/sci/ - Science & Math

because when you foil that shit, you get x^2+x-6
wrong sign on the linear term.

You'd have the x change signs, too

(guessing you attempted to multiply the answer by negative 1)

because -3 + 2 = -1, but -2 + 3=+ 1.
Think about it.

(x-3)(x+2)
= (x+(-3))(x+2)
= x*x+x*2+(-3)*x+(-3)*2
= x^2+(-1)x+(-6)
=x^2+(-x)+(-6)
=x^2-x-6

Fucking foiling how does it work

>>2250904
Do I have to foil every time I factor a quadratic equation?

>>2250904
>>2250866
>>2250889
>>2250865
Man, you guys are geniouses.

>>2250910
FOIL is a method for EXPANDING BRACKETS, not factorizing. It stands for first, outer, inner, last; where you multiply the FIRST terms of both sets of brackets, then the OUTER terms in both sets of brackets then the INNER terms in both sets of brackets then the LAST terms in both sets of brackets.

>>2250932
Yes. Do I have to foil every time I factor a quadratic equation to KNOW in what order I have to put the numbers in.

Such as in x^2-x-6 = 0
how can I know if it's (x-3)(x+2) or (x-2)(x+3) without foiling?

If you know the root, pluging in x must make it zero (duh!)

So having -2 for root gives you the factor (x+2) and having 3 as a root gives you (x-3)

Any ideas on air production rates from biomass introduced to the Martian environment, assuming 100,000 years of orbital decay producing a denser atmosphere and 50,000 years of automated water convoy from Earth?

>>2250972
I don't understand. Can you try to explain it in a different way?

>>2250945
0/10

>>2250860
>Why not (x-2)(x+3)

When you foil those 2 polynomials, it comes out as x^2+x-6. Different quadratic is different

>>2250999
But how can I predict the order of those numbers without having to foil every time?

the factors add to the middle coefficient and multiply to the last
-3 + 2 = -1 , the middle coefficient
-3 * 2 = -6 , the last coefficient

>>2251013
you sort of can't. But the more you do it, the faster each calculation goes. I don't even consciously figure out the individual multiplicative steps, I just know the result.

>>2250988

If you want further help, you need to clarify what your problem is.

Do you have both x^2-x-6=0
and (x-3)(x+2)=0 given and want to try if they are equal (in which case you just need to expand the brackets), or do you start with x^2-x-6=0 and want to get to the bracket-form (in which case you need to calculate the roots... p-q-formula or something)?

>>2251020
Thanks a lot.

>>2251013
For a quadratic, look at terms a and c. Multiply the two terms and try to find multiples that add up to equal the middle term.

For instance:x^2-x-6. -6 is the number that comes up after you multiply a and c. -3 and 2 are multiples of -6 and they add up to get the middle term.

>>2251035
you're welcome. that only works when the first coefficient is 1 and squared
if it was 2x^2 or x^3 that trick doesn't work quite the same