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

May 19, 2015

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

Using Bhaskara, let's find the roots:

$\frac{- 4 \pm \sqrt{16 - 4 \left(3\right) \left(1\right)}}{6}$
$\frac{- 4 \pm 2}{6}$

${x}_{1} = - 1$, which is the same as the factor $x + 1 = 0$
${x}_{2} = - \frac{1}{3}$, which is the same as the factor $3 x + 1 = 0$

Now, we can rewrite your function as $\left(x + 1\right) \left(3 x + 1\right)$

May 19, 2015

There are a few ways to factor $3 {x}^{2} + 4 x + 1$

One is essentially trial and error:

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

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

$\left(3 x \pm \text{some number")(x +- "another number}\right)$

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 $1 \times 1$

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

$\left(3 x + 1\right) \left(x + 1\right)$ Now is it crucial that we check this.

$\left(3 x + 1\right) \left(x + 1\right) = 3 {x}^{2} + 3 x + 1 x + 1 = 3 {x}^{2} + 4 x + 1$ Good! That works, so we can write the answer:

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

Why do we need to check?

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