How do you Factor Trinomials with First Numbers?

>>9151664
That's meant to say 4x squared

>Factoring Trinomials
I had figured out a method to use Lattice to calculate the multiplication of two polynomials of any size when I was a senior in High School. You are 18+, right? Just use the snowflake method. Or use the Quadratic Equation to pull out a derivative.

>>9151669
I have no idea what any of those words mean. and yes im 18

>>9151677
>I have no idea what any of those words mean.
Well then look them up you fuck

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/thread

>>9151664
There's a few things. In this case you could factor

2 (2x^2 -5x + 3)

You see that you need a -1 and -3

From here you see that -2 + -3 = -5
So it has to factor as
2 (2x - 3)(x - 1)

You can't always factor this way but basically you have to factor the x^2 term and factor the x^0 term and see what combination gives you the x^1 term.

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>>9151664
I love you.

>>9151689
That doesn't solve shit

>>9151690
>>9151702
I'm just as lost, thanks for the help but i probably am dropping this course.

>>9151707
what the hell do you think it's for?

>>9151711
It doesn't solve the question i origionally posted so i have no idea. I'm not solving for x im factoring.

>>9151716
solve for x and then rearrange so it equals 0.
Say it gives you x=y and x=z, make both of those equal to 0, so x-y and x-z and then multiply so:
(x-y)(x-z).
Are you retarded?

>>9151718
I literally don't understand what you are talking about. I know it's suppose to end up as something like (4x + 2) (x + 3) or something like that.

>>9151711
You know how to factor
(x^2 +bx + c) where b and c are integers yea?

You need two numbers that multiply to c and add to b.

It's literally exactly the same when the x^2 term has a coefficient other than 1.

In your case, you need two numbers that multiply to 6.

Your choices are 1 and 6 or 2 and 3
Since the x term is negative, your choices have to both be negative, or the x^2 coefficient has to be negative. Since the x^2 term is 4, this means that it has to be
-1 and -6 or -2 and -3.

Next, since the x term is 4, it has to be 1 and 4 or 2 and 2.

If it's 2 and 2 then you have
2×1 + 2×6 = 14
2x2 + 2x3 = 10

So 2 and 2 fits what we want.
(2x -2)(2x-3)

This whole thing can be done in your head once you do it enough but is difficult if you have non real roots. In which case you use>>9151689

>>9151728
Thanks, i understand it now.

>>9151731
I would have caught you had my friend not :)

Remember that always.

>>9151742
Yep. i will

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>>9151728
>Your choices are 1 and 6 or 2 and 3
That's assuming that all coefficients are integers.

What you do is factor out a, which gives:
>a[x^2 + (b/a)x + (c/a)]

You apply the quadratic formula to find the factors of what's on the inside, then you re-multiply everything by a.

Remember not to distribute the entirety of a across both factors. The expression a(x+N)(x+M) is NOT the same as (ax+aN)(ax+aM). It is the same as (ax+aN)(x+M), and if the numbers u and v multiply to equal a, it is also the same as (uv+uN)(vx+vN).

>>9151762
I don't have to assume. All coefficients were integers in this case.