# Factoring Quadratic equations

• Jan 25th 2011, 05:02 PM
SpencerMakesMath
I remember doing something like this in Algebra 2, but I am in PreCal so sorry if it is the wrong section..

The equation is:
2x^2+9x+4=0

When I plug this equation into the quadratic formula I get the same answers my teacher got: x=-1/2 & x=-4

This is how she factored it:
2x^2+9+4=0
(2x+1)(x+4)=0
2x+1=0 | x+4=0
2x=-1 | x=-4
x=-1/2

The problem I have w/ understanding this equation is the numbers inside the parentheses. We were taught to find factors of 4 that add up to 9 and 4+1 clearly doesn't equal 9. Can someone please explain to me how she got those numbers?

/first post. This looks like a cool site and I will be coming here for math help from now on. Hopefully I will be able to answer just as many questions as I propose :p
• Jan 25th 2011, 05:30 PM
Aryth
The two in the $\displaystyle x^2$ term is what throws everything off. Look at the equation with the two divided through it:

$\displaystyle x^2 + \frac{9}{2} + 2 = 0$

Now we need two numbers that add up to $\displaystyle \frac{9}{2}$ and multiply to be 2. Since the second term has a half in it, that's probably going to be part of it, we can try $\displaystyle \frac{1}{2}$

$\displaystyle \left(x + \frac{1}{2}\right)(x + a) = 0$

Now, what number can we pick? We can solve:

$\displaystyle \frac{1}{2} + a = \frac{9}{2}$

If you work it out you get that:

$\displaystyle a = 4$

So:

$\displaystyle \left(x + \frac{1}{2}\right)(x + 4) = 0$

Is your answer, which is exactly what you and your teacher got, and even though it took a little bit of extra work, just understand that if there is a number infront of the $\displaystyle x^2$ term then at least ONE x will have to be multiplied by something. Dividing through helps sometimes, but doing it the way your teacher did will be important for larger numbers.
• Jan 25th 2011, 06:33 PM
LoblawsLawBlog
Another method, which is practically the same but doesn't involve fractions, is to find two numbers that multiply to 8 (the 8 comes from multiplying the 4 and 2) and add to 9. 8 and 1 will do. Now break up the terms as $\displaystyle 2x^2+9x+4=(2x^2+8x)+(x+4)=2x(x+4)+(x+4)=(2x+1)(x+4 )$.
• Jan 26th 2011, 04:40 AM
HallsofIvy
Quote:

Originally Posted by SpencerMakesMath
I remember doing something like this in Algebra 2, but I am in PreCal so sorry if it is the wrong section..

The equation is:
2x^2+9x+4=0

When I plug this equation into the quadratic formula I get the same answers my teacher got: x=-1/2 & x=-4

This is how she factored it:
2x^2+9+4=0
(2x+1)(x+4)=0
2x+1=0 | x+4=0
2x=-1 | x=-4
x=-1/2

The problem I have w/ understanding this equation is the numbers inside the parentheses. We were taught to find factors of 4 that add up to 9 and 4+1 clearly doesn't equal 9. Can someone please explain to me how she got those numbers?

"Finding factors of c that add up to b" only works for quadratics of the form $\displaystyle x^2+ bx+ c$- that is, where the coefficient of $\displaystyle x^2$ is 1. If the coefficient of x is not 1 then you have to take factors of that coefficient into account also. Here the coefficient of $\displaystyle x^2$ is 2, which has factors of 2 and 1 and 2(4)+ 1(1) is equal to 9.

Quote:

first post. This looks like a cool site and I will be coming here for math help from now on. Hopefully I will be able to answer just as many questions as I propose :p
• Jan 27th 2011, 12:59 PM
sinx
There's a trick I picked up back in precalc that might help you. There's a billion ways to factor (bottom's up, etc) but I find this one the most intuitive, and I have yet to find a factoring problem that it doesn't work on (that has factors...of course).

Given:

$\displaystyle 2x^2+9x+4=0$

Step 1:

Multiple the 'C' term and the 'A' term together. In this example we'll get $\displaystyle 8x^2$.

Step 2:

Now we need to find factors of $\displaystyle 8x^2$ that add up to get $\displaystyle 9x$. Clearly, the only factors that will work here are $\displaystyle 8$ and $\displaystyle 1$.

Step 3:

Next, we need to rewrite the problem like so:

$\displaystyle (2x^2+8x)(x+4)$

How did we get these numbers? All I did was bring down the factors of $\displaystyle 8x^2$ and group them with the squared term $\displaystyle 2x^2$ and the constant $\displaystyle 4$.

Step 4:

Now we simplify. The idea is to get the simplest group of terms on both sides.

$\displaystyle 2x(x+4) 1(x+4)$

I have a bad habit of forgetting the understood one (even in university level math!) so I always make sure to write it.

Step 5:

Simply combine the outside terms into a group, and use the $\displaystyle (x+4)$ group as itself. With this we get:

$\displaystyle (2x+1)(x+4)$

Step 6:

Set it equal to zero and solve algebraically:

$\displaystyle (2x+1)(x+4)=0$
$\displaystyle 2x+1=0$
$\displaystyle x+4=0$
$\displaystyle x = -4, 1/2$

I hope this helps! I don't remember it's name, but if you can somehow find the name there's a few videos on youtube about it so you can see it worked on paper (which makes it even easier).