Question

3.Explain how a four-term polynomial is factored by grouping and when a quadratic trinomial can be factored using this method. Include examples in your explanation.

Answer 1

damn you are much smarter than i am sorry this is alittle bit above my math skills sorry :/

Answer 2

wow im so awesome at what i do lol

Answer 3

So say for example we have

\[\large x^5 - 3x^3 - 2x^2 + 6\]

We group the terms together by 2's

\[\large (x^5 - 3x^3) - (2x^2 \color \red{-} 6)\]
Notice how that highlighted '-' is there...it WAS '+' but since we have a '-' in front of the parenthesis...when that distributes..it turns that '+' '-' again

So now...we factor out of each parenthesis what we can...

x^5 - 3x^3 ...well looks like at least x^3 can come out of there

2x^2 - 6 ...well we can factor a 2 out of there..

so now we have

\[\large x^3(x^2 - 3) - 2(x^2 - 3)\]

Now notice how we have 2 parenthesis that have the same thing inside them...and we also have 2 coefficients of those parenthesis...so we put those together respectively

\[\large (\text{coefficients in here}) (\text{those things in the parentheis since they are the same})\]

\[\large (x^3 - 2)(x^2 - 3)\]

Everything make sense?

Answer 4

Thanks :) this helped a lot ^^^

Answer 5

Great :)

So now...we need an example of a trinomial right?

Answer 6

Well say we have \[\large 3x^2 + 10x - 8\]

We can actually break up that 10x ...and make it -2x + 12x right? since -2 + 12 =10

so we would then have

\[\large 3x^2 - 2x + 12x - 8\]

NOW we can group these up by 2's

\[\large (3x^2 - 2x) + (12x - 8)\]

Lets factor out an 'x' of the first set of parenthesis...and also lets factor out a 4 of the last set...

\[\large x(3x - 2) + 4(3x - 2)\]

Look the same thing as before...same thing in both parenthesis.and 2 coefficients outside...so again

\[\large (x + 4)(3x - 2)\]