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.

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- Jan 28th 2006, 12:16 PMArbiturFactoring By Grouping..
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. - Jan 28th 2006, 12:30 PMCaptainBlackQuote:

Originally Posted by**Arbitur**

we start by taking out the $\displaystyle x$ so:

$\displaystyle 2x^4-10x^3-x^2+5x=x(2x^3-10x^2-x+5)$

Now we observe that this is high-school maths, so if this

is factorisable the the factors will have integer numeric terms.

Now $\displaystyle 5$ is prime so that suggests we consider

factors of the form $\displaystyle ax \pm 5$, for $\displaystyle a= 2$ or $\displaystyle a=1$.

The first one we try is $\displaystyle x-5$, putting $\displaystyle x=5$

in the expression gives zero and so $\displaystyle x-5$ is a factor.

Then doing the divisions gives us:

$\displaystyle 2x^4-10x^3-x^2+5x=x(x-5)(2x^2-1)=$$\displaystyle x(x-5)(\sqrt 2 x-1)(\sqrt 2 x+1)$.

RonL - Jan 28th 2006, 12:38 PMticbolQuote:

Originally Posted by**Arbitur**

= (2x^4 -10x^3) -(x^2 -5x)

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

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

= (x-5)(2x^2 -1)(x)

= x(x-5)(2x^2 -1) -----------answer. - Jan 28th 2006, 12:53 PMtopsquark
First I'll repeat CaptainBlack's first step: Factor out an "x":

$\displaystyle 2x^4-10x^3-x^2+5x=x(2x^3-10x^2-x+5)$

Now look at the cubic factor. I'll regroup (hence "factoring by grouping") this as:

$\displaystyle 2x^3-10x^2-x+5=(2x^3-x)+(-10x^2+5)$

We can factor an x from the first pair of terms and a -5 from the second pair giving:

$\displaystyle (2x^3-x)+(-10x^2+5)=x(2x^2-1)-5(2x^2-1)$

Notice that we can now factor a common $\displaystyle 2x^2-1$ from both terms. So:

$\displaystyle x(2x^2-1) -5(2x^2-1)=(x-5)(2x^2-1)$

So $\displaystyle 2x^3-10x^2-x+5=(x-5)(2x^2-1)$

Finally, putting the now factored cubic term back into the original problem we get:

$\displaystyle 2x^4-10x^3-x^2+5x=x(x-5)(2x^2-1)$

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.) - Jan 28th 2006, 01:03 PMRich B.
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. - Jan 28th 2006, 01:21 PMticbolQuote:

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.