Factoring trinomials

helpneeded105:

What two numbers multiply to 600, and add to 17?

The way to approach these problems is to let the two numbers be called a and b (or any other two letters you like!). Then you write
a + b = 17
a*b = 600

Rearrange the first to get b = 17 - a by subtracting a from both sides. Substitute this into the second to get a*(17 - a) = 600.
This can be written as 17a - a² = 600, which can be rearranged as:
a² - 17a + 600 = 0.
This is a quadratic equation that can be solved to find two values for a.
However, neither value exists on the real number line; they are what are called complex numbers.
Unless you've dealt with complex numbers it's best not to take this any further here!
If you made a mistake with the 600, and meant 60, then the quadratic factorises nicely into two real positive integers.


Original problem:
30+17a-20a^2

Do you mean you want to set this equal to 0 and find the two values of a which satisfy this? If so, you will need to use the quadratic formula or the solver on the calculator here.

helpneeded105:

What two numbers multiply to 600, and add to 17?
Everything Alan said is correct - There are no (real) solutions. I just want to talk about it a little more.

Original problem:
30+17a-20a^2

I am going to look at your original problems some more.

30+17a-20a ² = -20a ² + 17a +30

If you want to factorise this you need to look for 2 numbers that add to 17 and multiply to -20*30= -600
These numbers are not easy to find because thay are not rational. I could talk about this a bit more but I suspect Alan is correct when he said maybe the original question was
30+17a-20a^2 =0
in which case we have
-20a ² + 17a +30 =0
You can solve this using the quadratic formula. This will help you remember it.
http://www.youtube.com/watch?v=O8ezDEk3qCg
For you substitute in, a=-20, b=17 and c=30.
The unknown in your question is an a - usually it is an x.
Please do not get confused, the a in the formula is the number in front of the squared unknown. I mean the a in the formula is -20. do you understand what I am trying to say?

There are 2 solutions. They will be