# Solving a Quadratic by factoring where the leading coefficient !-0

I am studying for the GRE and reviewing math I haven't seen in years. I've got the answer and have solved similar problems, but am having trouble understanding a specific step in the factoring. I realize asking you folks is a lot like using an AT4 to shoot an ant, but I would appreciate the assistance if you could help.

The problem is Solve 3x^2+11x−4=0

The solution to the factoring process shows the following 3 steps: Step 1: =3x^2+12x−x−4 Step 2: =3x(x+4)−1(x+4) Step 3: =(3x−1)(x+4)

I can figure out everything through step 2 and after step 3, but I can't figure out how they factored step 2 into step 3.

Thanks for the help. ​

## 3 Answers

There's a common factor so you can rewrite:

$$3x\color{red}{(x+4)}-1\color{red}{(x+4)}$$

The common factor between the two terms is $x+4$. That means you can pull a $x+4$ from both:

$$\color{red}{(x+4)}(3x-1)$$

And if you want to get rid of the $x$s and think about this with simple constants, consider:

$$5\color{red}{(4)}-3\color{red}{(4)} = \color{red}4(5-3) = 8$$

• Even after reading your answer with the common factor comment, I couldn't see it until I finally replaced (x+4) with p. Then it became embarrassingly obvious! It may take a while to clear the cobwebs from my brain... :) – Aaron Apr 9 '18 at 3:14

Let's take it realllly slow:

$3x^2+11x−4=$:: ($3x^2 = (3x)*x$ so let's "pull" a $3x$ of it)

$3x*x + 11x - 4=$:: (let's pull $3x$ from the $11x$)

$3x*x + (3*4 - 1)x - 4=$

$3x*x + 12x - x -4=$

$3x*x + 3x*4 - x- 4=$. (let's pull the $3x$ from both $3x^2$ and from $12x$. In other words let's factor.)

$3x(x + 4) - x-4=$. (We have pulled out an $x+4$ nugget. Let's see if we can pull a $x+4$ nugget, from what is left; the $-x - 4$.)

($-x - 4 = -1(x+ 4)$ so

$3x(x+4) - x-4=$

$3x(x+4) - 1(x+4) =$. (Now pull the $(x+4)$ nugget out.

$(x + 4)[3x - 1]=$. (And now we.... oh, wait. We are done.)

Note: it was trial and error and it worked. Not all will work so nicely.

• You were exactly right but made the same mistake the author made when printing the solution to the problem. It took me a really long time to figure out why the "nugget" of (x+4) could be taken out. I finally replaced it with p, making it 3xp -1p which could finally allowed me to see that I could then make it p(3x-1). Thanks for the help! – Aaron Apr 9 '18 at 3:26

The problem is Solve 3x^2+11x−4=0

This is one of those cases where I think brute force may work better/faster than clever grouping and factoring. Just using the quadratic formula:

$$x_{1,2} = \frac{-11 \pm \sqrt{11^2+4 \cdot 3 \cdot 4}}{2 \cdot 3} = \frac{-11 \pm 13}{6} = \begin{cases} -4 \\[5px] 1/3 \end{cases}$$

The factorization then follows as $\;3 \cdot \big(x-(-4)\big)\big(x-1/3\big)=(x+4)(3x-1)\,$.