1. ## Trinomials, factoring, grouping

New math class with some new questions. I have defined number one, but I need feedback for number two, please.

Thanks in advance for the assistance.

1. What is a trinomial?
A = In algebra, a trinomial is a polynomial consisting of three terms or three monomials.

2. When factoring a trinomial by grouping, why is it necessary to write the trinomial in four terms? Provide an example.

2. here is trinomial with a 4th term
$\displaystyle x^2 - 3x + 4x -12$

$\displaystyle -3x + 4x\$ could be just $\displaystyle \ x$

now we can factor out x and 4 and get groups
we need the 4th term in order to do this.
$\displaystyle x(x-3) +4(x-3)$

now we can factor out $\displaystyle (x-3)$ and get
$\displaystyle (x-3)(x+4)$

however you can factor $\displaystyle x^2 +x -12$
directly without doing the grouping

hope this helps??

3. Bigwave,

Part of the question is "why is it necessary to write the trinomial in four terms?" Is there some rule that gives the answer to this? I don't know the answer to this part of the question.

Thanks again.

4. Originally Posted by dkpeppard
Bigwave,

Part of the question is "why is it necessary to write the trinomial in four terms?" Is there some rule that gives the answer to this? I don't know the answer to this part of the question.

Thanks again.
I don't think it is, certainly not in all cases.

The only kind I can think of is when there is an unknown constant which is not a function of x. Namely something like

$\displaystyle x^2-2x+k-3 = 0$ which can be written as $\displaystyle x^2-2x+(k-3)$

5. i would agree certainly with this

but i think what the concern was about grouping in order to factor

but as mentioned it certainly is not necessary in all cases...

I see you point though

in many cases grouping doesn't need to be done at all

6. Originally Posted by e^(i*pi)
I don't think it is, certainly not in all cases.

The only kind I can think of is when there is an unknown constant which is not a function of x. Namely something like

$\displaystyle x^2-2x+k-3 = 0$ which can be written as $\displaystyle x^2-2x+(k-3)$
Thanks for your feedback and example.