errr.. didnt know where to put this but im reviewing for my calculus exam and this is the first unit
2x^4-10x^3-x^2+5x
help.
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:
.
RonL
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:
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.
-Dan
(Apologies to ticbol. I didn't see your post on the thread before I wrote this.)
Hi:
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).
Regards,
Rich B.
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.