errr.. didnt know where to put this but im reviewing for my calculus exam and this is the first unit
I don't know what you mean by factoring by grouping, but to factor thisOriginally Posted by Arbitur
we start by taking out the so:
Now we observe that this is high-school maths, so if this
is factorisable the the factors will have integer numeric terms.
Now is prime so that suggests we consider
factors of the form , for or .
The first one we try is , putting
in the expression gives zero and so is a factor.
Then doing the divisions gives us:
First I'll repeat CaptainBlack's first step: Factor out an "x":
Now look at the cubic factor. I'll regroup (hence "factoring by grouping" this as:
We can factor an x from the first pair of terms and a -5 from the second pair giving:
Notice that we can now factor a common from both terms. So:
Finally, putting the now factored cubic term back into the original problem we get:
just as CaptainBlack had. Usually in factoring by grouping we leave the quadratic factors alone if they don't factor over the integers, so I'll end the factorization here.
(Apologies to ticbol. I didn't see your post on the thread before I wrote this.)
Factoring GCF x from the initial expression gives 2x^4-10x^3-x^2+5x = x(2x^3 -10x^2 - x + 5). Now taking 2x^2 from the first two terms within parentheses gives x[(2x^2)(x-5) - x + 5] or, x[(2x^2)(x-5) - (x-5)]. Finally, we factor out the (x-5) common to both terms within brackets leaving x[(x-5)(2x^2 - 1)] or simply x(x-5)(2x^2 - 1).
No need to apologize.Originally Posted by topsquark
Yeah, same problem here. I also do not like to "repeat" another guy's solution that is why I avoid answering sometimes. So happens that after I post my reply, there is/are already posted replies to the same question. :-)
Some guys can type faster than my right pointing finger.
Me, who hates hints, never mind complete solutions, to solve a problem?
If a problem says, "here is a clue" or "Hint:", I will ignore that problem 99.99045 percent.
That is why also I like going home and finding most of the recent questions already answered. Then I can answer the ones left untouched----if I can solve them. :-).
No need to pump this here harassed lone finger typer.
But, if I see that posted answers are not to my liking, or if I can give another way, I will post another, different, answer.