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

2. 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.

3. 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 $\displaystyle 2x^2+9x+4=(2x^2+8x)+(x+4)=2x(x+4)+(x+4)=(2x+1)(x+4 )$.

4. 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.

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

5. 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).