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A isn't 1?
I ran into this problem-
5x^2 + 49x + 72
And I just don't remember how the heck to factor problems like that, and when I use what I remember I get (5x+40)(x+9) But the factor calculator I'm using says the 40 is actually an 8.
Where the heck am I malfunctioning? What do I do? Can anyone tell me what to when the A isn't 1 in a ax^2+bx+c expression? Am I supposed to divide the 40 by 5 to get 8?
Yes. And that's pretty much it.Quote:
Originally Posted by MerkaMerka
If you used the quadratic formula, you are pretty much correct.
http://www.daleheffernan.com/QuadraticFormula.jpg
The 8 one is actually correct because if you expand (5x+40)(x+9) you get 5x^2+45x+9x+360
hmm but I'm not even sure if I'm starting off right or not.
5x^2 + 49x + 72
And then from there I multiply the 72 by 5 and get 360, and I find that 40 x 9 = 360 so I get (5x + 40)(x + 9).... And that SEEMS right but then I have to divide the 40 by a 5? I don't get at what point I have to divide the 40 by the 5...
Can someone just explain me how to get around to the end factoring? because I know I'm screwing up and I just can't figure out how to even start the factoring..
Also Orpheus excuse me because I'm very math illiterate but how do you get around to using the quadratic equation anywhere in factoring the problem? I thought that was to solve it, which I'm only trying to factor...?
It's pretty much the same thing. When you factor quadratic equations you are pretty much getting the roots. The quadratic formula gives you the roots, you just have to put them in the form of factors.
(x + something)(x + somethingelse)
Where something and somethingelse are the 2 roots.
When A isn't 1 then you need to consider the factors of A. Since 5 is a prime number the only factors are 1 and 5 so you'll have something that looks like (1x+__)(5x+__). The blanks will be factors of C. Just FOIL it out to figure out of you have the correct factors.
I hate Algebra... (There's my helpful insight!)