Question

How do you factor `3x^2 + 4x + 1`?

Answer 1

You find its roots and then turn them into factors, as follows.

Using Bhaskara, let's find the roots:

`(-4+-sqrt(16-4(3)(1)))/6`
`(-4+-2)/6`

`x_1=-1`, which is the same as the factor `x+1=0`
`x_2=-1/3`, which is the same as the factor `3x+1=0`

Now, we can rewrite your function as `(x+1)(3x+1)`

Answer 2

There are a few ways to factor `3x^2+4x+1`

One is essentially trial and error:

We know how to multiply binomials. now we want to find two binomials whose product is `3x^2+4x+1`

The product of the `x^2` terms (the F in FOIL) needs to be `3x^2`. So, if we can factor using whole numbers, we must have:

`(3x +- "some number")(x +- "another number")`

The product (multiply) of the constants (the "other numbers" -- and the L in FOIL) must be `1`. The only way to get `1` by multiplying integers is `1xx1`

So if it can be factored using integers, the factoring must be:

`(3x+1)(x+1)` Now is it crucial that we check this.

`(3x+1)(x+1) = 3x^2 +3x+1x+1 = 3x^2+4x+1` Good! That works, so we can write the answer:

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

Why do we need to check?

We would have followed exactly the same reasoning to try to factor `3x^2+5x+1`. But this cannot be factored using integers. We would only discover this when we multiplied `(3x+1)(x+1)` and did not get what we hoped for. (We would still get `4x` in the middle instead of `5x`.)