x + 6 = 3(2x - 1)(3x - 2)`

Explanation:

The quadratic formula is an all purpose method that will allow the factorization of any quadratic expression, however rather than resort to that right off, let's see if we can factor the expression by simply working backwards.

Before anything, let's take out the common factor of `3` from the coefficients.

`18x^2 - 21x + 6 = 3(6x^2 - 7x + 2)`

This method only works well for factorizations using integers, so we are looking `a, b, c, d in ZZ` (the integers) such that

`3(ax + b)(cx + d) = 3(6x^2 - 7x + 2)`

`=> 3((ac)x^2 + (ad + bc)x + bd) = 3(6x^2 - 7x + 2)`

`=>{(ac = 6), (ad + bc = -7), (bd = 2):}`

By the first equation, we know that `a` and `c` form a pair from one of
`+-1` and `+-6`
`+-2` and `+-3`

By the third equation, we know that `b` and `d` form a pair from one of
`+-1` and `+-2`

A couple tries reveals that the system is solved by
`a = 2`
`b = -1`
`c = 3`
`d = -2`

Thus, substituting back in, we get

`18x^2 -21x + 6 = 3(2x - 1)(3x - 2)`